Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download geom practice worksheet answers
Document related concepts
Multilateration wikipedia , lookup
Noether's theorem wikipedia , lookup
Euler angles wikipedia , lookup
Geometrization conjecture wikipedia , lookup
Rational trigonometry wikipedia , lookup
Integer triangle wikipedia , lookup
History of trigonometry wikipedia , lookup
Trigonometric functions wikipedia , lookup
History of geometry wikipedia , lookup
Line (geometry) wikipedia , lookup
Transcript
Chapter 1 Answers
6.
2. 1.00001, 1.00001
3. 42, 54
4. -64, 128
6. 63.5, 63.75
7. Sample: 2 or 3
8. 51
or 49
9. 6 or 8
10. A or AA
11. D or G
12. Y or A
13. any hexagon
14. circle with 8 equally
spaced diameters
15. a 168.75° angle
16. 21 hand17. h = n(n 22 1)
18. 34
19. Sample: The
shakes
1
1. 47, 53
5. 22, 29
farther out you go, the closer the ratio gets to a number that is
approximately 0.618.
20. 0, 1, 1, 2, 3, 5, 8, 13
All rights reserved.
Rig
3
Front
Front
7. A, C, D
8. C
Top
9. D
10. B
Right
11. A
1. AC
2. any two of the following: ABD, DBC, CBE,
ABE, ECD, ADE, ACE, ACD
3. Points E, B, and D
are collinear.
4. yes
5. yes
6. no
7. no
8. yes
9. no
10. yes
11. yes
12. yes
* )
13. *no ) 14. yes
15. yes
16. *G ) 17. LM
18. PN
19. the empty set
20. KP
21. M
22. *Sample:
plane
ABD
23.
Sample:
plane
ABC
)
24. AB
25. yes
26. no
27. yes
28. the empty set
29. no
30. yes
31. yes
1.
nt
1
Practice
* ) 1-3
Practice 1-2
Fro
Right
Practice 1-1
ht
2.
Practice 1-4
Fro
nt
Rig
1. true
6. false
2. false
3. true
4. false
5. false
7. JK, HG
8. EH
9. any three of the
* ) * )* ) * )* ) * )
) * )* ) * )* ) * )* ) * )* )
HG and FK ; JK and EH ; JK and FG ; EJ and FG ; EH
* )* ) * )* ) * )* ) * )* )
and FK ; JE and KG ; EH and KG ; JH and KF ; JH and
* )
ht
following pairs: EF and JH ; EF and GK ; HG and JE ;
*
3.
10. planes A and B
11. planes A and C; planes
12. planes A and C
13. planes B and C
)
)
)
)
14. Sample: EG
15. 6
16. EF and ED or EG and
)
) )
) )
ED
17. FE , FD
18. GF , GD
19. yes
Fro
B and C
nt
Rig
1
1
1
1
1
1
20. Sample:
Right
4.
ht
S
T
Front
U
5.
2
2
1
Right
Top
Front
Q
Front
Front
21. Sample:
R
Right
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
GE
W
P
Top
Geometry Chapter 1
Right
Answers
41
Chapter 1 Answers
(continued)
5.
Practice 1-5
1. 4
2. 12
3. 20
4. 6
5. 22
6. -10 or 6
7. -1 or 1
8. 3; 4; no
9. 6; 6; yes
10. C or -2
11. 15
12. 31
13. 14
14. x = 11 32 ; AB = 31;
BC = 31
15. x = 35 23 ; AB = 103; BC = 103
45°
Practice 1-6
1. any three of the following: &O, &MOP, &POM, &1
2. &AOB
3. &EOC
4. &DOC
5. 51
6. 90
7. 17
8. 107
9. 141
10. 68
11. &ABD, &DBE,
&EBF, &DBF, &FBC
12. &ABF, &DBC
13. &ABE,
&EBC
14. x = 5; 8; 21; 13
15. x = 9; 85; 35; 120
90°
6.
MN
OP
B
Practice 1-7
7.
1.
All rights reserved.
A
OP
MN
X
B
Y A
K
L
8.
2.
2
1
A
Y
1
9.
3.
2
1 2
1
X
2
B
Y
10.
X
2 ? 2
2
4.
C
11.
X
Z
42
Answers
Geometry Chapter 1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
X
Chapter 1 Answers
(continued)
12.
Practice 1-9
1. 792 in.2
2. 3240 in.2
3. 2.4 m2
4. 32p
5. 16p
6. 7.8p
7. 26 cm; 42 cm2
8. 46 cm; 42 cm2
9. 29 in.; 42 in.2
10. 40 ft; 51 ft2
11. 40 m; 99 m2
2
12. 40 m; 91 m
13. 68; 285
14.3 26; 22
15. 30; 44
16. 156.25p
17. 10,000p
18. p16
19. 48; 28
20. 36
21. 26; 13
2 ? X
W
Reteaching 1-1
1.
13.
,
All rights reserved.
X
Z
14. true
19. false
15. false
20. true
2.
16. false
17. true
18. true
,
Practice 1-8
1.–5.
3.
y
6
A
,
C
4
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
6
D
4
E
4.
2
2
6 x
4
2
B
,
4
5.
6
6. 5"2 7.1
7. 2"17 8.2
8. 12
9. 8
10. 12
11. "26 5.1
12. (5, 5)
13. (21 , 1)
14. (10 12 , -5)
15. (-2 12 , 6)
16. (-0.3, 3.4)
5
7
17. (2 8 , -8 )
18. (5, -2)
19. (4, 10)
20. (-3, 4)
21. yes; AB = BC = CD = DA = 6
22. "401 20.025
23.
y
S
,
6.
,
Reteaching 1-2
1a.
6
Q
4
2
P
24. 24.7
R
2
4
6
Fro
nt
Rig
ht
8x
25. (3.5, 3)
Geometry Chapter 1
Answers
43
Chapter 1 Answers
1b.
(continued)
6.
Front
Top
Right
Fro
2a.
nt
Rig
ht
Reteaching 1-3
1.–6. Check students’ work.
Fro
nt
h
Rig
7.
t
6
y
S
2
Front
Top
6 4 2 O
P2
Right
4
3a.
R
Fro
nt
Rig
2
4
6x
T
6
8. Sample: R, Q, T
9. P, R, S
* ) * )
10. Sample: RS and QT
ht
Front
Top
Right
4a.
Samples:
* ) * )
* ) * )
* ) * )
1. AB and EF
2. AB and DH
3. AC and AE
4. plane ABD and plane EFH
5. plane CDH and plane
BFD
6.
plane
CDH,
plane
ACG,
and plane ABD
* ) * )
7. EG and BF
8.
Fro
nt
h
Rig
9.
t
4b.
Front
Top
Right
10.
5.
Front
Top
Answers
Right
Geometry Chapter 1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Reteaching 1-4
3b.
44
All rights reserved.
4 Q
2b.
Chapter 1 Answers
(continued)
4.
11.
D
12.
E
F
5.
T
13.
All rights reserved.
U
V
6.
J
Reteaching 1-5
1. 40 km
6. Ryan
2. 25 km 3. 30 km 4. 30 km 5. 60 km
7. 100 km; 25 + 15 + 30 + 30 = 100
Reteaching 1-6
K
L
Reteaching 1-8
1.
6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1a. Each angle is 108°.
1b. obtuse
2a. Each angle is
72°.
2b. acute
3a. Each angle is 108°.
3b. obtuse
4. 90°, right
5. 55°, acute
6. 110°, obtuse
7. 50°,
acute
8. 90°, right
9. 100°, obtuse
Reteaching 1-7
X
4
2
6
Samples:
y
4
1.
2
2
2
4
C
4
6x
Z
Y
6
A
B
2.
Q
2. XY 10.6, YZ 4.1, XZ 7.2
3. 4.1 + 7.2 = 11.3, 11.3 10.6
4. 12.2
5. 8.1
6. 9.9
7. 11.4
8. 9.2
9. 8.5
10. 18.0
11. 8.5
12. 17
Reteaching 1-9
R
3.
S
X
1a. 1 by 60, 2 by 30, 3 by 20, 4 by 15, 5 by 12, and 6 by 10;
Check students’ drawings.
1b. 122, 64, 46, 38, 34, and 32
2a. 1 by 36, 2 by 18, 3 by 12, 4 by 9 and 6 by 6; Check students’
drawings.
2b. 1 by 36
3. 3 cm by 6 cm
4. Sample:
diameter 8 units;
radius 4 units
4
Y
Z
Geometry Chapter 1
Answers
45
Chapter 1 Answers
5. Sample: 50 units2
7. They are close.
(continued)
6. Sample: 50.3 units2
6.
Enrichment 1-1
Top
2. 16
4.
Round
1
2
3
4
5
6
Round of
64
32
16
8
4
2
Games in
Round
32
16
8
4
2
1
Total Games
Played
32
48
56
60
62
63
Bottom Left side
Enrichment 1-2
Back
7.
Top
5. The number in the fourth column is equal to the sum
of the numbers in the third column up to the given row.
6. 16
7. 1 + 2 + 22 + c + 2 n = 2 n + 1 - 1
8. 127
9. 3
1. 4
Right side Front
3. 8
Right side Front
Bottom Left side
Back
Right side
Front
Bottom Left side
Back
All rights reserved.
1. 32; 32; 32
8.
2. 7
3.
Top
nt
Rig
ht
4.
9.
Top Right side
Front
Right side
Front
Bottom Left side
Back
Top
Bottom Left side
Back
5.
Top Right side
Bottom Left side
46
Front
Back
Answers
10. V = 9 units3; S.A. = 32 units2
11. V = 9 units3;
2
3
S.A. = 34 units
12. V = 9 units ; S.A. = 34 units2
13. V = 7 units3; S.A. = 28 units2
14. V = 11 units3;
S.A. = 34 units2
15. V = 14 units3; S.A. = 44 units2
16. Sample: The foundation drawings; you can find the sum of
the numbers on each square.
17. Sample: When there are
no valleys in the cube arrangement, you can count the total
number of squares used in all six views of an orthographic
projection to find the surface area.
Geometry Chapter 1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Fro
Chapter 1 Answers
(continued)
11.
Enrichment 1-3
1. No; they will be parallel.
2. No; they will be skew.
3. S1 and S3; T and B; S2 and S4
4. S3
5. S4
6. line l or L3
7. line b
8. line n or L2
9. line a
10. 1; S3
11. 0
12. 0
13. 1; S3
14. S1 and S4
15. T and S1
16. S2
17. S3
18. Check students’ work.
19. Check students’ work.
Enrichment 1-4
All rights reserved.
1. north–south
2. east–west
3. east–west
4. north–south
5. always
6. never
7. never
8. always
9. never
10. 1000 ft
11. Sample: If
one airplane is flying east–west and one is flying north–south,
then one was out of its proper altitude range. Sample: If both
are flying east–west at the same cruising altitude, then the air
traffic controller instructed one of the airplanes incorrectly.
12. First Avenue, Second Avenue, or Third Avenue
13. Park Street, First Avenue, Second Avenue, or Grove
Street
14. Second Avenue or First Avenue
15. No;
the streets are parallel.
16. Carter Street, Oak Street, Hill
Street, Pine Street, and Crosstown Road
Enrichment 1-5
1. 1 mi
2. 9 mi
3. 5 mi
4. ABDE, 12 mi; ABCFE,
14 mi
5. F; more shorter connections
6. I; only one
long connection
7. link E with I; it is short and would
or
Enrichment 1-8
1.–7., 10.–14., 16.–19.
C
B
D
E
F
A
H
G
8. "32
9. "32; Check students’ work.
20. triangle
15. trapezoid
Enrichment 1-9
increase the number of connections to I.
1.–4. Sample:
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Enrichment 1-6
1. -4
2. 1 12
3. 21
1
1
7. 4 2 units
8. 5 2 units
4. III
5. II
9. 1 unit
6. I
Enrichment 1-7
A
1.–6.
B
B
A
F
F
I
E
C
D
C
E
students’ work. Early mathematicians’ beliefs were incorrect
as stated in Exercise 5.
D
7.–8.
A
B
B
5. Answers may vary.
6.–12. Check students’ work.
13. They are all the same, about 3.142; p.
14. Check
Chapter Project
A
F
F
• Table: 4; 8; 16
• 32; 64; multiply by 2
I
E
C
C
D
Activity 1: Paper Folding
E
D
9. hexagon ABBAFF and quadrilateral ICDE
10. hexagon CDEEDC and quadrilateral AFIB
Activity 2: Creating
Check students’ work.
Activity 3: Writing
Check students’ work.
Activity 4: Researching
Check students’ work.
Geometry Chapter 1
Answers
47
Chapter 1 Answers
(continued)
✔ Checkpoint Quiz 1
1
1. 91 , 27
2. 49, 56
3. G
4. Any three points are
5. intersecting
6. skew
7. parallel
coplanar.
8.
9. E, G, O, and F* ) 10. Sample: A, D, F, and B
11a. O
11b. EF
11c. O
11d. G
12. Yes;
they are vertical angles.
13. No; there are no markings.
14. Yes; they are adjacent and m&AOD + m&DOC = 180.
15. No; you cannot conclude this from the diagram.
16a. 11
16b. 22
16c. 40
17. 56
18. 56
19. 34
20. 90
21. 146
22. 112
23. 63°; acute
24. 102°; obtuse
25. 90°; right
26.
9.
ht
Rig
nt
2
1
1
1
A
All rights reserved.
Fro
1
1
27.
1
P
Q
✔ Checkpoint Quiz 2
1. 25
2.
28.
A
B
29.
R
T
P
Q
A
1
– ? lP
2
P
4. 10
30.
34.
37.
39.
5. (3, 0)
6. B
7. 16p ft
8. 64p ft2
Chapter Test, Form B
Chapter Test, Form A
1 1
-21 ; -16
, 32.
1. Add 5; 32, 37.
8.1
31. 8.5
32. 19.7
33. (4, -11)
52 ft; 169 ft2
35. 38 cm; 70 cm2
36. 32 in.; 42 in.2
1
2
2
p ft; 4 p ft
38. 3p cm; 2.25p cm
22p in.; 121p in.2
2. Multiply by
3. Add
4, then 5, then 6, and so on; 29, 37.
4. Add another side
of a hexagon; check students’ work.
5. The relationship (x - 1)(x + 1) = x2 - 1 is true for all
whole numbers x. No counterexample exists.
6. Sample:
6 ft
1. Multiply by 3; 81, 243.
2. Add 11, add 9, add 11, add 9,
and so on; 44, 55.
3. Insert one more dot; a circle with
three dots, a circle with four dots
4. The relationship
(x 2 1) 1 (x 1 1)
=
2
x is true for all
whole numbers x. No counterexample exists.
5. Sample:
6. Sample:
2 cm
6 ft
10 cm
7 ft
4 cm
7. Sample:
3m
10 m
8. Possible answers: A, O, and B; E, O, and F; C, O, and D
48
8 in.
22 in.
Answers
2 cm
7. M, E, B, and D
8. C, M, and O or M, B, and E
9. Sample: A, F, O, and D
10a. M
10b. M
10c. F
10d. O
11. &AOB
12. &DOE
13. &COD
14. &AOE and &BOC or &AOB and &COE
15a. 5
15b. 9
15c. 48
16. 25
17. 25
Geometry Chapter 1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3.
Chapter 1 Answers
18. 130
19. 180
20. 140º; obtuse
(continued)
21. 63º; acute
22.
P
TASK 2: Scoring Guide
a. Sample:
23.
A
2 Student gives explanations, descriptions, and a drawing that
are generally clear but may contain a few errors.
1 Student makes significant errors in explanations and
descriptions.
0 Student makes little or no attempt.
B
C
P
A
All rights reserved.
24.
G
D
P
b. No; any three points are coplanar.
c. Sample: A, B, and C
25.
A
B
26. 5
27. 17.3
28. (9, 18)
29. 58 in.; 168 in.2
2
30. 38 cm; 78 cm
31. 8p in.; 16p in.2
2
32. 30p cm; 225p cm
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
B
E
3 Student gives an accurate diagram, correct answers, and a
valid explanation.
2 Student draws a diagram that contains some errors but gives
answers to parts b and c that are correct.
1 Student draws a diagram that contains significant errors and
gives answers to parts b and c that are incorrect.
0 Student makes little or no attempt.
TASK 3: Scoring Guide
Sample:
P
Alternative Assessment, Form C
TASK 1: Scoring Guide
Samples:
a. Exterior rays form a right angle for each figure. The total
number of rays doubles with each successive figure. Two
parallel segments connect the innermost rays.
b. Other answers are possible. For example, instead of
considering the total number of rays (2, 4, 8, 16), consider the
rays in the interior of the right
angle: 0, 2, 6, . . . In that case, you
can argue that 12 comes next,
giving a total of 14 rays, not 16,
in figure 4.
c. same as in part b, except with
32 rays (If considering the rays in
the interior of the right angle,
Figure 5 will have 20 rays.)
C
(90 + mlPQR)°
B
A
R
Q
3 Student constructs the figures correctly.
2 Student constructs the figures, but the constructions may
contain some minor errors.
1 Student misses some key ideas, resulting in significant errors
in the constructions.
0 Student makes little or no attempt.
3 Student gives a correct explanation, clear descriptions, and
an accurate drawing.
Geometry Chapter 1
Answers
49
Chapter 1 Answers
(continued)
21. Sample:
TASK 4: Scoring Guide
Foundation drawing:
1
1
2
2
2
2
1
E
F
All rights reserved.
Orthographic drawing:
22. Sample:
Front
Top
Right
3 Student draws completely accurate figures.
2 Student draws generally accurate figures, but these may
contain a minor error.
1 Student draws figures containing significant errors.
0 Student makes little or no attempt.
45°
23. (-4, -2)
24. (0, -3)
25. "29
Cumulative Review
2. F
3. D
4. G
5. D
6. G
7. C
9. C
10. J
11. B
12. G
13. B
14. Sample: 1, 2, 3, 4, 5, 6, c and 1, 2, 3, 1, 2, 3, 1, c The
first is the natural numbers. In the second, the numbers 1, 2,
and 3 repeat.
15. Yes; the markings show they are congruent.
16. No; there are no markings.
17. Yes; you can
conclude the angles are supplementary from the diagram.
18. sometimes
19. never
20. Not acceptable; too
general; Sample: a ruler is a tool used for measuring lengths
and is marked showing uniform units.
50
Answers
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. A
8. J
Geometry Chapter 1
Chapter 2 Answers
Practice 2-1
1. Sample: It is 12:00 noon on a rainy day.
2. Sample: The
car will not start because of a dead battery.
3. Sample:
6
4. If you are strong, then you drink milk.
5. If a
rectangle is a square, then it has four sides the same length.
6. If you are tired, then you did not sleep.
7. If x = 26,
then x - 4 = 22; true.
8. If x 0, then ∆x∆ 0;
true.
9. If m is positive, then m2 is positive; true.
10. If 2y - 1 = 5, then y = 3; true.
11. If x 0, then
point A is in the first quadrant; false; (4, -5) is in the fourth
quadrant.
12. If lines are parallel, then their slopes are
equal; true.
13. If you have a sibling, then you are a twin;
false; a brother or sister born on a different day.
All rights reserved.
14.
siblings
twin
15. Hypothesis: If you like to shop; conclusion: Visit Pigeon
Forge outlets in Tennessee.
16. Pigeon Forge outlets are
good for shopping.
17. If you visit Pigeon Forge outlets,
then you like to shop.
18. It is not necessarily true. People
may go to Pigeon Forge outlets because the people they are
with want to go there.
19. Drinking Sustain makes you
train harder and run faster.
20. If you drink Sustain, then
you will train harder and run faster.
21. If you train
harder and run faster, then you drink Sustain.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 2-2
1. Two angles have the same measure if and only if they are
congruent.
2. 2x - 5 = 11 if and only if x = 8.
3. The converse, “If ∆n∆ = 17, then n = 17,” is not true.
4. A figure has eight sides if and only if it is an octagon.
5. If a whole number is a multiple of 5, then its last digit is
either 0 or 5. If a whole number has a last digit of 0 or 5, then
it is a multiple of 5.
6. If two lines are perpendicular, then
the lines form four right angles. If two lines form four right
angles, then the lines are perpendicular.
7. If you live
in Texas, then you live in the largest state in the contiguous
United States. If you live in the largest state in the contiguous
United States, then you live in Texas.
8. Sample: Other
vehicles, such as trucks, fit this description.
9. Sample:
Other objects, such as spheres, are round.
10. Sample:
This is not specific enough; many numbers in a set could fit
this description.
11. Sample: Baseball also fits this
definition.
12. Sample: Pleasing, smooth, and rigid all
are too vague.
13. yes
14. no
15. yes
Practice 2-3
1. &A and &B are complementary.
2. Football practice
is canceled for Monday.
3. DEF is a right triangle.
4. If you liked the movie, then you enjoyed yourself.
5. If two lines are not parallel, then they intersect at a point.
6. If you vacation at the beach, then you like Florida.
7. not possible
8. Tamika lives in Nebraska.
Geometry Chapter 2
9. not possible
10. It is not freezing outside.
11. Shannon lives in the smallest state in the United States.
12. On Thursday, the track team warms up by jogging 2 miles.
Practice 2-4
1. UT = MN
2. m&QWR = 30 3. SB = MN
4. y = 51
5. JL
6. Given; Addition Property
of Equality; Division Property of Equality
7. Given;
Addition; Subtraction Property of Equality; Multiplication
Property of Equality; Division Property of Equality
8. Substitution
9. Substitution
10. Transitive
Property of Equality
11. Symmetric Property of
Congruence
12. Transitive Property of Congruence
13. Definition of Complementary Angles; 90, Substitution; 3x,
Simplify; 3x, 84, Subtraction Property of Equality; 28, Division
Property of Equality
14. Given; (2x - 4), Substitution;
6x - 12, Distributive Property; -x, Subtraction Property of
Equality; 12, Multiplication Property of Equality
Practice 2-5
1. 30
2. 15
3. 20
4. 6
5. 16
6. 9
7. m&A = 135, m&B = 45
8. m&A = 36, m&B = 144
9. m&A = 75, m&B = 15
10. m&A = 10, m&B = 80
11. m&PMO = 55; m&PMQ = 125; m&QMN = 55
12. m&BOD = m&COE = 90; m&BOC = m&COD
= 45; m&AOB = m&DOE = 45
13. m&BWC =
m&CWD, m&AWB + m&BWC = 180; m&CWD +
m&DWA = 180; m&AWB = m&AWD
Reteaching 2-1
1.–3. Check students’ work.
4. If you hear thunder,
then you see lightning; statement: false; converse: false.
5. If your pants are jeans, then they are blue; statement: false;
converse: false.
6. If you are eating a tangerine, then you
are eating an orange fruit; statement: false; converse: true.
7. If a number is an integer, then it is a whole number;
statement: true; converse: false.
8. If a triangle has one
angle greater than 90°, then it is an obtuse triangle; statement:
true; converse: true.
9. If n2 = 64, then n = 8; statement: true; converse: false.
10. If you got an A for the
quarter, then you got an A on the first test; statement: false;
converse: false.
11. If a figure has four sides, then it is a
square; statement: true; converse: false.
12. If x = 144,
then "x = 12; statement: true; converse: true.
Reteaching 2-2
1. If n = 15 or n = -15, then ∆n∆ = 15. If ∆n∆ = 15, then
n = 15 or n = -15.
2. If two segments are congruent,
then they have the same measure. If two segments have the
same measure, then they are congruent.
3. If you live in
California, then you live in the most populated state in the
United States. If you live in the most populated state in the
United States, then you live in California.
4. If an
integer is a multiple of 10, then its last digit is 0. If an
integer’s last digit is 0, then it is a multiple of 10.
Answers
29
5. No; counterexamples may vary. Sample: A giraffe is a large
animal.
6. Yes; two planes intersect if and only if they
form a line.
7. Yes; a number is an even number if and
only if it ends in 0, 2, 4, 6, or 8.
8. Yes; a triangle is a threesided figure if and only if the angle measures sum to 180°.
Reteaching 2-3
1. Law of Detachment; the police officer will give Darlene
a ticket.
2. Law of Syllogism; if two planes do not have
any points in common, then they are parallel.
3. Law of
Detachment; Landon has a broken arm.
4. Not possible;
a conclusion cannot be drawn from a conditional and its
confirmed conclusion. (Brad may live in Peoria, Illinois.)
5. Law of Detachment; the circumference is p.
Reteaching 2-4
1. Addition Property of Equality
2. Symmetric Property
of Equality
3. Distributive Property
4. Subtraction
Property of Equality
5. Addition Property of Equality
6. Multiplication Property of Equality
7. Substitution
8. Transitive Property of Equality
9. 4, 5, 1, 3, 2
Reteaching 2-5
1.–7. 5, 2, 4, 1, 6, 3, 7 or 5, 2, 1, 4, 6, 3, 7
8. Sample: m&10 + m&6 = 180 by the Angle Addition
Postulate. m&6 + m&7 + m&8 = 180 by the Angle Addition
Postulate. m&10 + m&6 = m&6 + m&7 + m&8 by the
Transitive Property of Equality. Subtract from both sides, and
you get m&10 = m&7 + m&8. Because &10 is a right
angle, m&7 + m&8 = 90. It is given that m&7 = m&8.
By Substitution, 2m&8 = 90. Divide both sides by 2, and
m&8 = 45.
9. Sample: Because &9 and &10 form a
straight angle, m&9 + m&10 = 180 by the Angle Addition
Postulate. By the definition of supplementary angles, &9 and
&10 are supplementary.
10. Sample: Because &10 and
&6 form a straight angle, m&10 + m&6 = 180 by the Angle
Addition Postulate. &10 is a right angle, so 90 + m&6 = 180.
Subtract 90 from both sides, and you get m&6 = 90. So m&6
is a right angle by the definition of right angles.
Enrichment 2-1
1. point
2. postulate
3. parallel
4. perpendicular
5. perimeter
6. plane
7. protractor
8. pentagon
9. Sample: W, L
10. LXCBR
11. Sample: If LWHJP stands for plane,
then LHXHWWPW must be parallel because of the
repetition of the letter W.
12. Check students’ work.
Enrichment 2-2
1. If a kitchen is well-designed, then the distance in a straight
line from the center point of the sink, refrigerator, or stove
to another is between 4 feet and 9 feet. If the distance in a
straight line from the center point of the sink, refrigerator, or
stove to another is between 4 feet and 9 feet, then the kitchen
30
Answers
is well-designed. 2. The statement can be written as a
biconditional because both the conditional and its converse
are true.
3. A kitchen is considered to be well-designed if
and only if the distance in a straight line from the front
center point of one of these three objects to another is
between 4 feet and 9 feet. Additionally, the sum of
the three triangle legs should be less than or equal to 26
feet.
4. The definition is a good definition. The terms
used are commonly understood. The definition is precise.
The definition is reversible: If the distances in a straight line
between the front center points of the sink, refrigerator, and
stove are between 4 feet and 9 feet, and the sum of these
distances is less than or equal to 26 feet, then the kitchen is
considered to be well-designed.
5. Yes; the distance from
the sink to the refrigerator is 6.5 feet, the distance from the
refrigerator to the stove is 6.25 feet, and the distance from the
stove to the sink is 9 feet. The sum of these distances is less
than 26 feet.
6. Check students’ work.
Enrichment 2-3
1. p: one is a student at Richmond High School; q: one takes
geometry class; r: one studies logic.
2. If someone is a
student at Richmond High School, then he or she takes
geometry class.
3. If someone takes geometry class, then
he or she studies logic.
4. All students at Richmond High
School study logic.
5. Logic can be studied in classes
other than geometry, such as algebra or trigonometry.
6. a; p: a figure is a square; q: a figure is a polygon; r: a figure
is closed; if p S q and q S r then p S r.
7. b; p: a figure
is a square; q: a figure is a rectangle; r: a figure has four
right angles; if p S q and q S r, then p S r.
8. All
mathematicians enjoy puzzles.
9. Triangles have an
angle sum of 180.
10. Mt. McKinley is 20,320 ft tall.
11. Nolan Ryan recorded 5714 strikeouts over his career.
12. Trapezoids are four-sided figures.
Enrichment 2-4
D
8
B
T
R
S
T
I
T
I
O
N
O
N
B
S
I
V
I
D
I
T
I
T
I
O
N
M
U
L
I
O
N
11
I
C
L
C
P
I
I
R
E
T
T
T
V
A
C
I
L
S
L
E
I
N
E
V
I
T
I
M
T
N
O
I
T
C
A
R
T
M
I
O
I
T
I
D
S
N
A
U
7
Y
A
9
10
U
U
1
6
5
R
2
4
C
O
U
T
N
T
I
G
I
O
R
N
U
T
R
O
I
A
N
P
N
D
3
A
12
13
E
Q
N
U
U
C
A
B
E
L
I
T
Y
Enrichment 2-5
1. They are complementary angles; if the exterior sides of two
adjacent acute angles are perpendicular, then the angles are
complementary.
2. &ECD
3. They are equal;
Geometry Chapter 2
All rights reserved.
(continued)
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Chapter 2 Answers
Chapter 2 Answers
(continued)
because &ACB and &DCB have equal measures, you can
use the definition of supplementary angles and the subtraction
property to show that they are equal.
4. 90
5. 135
6. 45
7. NO bisects TP; a segment that is perpendicular
to another segment at its midpoint is the perpendicular
bisector of the segment.
Chapter Project
Check students’ work.
All rights reserved.
✔ Checkpoint Quiz 1
1. hypothesis: x + 4 = 10; conclusion: x = 6
2. hypothesis: if you want to get good grades in school;
conclusion: you must study hard
3. Sample: It could be
May.
4. Sample: Corn is a vegetable.
5. No; the lines
must be in the same plane.
6. yes
7. not possible
8. X, Y, and Z are coplanar.
9. If you ran a good race,
then your coach is happy.
10. If the car is old, then it is
not efficient.
✔ Checkpoint Quiz 2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. 4x = 13
2. &TQM <S
3. Symmetric
Property
4. Substitution Property
5. Division
Property
6. Addition Property
7. &1 &3,
&2 &4; Vertical Angles Theorem
8. &2 &3,
&1 &4; Linear Pair and Transitive Property of Equality
9. &1 &4, &2 &3; Subtraction Property of Equality
10. &1 &2, &3 &4; Vertical Angles Theorem
Chapter Test, Form A
1a. If a polygon has three sides, then it is a triangle.
1b. true
2a. If George lives in the United States, then
he lives in Texas.
2b. false
3a. If two angles are
congruent, then they are vertical angles.
3b. false
4. Addition Property of Equality
5. Transitive Property
of Equality
6. Subtraction Property of Equality
7. Division Property of Equality
8. Reflexive Property
of Equality
9a. 90
9b. 58
9c. 148
9d. 90
9e. 122
9f. 148
10. 17
11. 6
12. 14
13. 11
14. not possible
15. We win.
16. If the
bus is late, then we will receive a tardy penalty.
17. The
sum of the measures of &A and &B is 90.
18. If a
quadrilateral is a rectangle, then it has four right angles. If
a quadrilateral has four right angles, then it is a rectangle.
19. A rhombus has four congruent sides.
20. good
definition
21. good definition
22. A bat has wings.
23a. Distributive Property
23b. Addition Property
of Equality
23c. Division Property of Equality
24. 25
25. 19
26. 17
27. Answers may vary.
Sample: (1, 2)
28. Answers may vary. Sample: (3, -4)
Chapter Test, Form B
1a. If a polygon has five sides, then it is a pentagon.
1b. true
2a. If Mary lives in Minnesota, then she lives in
Geometry Chapter 2
Minneapolis.
2b. false
3. Reflexive Property of
Equality
4. Substitution Property of Equality
5. Division Property of Equality
6. Distributive Property
7a. 38
7b. 128
7c. 26
7d. 154
8. 50
9. 13.5
10. 4
11. Martina is quick.
12. If I don’t
wear sunscreen while swimming, then I’ll be in pain.
13. If a quadrilateral is a parallelogram, then it has two pairs
of opposite sides parallel. If a quadrilateral has two pairs of
opposite sides parallel, then it is a parallelogram.
14. A rectangle has four right angles.
15. good definition
16. An octopus has eight legs.
17a. Division Property of
Equality
17b. Subtraction Property of Equality
17c. Division Property of Equality
18. 17
19. 28
20. Answers may vary. There are two possibilities. Samples:
(4, -2) or (-2, 5)
Alternative Assessment, Form C
TASK 1: Scoring Guide
a. Sample: If it is raining, then the lawn is wet.
b. Sample: If the temperature is below 32°F, then the
temperature is below freezing.
3 Student gives conditionals that meet the criteria and
provides clear and accurate explanations of why the
conditionals meet the criteria.
2 Student gives examples and explanations that are generally
clear but may contain errors.
1 Student makes significant errors in examples or explanations.
0 Student makes little or no attempt.
TASK 2: Scoring Guide
Sample: The definition cannot be written as a biconditional.
A school is a place where people are educated.
3 Student gives an explanation and a definition that are
accurate, clear, and complete.
2 Student gives an explanation and a definition that are
generally clear but may contain errors.
1 Student makes significant errors in explanation or definition.
0 Student makes little or no attempt.
TASK 3: Scoring Guide
a. Sample: If two angles have sides that are opposite rays,
then they are vertical angles. &CEF and &GED are angles
whose sides are opposite rays. Therefore, &CEF and &GED
are vertical angles.
b. Sample: &CEF and &GED are vertical angles. All vertical
angles are congruent. Therefore, &CEF &GED.
3 Student gives accurate and complete examples and
explanations of the Laws of Detachment and Syllogism.
2 Student gives examples and explanations that are generally
correct but may be unclear.
1 Student gives examples or explanations that contain major
errors.
0 Student makes little or no attempt.
Answers
31
Chapter 2 Answers
(continued)
Cumulative Review
TASK 4: Scoring Guide
Sample:
1. b
2. a
3. C
4. G
5. A
6. H
7. C
8. F
9. D
10. G
11. C
12. J
13. C
14. 17
15. If a polygon has three sides, then it is a
triangle; true.
16. 196p m2
17. Sample: Two
y
3
2 R
T
P
1
V 1
2 1
1
S
2
3
x
supplementary angles are not necessarily a linear pair.
* )
18. BC
19. Substitution Property of Equality
20. (-1, 2)
21. (-2, 2)
22. "145
23. 4
24. 60 m2
25. 380.13 cm2
26. Sample: A midpoint
must be on the segment. Any point that lies on the perpendicular bisector of a segment is equidistant from the endpoints.
a. &TPR and &RPV; &RPV and &VPS; &VPS and
All rights reserved.
&SPT; &SPT and &TPR; &RPT and &VPS; &TPS
and &RPV
b. &RPT and &VPS; &TPS and &RPV
c. &RPT is a right angle because the lines are perpendicular,
so m&RPT = 90. Therefore, if ray PY bisects &RPT,
m&YPR = 45.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3 Student draws a diagram and gives responses that indicate a
thorough understanding of the definitions involved. Student
shows logic in part c that is complete and accurate.
2 Student draws a diagram and gives responses that, while
basically accurate, contain some minor flaws, inaccuracies,
or omissions.
1 Student draws a diagram or gives responses that contain
significant inaccuracies or omissions. Student displays flaws in
logical argument in part c.
0 Student makes little or no attempt.
32
Answers
Geometry Chapter 2
All rights reserved.
Chapter 3 Answers
Practice 3-1
Practice 3-4
1. corresponding angles
2. alternate interior angles
3. same-side interior angles
4. alternate interior angles
5. same-side interior angles
6. corresponding angles
7. 1 and 5, 2 and 6, 3 and 8, 4 and 7
8. 4 and 6, 3 and 5
9. 4 and 5, 3 and 6
10. m1 = 100, alternate interior angles; m2 = 100,
corresponding angles or vertical angles
11. m1 = 75,
1. 125
2. 69
3. 143
4. 129
5. 140
6. 136
7. x = 35; y = 145; z = 25
8. a = 55; b = 97; c = 83
9. v = 118; w = 37; t = 62
10. 50
11. 88
12. m3 = 22; m4 = 22; m5 = 88
13. 57.1
14. 136
15. m1 = 33; m2 = 52
16. isosceles
17. obtuse scalene
18. right scalene
19. obtuse
isosceles
20. equiangular equilateral
alternate interior angles; m2 = 75, vertical angles or
corresponding angles
12. m1 = 135, corresponding
angles; m2 = 135, vertical angles
13. x = 103; 77°, 103°
14. x = 24; 12°, 168°
15. x = 30; 85°, 85°
16a. Alternate Interior Angles Theorem
16b. Vertical
angles are congruent.
16c. Transitive Property of
Congruence
Practice 3-2
* )
* )
1a. same-side interior
1b. QR
1c. TS
1d. * same-side
interior
1e.
Same-Side
Interior Angles
)
1f. TS
1g. 3-5
2. l and m, Converse of Same-Side
Interior Angles Theorem
3. none
4. BC and AD ,
Converse of Same-Side Interior Angles Theorem
5. RT
and HU , Converse of Corresponding Angles Postulate
6. BH and CI, Converse of Corresponding Angles Postulate
7. a and b, Converse of Same-Side Interior Angles Theorem
8. 43
9. 90
10. 38
11. 100
12. 70
13. 48
Practice 3-5
1. x = 120; y = 60
2. n = 5137
3. a = 108; b = 72
4. 109
5. 133
6. 129
7. 129
8. 47
9. 127
10. 30
11. 150
12. 6
13. 5
14. 8
15. BEDC
16. FAE
17. FAE and BAE
18. ABCDE
19. 20
Practice 3-6
1. y = 31 x - 7
4. y
9.
1. True. Every avenue will be parallel to Founders Avenue, and
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
= 16 x - 32
y
6
4
2
y 4x 1
2 4 6 x
y 1 x 2
3
2 4 6 x
–6–4–2
–2
–4
–6 y 1 x 5
4
x
4
2
–6–4 –2
–2
–4
–6
2 4 6 x
–6–4–2
–4
–6
y
15.
6
4
2
2 4 6
6 y 2x 7
4
2
–6–4 –2
–2
–4
–6
y yx1
12.
y
6
4
2
–6–4–2
–2
–4
–6
13.
6. y = 45 x - 2
10.
–6
11.
3. y = 7x - 18
8. y = -x + 6
y
6
4
2
–6–4–2
–2
l2
5. y
3
7. y = 4x - 13
Practice 3-3
therefore every avenue will be perpendicular to Center City
Boulevard, and therefore every avenue will be perpendicular
to any street that is parallel to Center City Boulevard.
2. Not necessarily true. No information has been given about
the spacing of the streets.
3. True. The fact that one
intersection is perpendicular, plus the fact that every street
belongs to one of two groups of parallel streets, is enough to
guarantee that all intersections are perpendicular.
4. True.
Opposite sides of each block must be of the same type (avenue
or boulevard), and adjacent sides must be of opposite type.
5. Not necessarily true. If there are more than three avenues
and more than three boulevards, there will be some blocks
bordered by neither Center City Boulevard nor Founders
Avenue.
6. a ⊥ e
7. a || e
8. a || e
9. a ⊥ e
10. a ⊥ e
11. a || e
12. If the number of ⊥ statements is
even, then 1 || n. If it is odd, then 1 ⊥ n.
13. The proof
can instead use alternate interior angles or alternate exterior
angles (if the angles are congruent, the lines are parallel) or
same-side interior or same-side exterior angles (if the angles
are supplementary, the lines are parallel).
14. It is possible.
2. y = -2x + 12
1
= -2 x -
14.
2 4 6 x
y
6 y 2x 3
4
2
–6–4–2
–2
y
16.
y 3x 5
2 4 6 x
y
2
6 y x2
3
4
4
x
2 4 6
–6
–2
–2
–4
–6
2 4 6 x
l1
l3
Geometry Chapter 3
Answers
39
Chapter 3 Answers
2 4 6 x
–6–4–2
20.
y
6
4
2
x 2
–6–4
2 4 6 x
–2
–4
–6
21.
22.
y
–4
–6
–6
y
23.
y
6
4
2
y – 3x 2 2
–6–4–2
–2
–4
–6
yx
2 4 6 x
–4
–6
24.
2 4 6 x
x 2.5
2 4 6 x
–6–4–2
–2
–4
–6
25. y = –3x + 13
26. y = x + 4
27. y = 12 x - 3
1
1
28. y = -2 x - 2
29. y = 2x + 4
30. y = 13 x + 4
31. y = -51 x - 65
32. y = -6x + 45
33. x = 2;
y = -11
34. x = 0; y = 2
35. x = -4; y = -4
36. x = -1; y = 8
37.
y
6
4 (4, 0)
2
38.
y
(–2, 0) 6
4
2
–6–4–2 2
6 x
–2
–4
–6
–8
–10
–12 (0, 12)
–6 –4–2
–2
(0, 6)
(9, 0)
x
2 4 6 8
–4
–6
–2
–2 2 4 6 x
–4
–6
45a. m = $0.10
45b. the amount of money the worker is
paid for each box loaded onto the truck
45c. b = $3.90
45d. the base amount the worker is paid per hour
46. y = -25 x + 8
Practice 3-7
y
6
4
2
2 4 6 8 x
y
6
4
2
6
4
(–3, 0) 2
2 4 6 x
–6–4–2
y 5
44.
12 (0, 12)
10
6
4
2
2 4 6 x
–6–4–2
–2
–6–4–2
–2
–4
–6
y
6
4
–6–4–2
–2
–4
–6
6
4
2
y
6
4
2 (0, 1) (8, 0)
2 4 6 x
–6–4–2
–2
–4 (0, –4)
–6
43.
y
y 2x
42.
y
6
4
2 (4, 0)
2
6x
–6–4–2
–2
1
–4 y 2 x 3
–6
–4
–6
19.
41.
y
6
4
2
–6 –4 –2 2 4 6 x
–2
–4
–6 (0, –1)
1. neither; 3 2 13, 3 ? 13 2 -1
2. perpendicular;
? -2 = -1
3. parallel; -32 = -23
4. parallel;
-1 = -1
5. perpendicular; y = 2 is a horizontal line,
x = 0 is a vertical line
6. parallel; -21 = -21
7. neither; 1 2 18, 1 ? 18 2 -1
8. parallel; -32 = -23
9. perpendicular; -1 ? 1 = -1
10. neither; 12 2
7
5 1
5
7
2
-3, 2 ? –3 2 -1
11. neither; -3 2 -12
, -23 ? -12
2 -1
1
1
12. neither; 6 2 -5, 6 ? -5 2 -1
13. neither; 92 2 4, 92 ? 4 2 -1
14. parallel; 21 = 12
15. y = 32 x
4
16. y = -3 x + 24
y
17. y = -x - 3
6
4
18. y = 35 x + 6
2
19. y = 0
4 6 x
–6–4 L
20. y = 2x - 4
–4
21. y = 2x
1
2
Practice 3-8
1.–3.
4.–6.
Q
39.
y
6 (0, 6)
4
2
–6–4–2
–2
–4
–6
40
2 4 6 x
(6, 0)
Answers
40.
y
6
4
1
(– 2, 0) 2 (0, 2)
–6–4–2
All rights reserved.
18.
y
6
4 y 5x 4
2
T
2 4 6 x
–4
–6
Geometry Chapter 3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
17.
(continued)
Chapter 3 Answers
(continued)
14. Sample:
7.–9.
K
b
c
10. Sample:
15.
All rights reserved.
b
a
a
11. Sample:
a
b
Reteaching 3-1
1a.–1b. Sample:
c
60
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
12.
1c.–1d. Sample:
b
b
120 60
60 120
b
120 60
60 120
b
2. 110
7. 70
3. 70
8. 110
4. 110
5. 110
6. 70
Reteaching 3-2
13.
1.
l6m
1 3
Given
If 6 lines, then
corresponding s are .
a
m3 = m2 = 180
Substitution
c
2 and 3 are
supplementary.
Definition of
supplementary
m1 + m2 = 180
Angle Addition Postulate
Geometry Chapter 3
Answers
41
(continued)
2. 2 3
Given
3 1
a6b
Substitution If corresponding s,
then lines are parallel.
2 1
Vertical s are .
15. y = 14 x + 21
17. y = -x + 1
Reteaching 3-7
1a. -2
1b. 12
2a. 14
3b. 0
4a. y = -2x + 4
4c.
2b. -4
3a. undefined
4b. y = 12 x + 32
y
Reteaching 3-3
4
1. Draw a temporary line through A that is perpendicular to ,
then draw a second, permanent line through A that is
perpendicular to the first. This line will be parallel to . Repeat
for B. The two permanent lines will be parallel not only to but to each other (Theorem 3-9).
2. Draw two
perpendicular lines through C, then for each one draw a line
perpendicular to it that runs through D. The four lines will
define a rectangle.
3. Draw a line that runs through both E
and F, and then a second line at an acute or obtuse angle that
runs through E. Draw a third line, parallel to the second and
running through F, using the same procedure as in the Example
and in Exercise 1. In the same way, draw a fourth line parallel to
the first (at any desired distance from it). The four lines will
define a parallelogram.
16. y = -34 x + 4
18. y = 1
2
2
O
2
4
x
All rights reserved.
Chapter 3 Answers
2
4
5b. y = 32 x - 4
5a. y = -32 x - 4
5c.
y
4
Reteaching 3-4
2
1. ABD: mABD = 120, mADB = 30; CBE:
4
m3 = 60; m4 = 120
2
4
x
2
4
6a. y = 3x + 7
6c.
6b. y = -13 x - 3
y
4
2
Reteaching 3-5
1. 1 and 2 are interior angles; 3 and 4 are exterior
angles.
2. m1 = 135; m2 = 90; m3 = 45;
m4 = 90
3. 1 is an interior angle; 2 and 4 are
exterior angles; 3 is neither.
4. m1 = 60; m2 = 120;
2 O
4
2 O
2
4
x
2
4
Reteaching 3-6
Check students’ graphs.
1. y = 2x - 6
2. y = 13 x
3. y = -x - 3
5
4. y = 6 x + 2
5. y = -21 x + 1
6. y = 1
7. y = -72 x + 10
8. y = -x + 1
9. y = 25 x + 1
10. y = 1
11. y = -2x - 6
12. x = -3
13. y = -3x + 10
14. y = 3x - 10
42
Answers
7. mJK = -1; mLM = -1; parallel
8. mJK = 23 ; mLM =
3
1
9. mJK = -6; mLM = -15 ; neither
-2 ; perpendicular
3
10. mJK = -2 ; mLM = 54; neither
11. mJK = 2; mLM =
1
1
12. mJK = 5; mLM = 5; neither
-2; perpendicular
13. mJK = 14; mLM = -14; parallel
14. mJK undefined;
mLM = 0; perpendicular
Geometry Chapter 3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
mCBE = 120, mCEB = 30, mBCE = 30; BDE:
mBDE = 60, mDBE = 60, mBED = 60
2. DBE and ABC are acute, equiangular, and
equilateral; ABD and CBE are isosceles and obtuse;
ACE, ADE, CED, and CAD are right and scalene.
3. PQT: mPTQ = 45, mPQT = 90; PQR:
mPQR = 90, mQPR = 45, mQRP = 45; RQS:
mRQS = 90, mQSR = 45; SQT: mSQT = 90,
mQST = 45, mSTQ = 45
4. PT = TS = RS =
PR = 40 mm; PQ = QT = QR = QS = 28 mm
5. PQT, PQR, RQS, SQT, PRS, PTS, PRT,
and RST are right and isosceles.
Chapter 3 Answers
(continued)
ones, with precise control of the slant.
8. Figures A and B
would be easy, because the lines have no more than two or
three different slants. Figure C would be harder, because lots
of different slants are used to achieve the perspective effect.
Reteaching 3-8
1. Sample:
Z
Enrichment 3-4
b
1. 48
X
a
2. 2880
3. 4320
4. Angles have measures of
20, 70, or 90; AC 6 MD 6 LE 6 KF 6 JG; BM6 CH 6 NK;
AH 6 PG; CM 6 DL 6 EK 6 JF 6 IG.
Y
Enrichment 3-5
All rights reserved.
1. 2
2. 5
7. Number
2.–4. Check students’ work.
Enrichment
)
* 3-1
)
1. OE * is #) to AB .
2. 3, 5, 1, 4, 2, 6 or 4, 5, 1, 3, 2, 6; OE
)
is # to AB ; if two angles are congruent and supplementary,
then each measures 90°.
3. 1 > 2; Law of Reflection
4. 2 > 3; Alternate Interior Angles Theorem
5. 3 > 4; Law of Reflection
6. 1 > 4; Transitive
Property of Congruence
3. 9
1.–8.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y
3
1
5 A
4
T
2
D
3. 33
4. 41
Enrichment 3-3
1. If the paper shifts during drawing, lines meant to be parallel
will not be parallel, and lines meant to be perpendicular will
not be perpendicular.
2. The rectangular shape gives the
T-square two perpendicular guide edges on which to line up.
3. Theorem 3-10, which says that if two lines are perpendicular
to the same line, they are parallel to each other. A line drawn
with the T-square is perpendicular to the edge on which the Tsquare is lined up. Two lines drawn with the T-square lined up
on the same edge will be parallel to each other.
4. a ⊥ d. If the T-square is used to draw a line perpendicular to
a vertical edge of the rectangular backing surface, and then is
used to draw a line perpendicular to a horizontal edge of the
backing surface, then the two lines will be perpendicular.
5. Figure A would be easy; all the lines are parallel or
perpendicular. Figures B and C would be hard to draw, since
the lines are not all parallel or perpendicular to each other.
6. They will all be parallel to .
7. A drafting machine can
be used to draw slanted lines, not just vertical and horizontal
6
L
5. Sample:
H
Because mBAC = 41, mCAF = 180 - 41 = 139;
l 6 m because a pair of alternate interior angles are congruent.
6. CD and EF ; AK
7. Sample: CAB and IGH are
corresponding angles related to parallel segments AB and GH .
AK is the related transversal.
8. Sample: A, ADE,
and AED form a triangle, so 180 - (43 + 76) = 61, so
mAED = 61. Because AED and C are congruent
corresponding angles, DE 6 BC by the Converse of the
Corresponding Angles Postulate.
Geometry Chapter 3
6. 27
Enrichment 3-6
Enrichment 3-2
2. 106
5. 20
3
4
5
6
7
8
9
...
n
of sides
Total
180 360 540 720 900 1080 1260
(n 2)180
degree
measure
Number of
n(n 3)
0
2
5
9 14 20 27
2
diagonals
8
1. x = 11
4. 14
N
B
M
x
E
K
Y
P
C
S
7
R
RENE DESCARTES
Enrichment 3-7
1. (-3, 4)
2. (-2, 3)
3. (3, 3)
4. (-2, -2)
5. (-1, -1)
6. (4, -1)
7. (1, 0)
8. (2, -3)
9. (-1, 2)
10. (1, -4)
11. (-3, -3)
12. (2, 3)
13. (3, -2)
14. (5, 0)
15. (0, 1)
16. (4, 3)
y-axis
4
A
3
N
2
G
1
L
0
1
E
2
N
I
3
4 L
4 3 2 1 0
R
A
Y
E
T
N
I
O
P
1
2
3
4
Answers
5 x-axis
43
Chapter 3 Answers
(continued)
pentagons
Enrichment 3-8
1., 3., and 4.
hexagons
A
C
2. scalene, acute triangle
Exercises 1, 3, and 4.
All rights reserved.
B
Activity 3: Analyzing
5. No; refer to the answer to
6.
E
Activity 4: Modeling
F
7. isosceles, obtuse triangle
Chapter Project
Activity 1: Paper Folding
5; all triangles are right isosceles; yes.
Activity 2: Exploring
triangle
✔ Checkpoint Quiz 1
1. Converse of Corresponding Angles Postulate
2. Alternate Interior Angle Theorem
3. Same-Side
Interior Angles Theorem
4. Corresponding Angles
Postulate
5. Converse of Alternate Interior Angle
Theorem
6. Vertical Angles Theorem
7. Converse
of Corresponding Angles Postulate
8. Corresponding
Angles Postulate
9. Converse of Same-Side Interior
Angles Theorem
10. x = 50, y = 30, z = 65
✔ Checkpoint Quiz 2
1. pentagon; 80
2. hexagon; x = 110; y = 98
3. quadrilateral; x = 94; y = 105
y
y
4.
5.
quadrilaterals
8
6
4
2
(2,0)
4 6 8 x
8642
2 2
4
6
8 (0,8)
44
Answers
8
6
4 (0,3)
(1.2,0) 2
2 4 6 8 x
8642
2
4
6
8
Geometry Chapter 3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
D
Chapter 3 Answers
(continued)
y
6.
Chapter Test, Form B
8
6
4
2
42
2
4
6
8
1. true 2. true 3. false 4. true 5. false
6. m1 = 62, m2 = 62, alternate interior angles
7. m1 = 85, m2 = 95, same-side interior angles
8. m1 = 75, m2 = 75, alternate interior angles
y
y
9.
10.
(12,0)
2 4 6 8 10 12 x
(0,8)
7. Parallel; slopes are the same.
8. neither
9. Perpendicular; the product of the slopes is -1.
10. neither
Chapter Test, Form A
All rights reserved.
1. true
6. false
2. true
7. true
3. false
8. true
4. false
5. true
9. Answers may vary.
Sample: m1 = 125, Same-Side Interior Angles Theorem; m2
= 55, Alternate Interior Angles Theorem
10. Answers may
vary. Sample: m1 = 60, Corresponding Angles Postulate then
Angle Addition Postulate; m2 = 60, Same-Side Interior
11. Answers may vary. Sample: m1 = 85,
Angles Theorem
Alternate Interior Angles Theorem; m2 = 95, Same-Side
12. Answers may vary. Sample:
Interior Angles Theorem
m1 = 75, Corresponding Angles Postulate; m2 = 105,
13. Answers may vary. Sample:
Angle Addition Postulate
m1 = 91, Corresponding Angles Postulate and Same-Side
Interior Angles Theorem; m2 = 89, Corresponding Angles
14. Answers may vary. Sample: m1 = 60,
Postulate
Alternate Interior Angles Theorem; m2 = 115, Same-Side
Interior Angles Theorem
15.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
X
4
4
4
2
2
2
2
4 x
2
4 x
2
4
4
11. AD and WZ 12. AD and WZ 13. AD and WZ
14. AW and DZ 15. x = 92, y = 88 16. w = 31,
x = 65, y = 65, z = 115 17. 1440 18. 45
19. neither 20. perpendicular 21. parallel
22. y = 3x - 1 23. y = 14 x - 3 24. y = -2x - 2
Alternative Assessment, Form C
TASK 1: Scoring Guide
a.
t
1 2
4 3
5 8
6 7
a
b
b. Sample: 8 2; 8 4; 8 and 3 are
supplementary.
c. m2 = 75; m3 = 105, m4 = 75,
m5 = 105, m6 = 75, m7 = 105, m8 = 75
m
d. 7 and 4 are supp.
Given
16.
x
4 8
a6b
Supp. of the
same are .
If alt. int s
are , then
lines are 6.
7 and 8 are supp.
y
17. WA and XB
18. none
19. WZ and AB
20. none
21. WZ and AB
22. WZ and AB; AX
and BY
23. x = 22; y = 120
24. x = 70; y = 60;
z = 120
25. x = 35; y = 35; z = 55
26. 5940
27. 18
28. perpendicular
29. neither
30. parallel
31. y = 6x + 23
32. y = -21 x + 3
33. y = 13 x + 2
Geometry Chapter 3
Angle Add. Post.
3 Student draws an accurate diagram and supplies correct
answers and a complete and accurate flow proof.
2 Student draws a figure or gives answers that contain
minor errors.
1 Student draws a figure or gives answers that contain
significant errors or omissions.
0 Student makes little or no attempt.
Answers
45
(continued)
TASK 2: Scoring Guide
TASK 4: Scoring Guide
Sample:
Sample:
Q
A
P
Isosceles
Triangles
B
Equilateral
Triangles
3 Student constructs an accurate figure.
2 Student constructs a figure that contains minor errors
or omissions.
1 Student constructs a figure that contains significant errors
or omissions.
0 Student makes little or no attempt.
TASK 3: Scoring Guide
Sample:
Triangles
R
3 Student draws an accurate diagram.
2 Student draws a diagram that contains minor errors
or omissions.
1 Student draws a diagram that contains significant errors
or omissions.
0 Student makes little or no attempt.
All rights reserved.
Chapter 3 Answers
Cumulative Review
B
A
D
C
1. D
2. G
3. C
4. F
5. D
6. J
7. A
8. F
9. C
10. H
11. D
12. J
13. B
14. G
15. C
16. G
17. C
18. Given
19. Same-Side Interior Angles Theorem
20. Corresponding Angles Postulate
21. Corresponding
Angles Postulate
22. substitution
23. Check students’
work.
24. Sketches may vary. The right angle must be
between the equal sides.
25. No; a triangle cannot have
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
two sides equal and no sides equal at the same time.
a. For the figure given, y = x and y = x + 3. The lines are
parallel because both lines have a slope of 1.
c. For ABC, the sum of the three angles is 180°. Minor
discrepancies are the result of measurement error and
rounding error.
d. For the figure given, ABDC is not regular. By the
distance formula, the sides are not congruent.
3 Student draws the figure accurately, writes correct equations,
and reasons logically.
2 Student draws a figure, gives arguments, and writes
equations that are mainly correct but may contain minor
errors.
1 Student presents work with significant errors.
0 Student makes little or no attempt.
46
Answers
Geometry Chapter 3
Chapter 4 Answers
Practice 4-1
1. m1 = 110; m2 = 120
2. m3 = 90; m4 =
135
3. m5 = 140; m6 = 90; m7 = 40; m8 = 90
4. CA JS , AT SD , CT JD
5. C J,
A S, T D
6. WZ JM , WX JK ,
XY KL , ZY ML
7. W J, X K,
Y L, Z M
8. Yes; GHJ IHJ by
All rights reserved.
Theorem 4-1 and by the Reflexive Property of . Therefore,
GHJ IHJ by the definition of triangles.
9. No;
QSR TSV because vertical angles are congruent, and
QRS TVS by Theorem 4-1, but none of the sides are
necessarily congruent.
10a. Given
10b. Vertical
angles are .
10c. Theorem 4-1
10d. Given
10e. Definition of triangles
Practice 4-2
1. ADB CDB by SAS
2. not possible
3. not
possible
4. TUS XWV by SSS
5. not possible
6. DEC GHF by SAS
7. MKL KMJ by
SAS
8. PRN PRQ by SSS
9. not possible
10. C
11. AB and BC
12. A and B
13. AC
14a. Given
14b. Reflexive Property of
Congruence
14c. SAS Postulate
15. Statements
Reasons
1. EF FG , DF FH
1. Given
2. DFE HFG
3. DFE HFG
2. Vertical s are .
3. SAS Postulate
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. not possible
2. ASA Postulate
3. AAS Theorem
4. AAS Theorem
5. not possible
6. not possible
7. ASA Postulate
8. not possible
9. AAS Theorem
10. Statements
Reasons
11.
so UVX WVX by SSS, and U W by CPCTC.
6. ZAY and CAB are vertical angles, so ABC AYZ by ASA. ZA AC by CPCTC.
7. EG is a
common side, so DEG FEG by SAS, and FG DG
by CPCTC.
8. JKH and LKM are vertical angles, so
HJK MLK by AAS, and JK KL by CPCTC.
9. PR is a common side, so PNR RQP by SSS, and
N Q by CPCTC.
10. First, show that ACB and
ECD are vertical angles. Then, show ABC EDC by
ASA. Last, show A E by CPCTC.
11. First, show
FH as a common side. Then, show JFH GHF by
SAS. Last, show FG JH by CPCTC.
Practice 4-5
1. x = 35; y = 35
2. x = 80; y = 90
3. t = 150
4. r = 45; s = 45
5. x = 55; y = 70; z = 125
6. a = 132; b = 36; c = 60
7. x = 6
8. a = 30; b = 30; c = 75
9. z = 120
10. AD ; D F
11. GA ; ACG AGC
12. KJ ; KIJ KJI
13. DC ; CDE CED
14. BA ; ABJ AJB
15. CB ; BCH BHC
16. 130
17. 65
18. 130
19. 90
20. x = 70; y = 55
21. x = 70; y = 20
22. x = 45; y = 45
Practice 4-6
Practice 4-3
1. K M, KL ML
2. JLK PLM
3. JKL PML
3. KLJ PMN by ASA, so K P by CPCTC.
4. QS is a common side, so QTS SRQ by AAS.
QST SQR by CPCTC.
5. VX is a common side,
1. Given
2. Vertical s are .
3. ASA Postulate
Q S
Given
TRS RTQ
QRT STR
Given
AAS Theorem
RT TR
Reflexive Property of 12. BC EF
13. KHJ HKG or KJH HGK
14. NLM NPQ
Practice 4-4
1. BD is a common side, so ADB CDB by SAS, and
A C by CPCTC.
2. FH is a common side, so
1. Statements
1. AB # BC , ED # FE
2. B, E are right s.
3. AC FD , AB ED
4. ABC DEF
2. Statements
1. P, R are right s.
2. PS QR
3. SQ QS
4. PQS RSQ
3.
Reasons
1. Given
2. Perpendicular lines
form right s.
3. Given
4. HL Theorem
Reasons
1. Given
2. Given
3. Reflexive
Property of 4. HL Theorem
MJN and MJK
are right s.
Perpendicular lines
form right s.
MN MK
MJN MJK
Given
HL Theorem
MJ MJ
Reflexive Property of FHE HFG by ASA, and HE FG by CPCTC.
Geometry Chapter 4
Answers
35
Chapter 4 Answers
(continued)
GI JI
Reteaching 4-2
HI HI
IHG IHJ
Reflexive Property of HL Theorem
GHI JHI
GHI and JHI
are right s.
Given
Theorem 2-5
mGHI + mJHI = 180
Practice 4-7
1. ZWX YXW; SAS 2. ABC DCB; ASA
3. EJG FKH; ASA
4. LNP LMO; SAS
5. ADF BGE; SAS
6. UVY VUX; ASA
B
C B
C
7.
common side: BC
D
8. F
G
E
G
common side: FG
F
H
9.
common angle: L
L
L
N
K
J
M
10. Sample:
Statements
1. AX AY
2. CX # AB, BY # AC
3. mCXA and
mBYA = 90
4. A A
Reasons
1. Given
2. Given
3. Perpendicular lines
form right s.
4. Reflexive Property
of 5. BYA CXA
5. ASA Postulate
11. Sample: Because FH GE, HFG EGF,
and FG GF, then FGE GFH by SAS. Thus, FE
GH by CPCTC and EH HE, then GEH FHE
by SSS.
Reteaching 4-1
1. b
36
2. c
3. a
4. 117
Answers
5. 119
3. SSS; AEB CDB
5. SSS; PRQ VUT
7. SAS; QSP QSR
Reteaching 4-3
1.
2.
Angle Addition Postulate
5. RS VW
6. none
7. mC and mF = 90
8. GH JH
9. LN PR
10. ST UV or
SV UT
11. mA and mX = 90
12. mF
and mD = 90
13. GI # JH
A
1.–2. Check students’ work.
4. SAS; MNQ ONP
6. SSS: JMK LMK
8. SAS; YTX WXT
or
3. Check students’ work.
4. ABD CBD
5. JMK LKM or JKM LMK
6. UZ YZ
7. DY DO
8. P A
9. CYL ALY
All rights reserved.
Given
Reteaching 4-4)
1a. QK QA; QB bisects KQA
1b. definition
1c. BQ BQ
1d. SAS Postulate
of bisector
1e. CPCTC
2. Statements
Reasons
1. MN MP,
1. Given
NO PO
2. MO MO
2. Reflexive Property
of 3. MPO MNO
3. SSS Postulate
4. N P
4. CPCTC
3. Statements
Reasons
1. ON bisects JOH,
1. Given
J H
2. JON HON
2. Definition of bisector
3. ON ON
3. Reflexive Property
of 4. JON HON
4. AAS Theorem
5. JN HN
5. CPCTC
Reteaching 4-5
1. Each angle is 60°.
2. 120
3. 120
4. 50
5. 70
6. 60
7. 65
8. 115
9. 55
10. 120
11. 60
Reteaching 4-6
1. Sample: RS = 1.3 cm, ST = 1.6 cm, QT = 2.5 cm, QR
= 2.3 cm; not congruent
2. Sample: NT = 2.3 cm, TG
= 2.3 cm, AT = 1.9 cm; NAT GAT
3. Sample:
TO = 3.3 cm, TR = 2.8 cm, MO = 2.8 cm; TOM OTR
4. HL Theorem can be applied; BDA CAD.
5. HL Theorem cannot be applied.
6. HL
Theorem can be applied; MUN MLN.
7. HL
Theorem can be applied; THF FET or THF TEF.
8. HL Theorem can be applied; OKR AHR.
9. HL Theorem cannot be applied.
Geometry Chapter 4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4.
Chapter 4 Answers
(continued)
Reteaching 4-7
All rights reserved.
1. Statements
1. PSR and PQR
are right s;
QPR and SRP
2. PSR and PQR
2. Right s are congruent.
3. PR PR
3. Reflexive Property of 4. QPR SRP
4. AAS Theorem
5. STR QTP
5. Vertical s are .
6. PQ RS
6. CPCTC
7. STR QTP
7. AAS Theorem
2. Sample: Prove MLP QPL by the AAS Theorem.
Then use CPCTC and vertical angles to show MLN QPN by the AAS Theorem.
3. Sample: Prove ACD
ECB by the SAS Postulate. Then use CPCTC and vertical angles to show ABF EDF by the AAS Theorem.
Enrichment 4-1
Check students’ work. Samples shown.
1.
2.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Enrichment 4-2
Reasons
1. Given
3.
4.
1a. Definition of perpendicular lines
1b. AKF GEL
1c. SAS
2a. Segment Addition Postulate
2b. LR + RG = TF + TA
2c. RG TA
2d. Alternate Interior Angles
2e. Corresponding Angles
2f. DAT JGR
2g. SAS
Enrichment 4-3
1.–11. Check students’ work.
2a. ASA
2b. The top
angles are congruent because the fold bisected the right angles
formed by the folds in steps 1 and 3. The corners of the paper
are right angles; therefore, those angles are congruent. The
included sides are congruent because the fold in step 1 found
the midpoint of the width of the paper, thus creating two equal
segments.
3a. ASA
3b. The top angles are congruent
because the fold bisected the right angles formed by the folds
in steps 1 and 2. The upper corners that became inside angles
along the center line are right angles; therefore, those angles
are congruent. The included sides are congruent because the
fold in step 1 found the midpoint of the width of the paper,
thus creating two equal segments.
4a. The top angles are
congruent because the fold bisected the right angles formed by
the fold in step 1. The inside angles along the center line are
right angles because the horizontal fold that formed them is
perpendicular to the original fold in step 1.
4b. The
included sides are congruent because the fold in step 1 found
the midpoint of the width of the paper, thus creating two equal
segments.
8a. ASA
8b. The top angles are congruent
because the fold bisected the right angles formed by the fold in
step 7. The inside angles along the center line are congruent
because of the Angle Addition Postulate. The included sides are
congruent because the fold in step 7 found the midpoint of the
width of the paper, thus creating two equal segments.
Enrichment 4-4
1. ABT
2. ACT
3. 45; 45; ABT; ACT
4. 30; 30;
ATB; ATC
5. Reflexive Property of 6. AAS
Theorem
7. CPCTC
8. Definition of segments
9. Definition of segments
10. Definition of segments
11. SSS Postulate
12. CPCTC
13. 60
Enrichment 4-5
1. 60
2. 60
3. 60
4. 70
5. 70
6. 40
7. 72
8. 72
9. 36
10. 30
11. 30
12. 120
13. 80
14. 80
15. 20
16. 80
17. 80
18. 20
19. 41
20. 23
21. 116
22. 23
23. 41
24. 116
25. 80
26. 80
27. 20
28. 82
29. 82
30. 16
31. 78.5
32. 78.5
33. 37.5
34. 40
35. 40
36. 100
5.
Geometry Chapter 4
Answers
37
Chapter 4 Answers
(continued)
Enrichment 4-6
Chapter Project
1
Activity 1: Modeling
I
H
E
G 5H
O
S
Y
N
O
P
T
Activity 2: Observing
G
S
O
E
Check students’ work.
9A
R
C
U
12
V
E
U
E
R
T
I
C
A
14
A A
B
16
A C
E
S
15
R E
A
C
O
T
N
E
F
L
E
E
X
I
Activity 3: Investigating
11
N
N
N M P
U
M
S
L
R
10
E
E
T
7H
13
L
N
T
6A
T
tetrahedron
Sample: You could add in diagonals of the cube.
Check students’ work.
V
V
E
N
R
T
17
E
E
S A
Finishing the Project
S
R
I
O
R
Enrichment 4-7
1. Sample: ABD AEC
2.
A
A
E D
5. 8
7.
K
H
J
L
N
J
H
M
common side: IM
M
11. 4
12. PUV TWV by AAS; PSV
TQV by AAS; PVW TVU by SSS; WQV USV by AAS.
13. 14
14. Check students’ work.
15. Check students’ work.
10. 4
38
Answers
1. ABC, ABD
2. Hypotenuse-Leg Theorem
3. CPCTC
4. S Q, RQ TS, STR QRT
5a. definition of a bisector
5b. Reflexive
5c. ASA
5d. CPCTC
5e. definition
1. x = 50; y = 65
2. a = 118; b = 62; c = 59
3. HL
4. not possible
5. SAS
6. AAS
7. ASA
8. SSS
9. not possible
10. SSS
11. not possible
12. Check students’ work; J P,
6. Sample: GNJ KLH
8. Sample: HIM JIM
I
I
9.
LN TR, L T, M Q, N R
8. Alternate Interior Angles Theorem
9. Alternate
Interior Angles Theorem
10. ASA
Chapter Test, Form A
C
G
2. SAS
3. SSS
4. not possible
6. not possible
7. LM TQ, MN QR,
common side: DC
C
D
1. AAS
5. AAS
✔ Checkpoint Quiz 2
C
3. Sample: DEC CBD
4.
B
✔ Checkpoint Quiz 1
common angle: A
B E
D
Check students’ work.
All rights reserved.
8L
triangle
Yes; the brace makes two rigid triangles.
3R 4 I
K Q,L R, JK PQ , KL QR ,
JL PR .
13. B
14. F
15. C
16a. Given
16b. Given
16c. Converse of Isosceles Triangle Theorem
16d. SAS Postulate
16e. CPCTC
16f. Isosceles
Triangle Theorem
17. Sample:
Statements
Reasons
1. BD # AC ; D is
1. Given
midpoint of AC
2. BDC BDA
2. Perpendicular lines
form right s.
3. AD CD
3. Definition of midpoint
4. BD BD
4. Reflexive Property
of 5. BAD BCD
5. SAS Postulate
6. BC BA
6. CPCTC
18. Sample: Given that X is the midpoint of AD and BC,
AX DX and BX CX by the definition of midpoint.
AXB DXC because all vertical angles are congruent.
AXB DXC by the SAS Postulate, and therefore AB
DC by CPCTC.
Geometry Chapter 4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2C
Chapter 4 Answers
(continued)
Chapter Test, Form B
1. x = 58, y = 64 2. a = 40, b = 70, c = 70 3. SSS
4. SAS 5. HL 6. AAS 7. not possible 8. not possible
9. SSS or SAS 10. SSS or SAS 11. ASA
12.
A
D
TASK 2: Scoring Guide
a. Sample:
L
5 cm
4 cm
O
K
4 cm
M
5 cm
N
C
B F
E
b. Sample: Using the Pythagorean theorem, show that
All rights reserved.
kABC O kDEF
A D; B E; C F; AB DE;
AC DF; BC EF 13. B 14. AB XY;
BC YZ; AC XZ 15. A X; B Y;
C Z 16. C 17. J 18a. Given 18b. Given
18c. Def. of equilateral 18d. Reflexive Property of 18e. SSS Postulate
Alternative Assessment, Form C
N
kMNO O kOPM
P
O
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
ASA:
kABE O kACD
D
E
B
C
P
SAS:
kPOS O kPOR
S
O
Sample:
Statements
1. AE AD, B C
2. A A
3. ABD ACE
4. AB AC
5. EB DC
Reasons
1. Given
2. Reflexive Property of 3. AAS Theorem
4. CPCTC
5. Segment Addition Postulate
3 Student gives a proof that is accurate and complete.
2 Student gives a proof that contains minor errors or
omissions.
1 Student gives a proof that contains significant errors or
omissions.
0 Student makes little or no attempt.
TASK 4: Scoring Guide
Sample: The SSS, ASA, and SAS Postulates are statements
that are accepted as true without proof. The HL and AAS
Theorems, on the other hand, can be proved true, using
postulates, definitions, and previously proved theorems.
R
3 Student’s figures and information are clear and accurate.
2 Student’s figures and information contain minor errors or
omissions.
1 Student’s figures and information contain significant errors
or omissions.
0 Student makes little or no attempt.
Geometry Chapter 4
3 Student’s figures and explanation are accurate and clear.
2 Student’s figures and explanation contain minor errors or
omissions.
1 Student’s figures and explanation contain significant errors
or omissions.
0 Student makes little or no attempt.
TASK 3: Scoring Guide
TASK 1: Scoring Guide
Sample:
SSS: M
KO = MO. Then KOL MON by SAS Postulate or
SSS Postulate.
3 Student gives an explanation that is thorough and correct.
2 Student gives an explanation that is partially correct.
1 Student gives an explanation that lacks demonstrated
understanding of the difference between a theorem and
a postulate.
0 Student makes little or no attempt.
Answers
39
Chapter 4 Answers
(continued)
Cumulative Review
1. D
7. A
13.
2. G
8. G
3. D
9. C
C
4. H
10. G
5. D
6. G
11. A
12. J
D
14. x = 102; y = 102
16. AE CD
15. c, e, a, b, d or e, c, a, b, d
Given
AED CDE
SAS Theorem
CED ADE
CPCTC
All rights reserved.
AED CDE
Given
ED DE
Reflexive Property
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
17. Sample: The alarm sounds if and only if there is smoke. If
the alarm sounds, then there is smoke. If there is smoke, then
the alarm sounds.
18. AAA, SSA
40
Answers
Geometry Chapter 4
Chapter 5 Answers
Practice 5-1
Practice 5-5
1a. 8 cm
1b. 16 cm
1c. 14 cm
2a. 22.5 in.
2b. 15.5 in.
2c. 15.5 in.
3a. 9.5 cm
3b. 17.5 cm
3c. 14.5 cm
4. 17
5. 20.5
6. 7
7. 128
19
8. 42
9. 16.5
10a. 18
10b. 61
11. GH 6 AC,
HI 6 BA, GI 6 BC
12. PR 6 YZ, PQ 6 XZ,
XY 6 RQ
1. M, N
2. C, D
3. S, Q
4. R, P
5. A, T
6. S, A
7. yes; 4 + 7 8, 7 + 8 4, 8 + 4 7
8. no; 6 + 10 17
9. yes; 4 + 4 4
10. yes; 1 + 9 9, 9 + 9 1, 9 + 1 9
11. yes; 11 + 12 13, 12 + 13 11, 13 + 11 12
12. no; 18 + 20 40
13. no; 1.2 + 2.6 4.9
14. no; 8 21 + 9 14 18
15. no; 2. 5 + 3.5 6
16. BC, AB, AC
17. BO, BL, LO
18. RS, ST, RT
19. D, S, A
20. N, S, J
21. R, O, P
22. 3 x 11
23. 8 x 26
24. 0 x 10
25. 9 x 31
26. 2 x 14
27. 13 x 61
All rights reserved.
Practice 5-2
1. WY is the perpendicular bisector of XZ.
2. 4
3. 7.5
4. 9
5. right triangle
6. 5
7. 17
8. 17
9. equidistant
10. isosceles triangle
11. 3.5 ) 12. 21
13. 21
14. right triangle
15. JP is the bisector of LJN.
16. 9
17. 45)
18. 45
19. 14
20. Sample: Point M lies on JP .
21. right isosceles triangle
Practice 5-3
1. (-2, 2)
2. (4, 0)
3. (2, 1)
4. Check students’ work. The final result of the construction
is shown.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Triangles will vary.
1a. PQ = 8
1b. MN = 16
1c. YZ = 32
2a. NO = 2.5
2b. ST = 10
2c. UV = 20
3a. QR = 4
3b. ST = 8
3c. UV = 16
Reteaching 5-2
B
A
Reteaching 5-1
1. no
2. yes
3. yes
4. no
5. B
6. C
Reteaching 5-3
C
1. Sample:
5. altitude
6. median
7. none of these
8. perpendicular bisector
9. angle bisector
10. altitude
11a. (2, 0)
11b. (-2, -2)
12a. ( 0, 0)
12b. (3, -4)
13a. (0, 0)
13b. (0, 3)
Practice 5-4
1. I and III
2. I and II
3. The angle measure is
not 65.
4. Tina does not have her driver’s license.
5. The figure does not have eight sides.
6. The restaurant
is open on Sunday.
7. ABC is congruent to XYZ.
8. mY 50
9a. If two triangles are not congruent,
then their corresponding angles are not congruent; false.
9b. If corresponding angles are not congruent, then the
triangles are not congruent; true.
10a. If you do not live
in Toronto, then you do not live in Canada; false.
10b. If
you do not live in Canada, then you do not live in Toronto;
true.
11. Assume that mA 2 mB.
12. Assume
that TUVW is not a trapezoid.
13. Assume that LM
does not intersect NO.
14. Assume that FGH is not
equilateral.
15. Assume that it is not sunny outside.
16. Assume that D is obtuse.
17. Assume that
mA 90. This means that mA + mC 180. This,
in turn, means that the sum of the angles of ABC exceeds
180, which contradicts the Triangle Angle-Sum Theorem. So
the assumption that mA 90 must be incorrect. Therefore,
mA 90.
Geometry Chapter 5
2. Sample:
Reteaching 5-4
1. Step 1: B; Step 2: D; Step 3: F; Step 4: E; Step 5: A; Step 6: C
2. Step 1: Assume EF DE. Step 2: If EF DE, then
by the Isosceles Triangle Theorem, D F. Step 3: But
D F. Step 4: Therefore, EF DE.
Reteaching 5-5
1. Check students’ work. The longest side will be opposite the
largest angle. The shortest side will be opposite the smallest
angle.
2. largest: DEF; smallest: DFE
3. largest:
PQR; smallest: PRQ
4. largest: ACB; smallest:
CBA
5. longest: DF; shortest: FE
6. longest: PQ;
shortest: RQ
7. longest: SV; shortest: ST
Answers
29
Chapter 5 Answers
(continued)
3. centroid
4. interior
5. It is not possible for the
centroid to be on the exterior because all the medians are in
the interior of the triangle.
Enrichment 5-1
1., 5.
y
A
6.
4
E
F
D
2
H
F
4 C6 x
2E
B
2
2. D(0, 3)
3. E(2.5, 0)
4. F(2.5, 3)
6. 3.75 square units
7. They are equal.
yQ
9., 13.
8. 1 : 4
2
All rights reserved.
4
V
T
2
4
6
8
10 12 x
I
2
R
4
U
S
7. orthocenter
10. T(3.5, 0.5)
11. U(9, -3)
12. V(6.5, 0.5)
14. 21 square units
15. 5.25 square units
8. It is possible for the orthocenter
to be in the interior, the exterior, or on the triangle itself.
Explanations may vary.
9.
X
Center
of Gravity
Enrichment 5-2
1.–10. Check students’ work.
Y
10. Check students’ work.
11. centroid
Enrichment 5-4
11. Check students’ work. Give hint: mAOB = 45°.
A
O
1. If a fish is a salmon, then it swims upstream.
2. If a fish
swims upstream, then it is a salmon.
3. If a fish is not a
salmon, then it does not swim upstream.
4. If a fish does
not swim upstream, then it is not a salmon.
5. Exercises 1
and 4
6. the salmon
7. the fish that swim upstream
8.
9.
U
B
Q
P
P
10.
Enrichment 5-3
1.–2. A
U
Q
U
11.
U
Q
Q
P
P
D
B
30
C
Answers
Geometry Chapter 5
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Z
Chapter 5 Answers
(continued)
12. If a quadrilateral is a rectangle, then the angles of the
quadrilateral are congruent.
13. If the angles of a
quadrilateral are not congruent, then the quadrilateral is
not a rectangle.
14. If a quadrilateral is not a rectangle,
then the angles of the quadrilateral are not congruent.
15. All are true (12, 13, and 14).
16. the contrapositive
Enrichment 5-5
1. U
8. T
14. F
2. T
3. F
4. T
5. U
6. F
7. T
9. T
10. F
11. T
12. U
13. T
15. F
16. U
17. T
18. T
The name of the mathematician is THALES (th –a l –ez).
All rights reserved.
Chapter Project
Activity 1: Creating
isoceles; median, angle bisector, perpendicular bisector
1. 4 2. 4.5 3. 6.5 4. 5 5. Sample: VP = RT 6. D,
B, C 7. C, D, B 8. C 9. 3; 21 10. inside
11. inside 12. on 13. BC 14. LO, LM, MO, MN, NO
15. 3 16. 25 17. 6
Check students’ work.
Activity 3: Designing
Check students’ work.
✔ Checkpoint Quiz 1
1. angle
) bisector, since TM NM
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
angle. This means that one angle measures 90. This, in turn,
means that each of the three angles is 90 because the triangle
is equilateral and also equiangular. This means that the angle
sum of the triangle is 270, contradicting the Triangle AngleSum Theorem. So the assumption that an equilateral triangle
can have a right angle must be incorrect. Therefore,
an equilateral triangle cannot have a right angle.
14. Each leg must be greater than 5 units in length.
15. 6; 30
16. on
17. inside
18. outside
19. AB
20. XD, XC, XB
21. (5, 4)
22. (1, -2)
23. 2
24. 15
25. 24
Chapter Test, Form B
Activity 2: Experimenting
QM is the angle bisector
4. 6
5. 4
6a. 18
8. congruent
3a. If two lines are not parallel, then they intersect; false.
3b. If two lines intersect, then they are not parallel; true.
4. x = 9.5
5. x = 5
6. Sample: EF HG
7. I and III
8. II and III
9. D, B, C
10. B, D, C
11. C, D, B
12. C
13. Sample: Assume that an equilateral triangle has a right
2. congruent, since
3. 20, since TM NM
6b. 42
7. altitude
✔ Checkpoint Quiz 2
1. Inv: If you don’t eat bananas, then you will not be healthy.
Contra.: If you are not healthy, then you don’t eat bananas.
2. Inv.: If a figure is not a triangle, then it does not have three
sides. Contra.: If a figure does not have three sides, then it is
not a triangle.
3. Suppose that there are two obtuse angles
in a triangle. This would mean that the sum of two angles of a
triangle would exceed 180°. This contradicts the Triangle-Sum
Theorem. Therefore, there can be only one obtuse angle in a
triangle.
4. Suppose that the temperature is below 32°F
and it is raining. Because 32°F is the freezing point of water,
any precipitation will be in the form of sleet, snow, or freezing
rain. This contradicts the original statement. Therefore,
the temperature must be above 32°F for it to rain.
5. yes; 6 + 4 8, 8 + 6 4
6. no; 2.6 + 4.1 6.7
7. B, C, A
8. D, F, E
Alternative Assessment, Form C
TASK 1: Scoring Guide
Sample: Assume that Harold can create a triangular garden
plot with sides of 6 ft, 6 ft, and 13 ft. Then a triangle is possible
with sides 6, 6, and 13. But 6 + 6 13, so by the Triangle
Inequality Theorem, no such triangle exists. Therefore, the
plan is impossible.
3 Student gives a clear and accurate explanation.
2 Student gives an explanation that contains minor
inaccuracies.
1 Student gives an explanation that contains significant
flaws in logical reasoning.
0 Student makes little or no attempt.
TASK 2: Scoring Guide
Sample: The locus of points in a plane equidistant from a
single point forms a circle, as shown in the first figure. The
locus of points in a plane equidistant from the endpoints
of a segment forms a perpendicular bisector of the segment,
as shown in the second figure.
Chapter Test, Form A
1a. If a triangle does not have three congruent sides, then it is
not equiangular; true.
1b. If a triangle is not equiangular,
then it does not have three congruent sides; true.
2a. If
an isosceles triangle is not obtuse, then the vertex angle is
not obtuse; true.
2b. If the vertex angle of an isosceles
triangle is not obtuse, then the triangle is not obtuse; true.
Geometry Chapter 5
Answers
31
(continued)
3 Student gives an accurate explanation and drawings.
2 Student gives a drawing or an explanation that contains
minor errors.
1 Student gives an explanation or a drawing that contains
significant errors.
0 Student makes little or no attempt.
TASK 3: Scoring Guide
Expressed as a conditional: If ABC is an equilateral triangle
with midpoints D, E, and F on sides AB, BC, and AC
respectively, then DEF forms an equilateral triangle.
Sample: DEF is an equilateral triangle. Each of DEF’s
sides is exactly 12 the corresponding side of ABC. If ABC’s
sides are all the same length, then all of DEF’s sides are all
the same length as well.
Cumulative Review
1. A
8. H
14. F
2. G
3. D
4. H
5. B
6. G
7. D
9. D
10. F
11. A
12. J
13. B
15. 24
16. 28
17. Sample: The principal is
giving a student an award.
18. Suppose that an equilateral
triangle has an obtuse angle. Then one angle in the triangle
has a measure greater than 90. But this contradicts the fact
that in an equilateral triangle, the measure of each angle is 60.
Therefore, the equilateral triangle does not have an obtuse
angle.
19. Sample:
8
8
6
4
8
10
3 Student accurately states the conditional and the proof.
2 Student has minor errors in the statement of the conditional
or in the proof.
1 Student gives a statement of conditional or a proof that
contains significant errors.
0 Student makes little or no attempt.
20. (-1, 2)
21.
4
All rights reserved.
Chapter 5 Answers
T
Q
M
TASK 4: Scoring Guide
Sample:
22. C, B, A
B
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
D
C
ADC is isosceles. BAC = BCA because ABC
is an isosceles triangle with AB = BC. The angle bisectors
create two new angles at each vertex of ABC, each of
1
which has a measure 2 that of the angle bisected. Thus
1
mDAC = 2 mBAC and mDCA = 12 mBCA.
Because mBAC = mBCA, mDAC = mDCA,
which makes ADC an isosceles triangle.
3
2
1
0
Student gives accurate answers and justification.
Student gives answers containing minor errors.
Student gives a justification containing major logical gaps.
Student makes little or no effort.
32
Answers
Geometry Chapter 5
Chapter 6 Answers
Practice 6-1
Practice 6-5
1. parallelogram
2. rectangle
3. quadrilateral
4. parallelogram, quadrilateral
5. kite, quadrilateral
6. rectangle, parallelogram, quadrilateral
7. trapezoid,
isosceles trapezoid, quadrilateral
8. square, rectangle,
parallelogram, rhombus, quadrilateral
9. x = 7;
AB = BD = DC = CA = 11
10. m = 9; s = 42;
ON = LM = 26; OL = MN = 43
11. f = 5; g = 11;
FG = GH = HI = IF = 17
12. parallelogram
13. rectangle
14. kite
15. parallelogram
1. 118; 62
5. 101; 79
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
Practice 6-2
1. 15
2. 32
3. 7
4. 8
5. 12
6. 9
7. 8
8. 358
9. 54
10. 34
11. 54
12. 34
13. 100
14. 40; 140; 40
15. 70; 110; 70
16. 113; 45; 22
17. 115; 15; 50
18. 55; 105; 55
19. 61; 72; 108; 32
20. 32; 98; 50
21. 16
22. 35
23. 28
24. 4
2. 99; 81
3. 59; 121
4. 96; 84
6. 67; 113
7. x = 4
8. x = 16;
y = 116
9. x = 1
10. 105.5; 105.5
11. 90; 25
12. 118; 118
13. 90; 63; 63
14. 107; 107
15. 90; 51; 39
16. x = 8
17. x = 7
18. x = 28; y = 32
Practice 6-6
1. (1.5a, 2b); a
2. (1.5a, b); " a2 1 4b2
3. (0.5a, 0); a
2
4. (0.5a, b); " a 1 4b2
5. 0
6. 1
7. -21
8. 2
9. 2b
10. -2b
11. 2b
12. -2b
3a
3a
3a
3a
13. E(a, 3b); I(4a, 0)
14. O(3a, 2b); M(3a, -2b);
E(-3a, -2b)
15. D(4a, b); I(3a, 0)
16. T(0, 2b);
A(a, 4b); L(2a, 2b)
17. (-4a, b)
18. (-b, 0)
Practice 6-7
p
q
p
= qx
Practice 6-3
1a.
1. no
2. yes
3. yes
4. no
5. yes
6. yes
7. x = 2; y = 3
8. x = 6; y = 3
9. x = 64;
y = 10
10. x = 8; the figure is a because both pairs
of opposite sides are congruent.
11. x = 40; the figure is
y
not a because one pair of opposite angles is not congruent.
12. x = 25; the figure is a because the congruent opposite
sides are 6 by the Converse of the Alternate Interior Angles
Theorem.
13. Yes; the diagonals bisect each other.
14. No; the congruent opposite sides do not have to be 6.
15. No; the figure could be a trapezoid.
16. Yes; both
pairs of opposite sides are congruent.
17. Yes; both pairs
of opposite sides are 6 by the converse of the Alternate
Interior Angles Theorem.
18. No; only one pair of
opposite angles is congruent.
19. Yes; one pair of opposite
sides is both congruent and 6.
20. No; only one pair of
opposite sides is congruent.
Practice 6-4
1a. rhombus
1b. 72; 54; 54; 72
2a. rectangle
2b. 72; 36; 18; 144
3a. rectangle
3b. 37; 53; 106; 74
4a. rhombus
4b. 59; 90; 90; 59
5a. rectangle
5b. 60; 30; 60; 30
6a. rhombus
6b. 22; 68; 68; 90
7. Yes; the parallelogram is a rhombus.
8. Possible;
opposite angles are congruent in a parallelogram.
9. Impossible; if the diagonals are perpendicular, then the
parallelogram should be a rhombus, but the sides are not of
10. x = 7; HJ = 7; IK = 7
equal length.
11. x = 7; HJ = 26; IK = 26
12. x = 6;
HJ = 25; IK = 25
13. x = -3; HJ = 13; IK = 13
14a. 90; 90; 29; 29 14b. 288 cm2
15a. 70; 90; 70; 70
15b. 88 in.2
16a. 38; 90; 90; 38 16b. 260 m2
17. possible
18. Impossible; because opposite angles are
congruent and supplementary, for the figure to be a
parallelogram they must measure 90, the figure therefore must
be a rectangle.
Geometry Chapter 6
p
1b. y = mx + b; q = q (p) + b; b + q p2
p2
p
1c. x = r + p
+q- q
p2
rp
p2
q;
1d. y = q (r + p)
p2
rp
+ q - q ; y = q + q + q - q ; y = q + q;
rp
1e.
intersection at (r + p, q + q)
1g. y = mx + b; q =
r
q (r) +
r
q
b; b = q -
1f. (r, q)
r2
q;
2
y = qx + q - q
1h. y = qr (r + p) + q - rq ;
2
2
rp
rp
y = rq + q + q - rq ; y = q + q; intersection at
rp
rp
(r + p, q + q)
1i. (r + p, q + q)
2a. (-2a, 0)
b
3a b
2b. (-a, b)
2c. (- 2 , 2 )
2d. a
3a. (-4a, 0)
3b. (-2a, 3a)
3c. 32
3d. (2a, -a)
1
3e. 2
4. The coordinates for D are (0, 2b). The
r
r2
coordinates for C are (2a, 0). Given these coordinates, the
lengths of DC and HP can be determined:
DC = " (2a 2 0) 2 1 (0 2 2b) 2 = " 4a2 1 4b2;
HP = " (0 2 2a) 2 1 (0 2 2b) 2 = " 4a2 1 4b2;
DC = HP, so DC HP.
Reteaching 6-1
1.–4. Samples:
1.
trapezoid
2.
parallelogram
3.
rectangle
Answers
35
Chapter 6 Answers
(continued)
Reteaching 6-3
5. parallelogram
6. rectangle
7. isosceles trapezoid
8. rhombus
9. trapezoid
10. kite
11. rectangle
12. square
3. no
4. yes
1. BD CD, AE BD,
AE 6 CD
2. AE CD
3. ACDE is a
parallelogram.
Reteaching 6-2
1. Statements
1. Parallelogram ABCD
2. AB CD, BC DA
3. BD DB
4. ABD CDB
5. A C
2. Statements
1. Parallelogram ACDE;
CD BD
2. C E
3. CBD C
4. CBD E
3. Statements
1. Parallelogram ACDE;
AE BD
2. AE CD
3. CD BD
4. CBD C
4. Statements
1. Parallelogram ACDE;
CBD E
2. E C
3. CBD C
4. CD BD
5. BDC is isosceles.
5. Statements
1. Isosceles trap. ABDE;
C E
2. AE BD;
E BDE
3. C BDE
4. CBD BDE
5. C CBD
6. BCD is isosceles.
7. BD CD
8. AE CD
36
Answers
Reasons
1. Given
2. Opposite sides of a
parallelogram are
congruent.
3. Reflexive Prop. of 4. SSS
5. CPCTC
Reasons
1. Given
2. Opposite angles of a
parallelogram are .
3. Isosceles Triangle
Theorem
4. Substitution
Reasons
1. Given
2. Opposite sides of a
parallelogram are .
3. Substitution
4. Isosceles Triangle
Theorem
Reasons
1. Given
2. Opposite angles of a
parallelogram are .
3. Substitution
4. If 2 s of a are ,
sides opposite them
are .
5. Def. of isosceles
triangle
Reasons
1. Given
2. Definition of isosceles
trapezoid
3. Transitive Property
4. Alt. int. ’s are .
5. Transitive Property
6. Definition of isosceles
triangle
7. Definition of isosceles
triangle
8. Transitive Property
6. Statements
1. CBD C,
AE BD, AC ED
2. BD CD
3. AE CD
4. ACDE is a
parallelogram.
Reasons
1. Given
2. Substitution
3. If one pair of opposite
sides is both congruent
and parallel, then the
quadrilateral is a
parallelogram.
Reasons
1. Given
2. If 2 s of a are ,
sides opposite them
are .
3. Substitution
4. If both pairs of
opposite sides are ,
then the quad. is a
parallelogram.
All rights reserved.
1. no
2. no
5. Statements
rhombus
Reteaching 6-4
1. m1 = 60; m2 = 30; m3 = 90
m2
m2
m2
m3
=
=
=
=
50; m3 = 50; m4 = 100
100; m3 = 40; m4 = 40
60; m3 = 60
5. m1 =
15; m4 = 90
6. m1 =
2. m1 = 80;
3. m1 = 80;
4. m1 = 60;
75; m2 = 75;
45; m2 = 45
Reteaching 6-5
1. 90
2. 52
7. 27
8. 63
10. Sample:
3. 90
9. 27
4. 38
5. 52
6. 90
Statements
Reasons
1. LP MN
1. Given
2. LPN MNP
2. Theorem 6-15
3. PN PN
3. Reflexive Prop. of 4. LNP MPN
4. SAS Postulate
5. PM LN
5. CPCTC
6. LM LM
6. Reflexive Prop. of 7. PLM NML
7. SSS Postulate
8. LPQ MNQ
8. CPCTC
9. LQP MQN
9. Vertical angles are .
10. LQP MQN
10. AAS Theorem
11. 87
12. 48
13. 45
14. 48
15. 24
16. 24
17. 132
18. 24
19. 132
20. Sample:
Both LMQ and PNQ have the same angle measures, but
their sides have different lengths.
Reteaching 6-6
1. Q(x + k, m) 2. X(-a, 0); W(0, -b) 3. S(a, -a);
T(0, -a) 4. Each side has length a" 2, so it is a rhombus.
One pair of opposite sides has slope of 1, and the other pair
has slope of -1. Therefore, because (1)(-1) = -1, the
rhombus has four right angles and is a square. 5. Each side
Geometry Chapter 6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4.
(continued)
has length of " 2a 2 1 2a 1 1. Therefore, the figure is a
rhombus.
6. C(x - k, m)
Reteaching 6-7
All rights reserved.
1. Each diagonal has length " c2 1 (b 1 a) 2.
2. The
a c
midpoints are (b2, 2c ) and (b 1
,
).
The
line
connecting
the
2
2
midpoints has slope of 0 and is therefore parallel to the third
side.
3. The midpoints are (a2, 0), (a, b2), (a2, b) and (0, b2).
The segments joining the midpoints each have length
1
2
2
4. The midpoints are (a2, b2), (-a2, b2),
2 "a 1 b .
a
b
a
(-2, -2 ), and (2, -b2). The quadrilateral formed by these
points has sides with slopes of 0, 0, undefined, and undefined.
Therefore, the sides are vertical and horizontal, and consecutive sides are perpendicular.
5. The median meets the
base at (0, 0), the midpoint of the base. Therefore, the median
has undefined slope; i.e., it is vertical. Because the base is a
horizontal segment, the median is perpendicular to the base.
6. The midpoints are (a2, 0), (a 12 d, 2e ), (b 12 d, c 12 e),
and (b2, 2c ). One pair of opposite sides has slope of de ,
c
and the other pair of opposite sides has slope of b 2
a.
Therefore, the figure is a parallelogram because opposite
sides are parallel.
1. Some
2. No
3. Some
6. Some
7. No
8. Some
11. Some
12. No
13. No
16. All
17. Some
18. No
4. All
5. No
9. All
10. Some
14. Some
15. All
19. Some
20. All
Enrichment 6-2
1. ABED, BCFE, DEHG, EFIH, ACIG, DBFH
2. ABED, BCFE, DEHG, EFIH, ACIG, DBFH, ACFD,
DFIG, ABHG, BCIH
3. ABED, BCFE, DEHG,
EFIH, ACIG, DBFH, ACFD, DFIG, ABHG, BCIH,
AEHD, DBEG, BFIE, ECFH, BFEA, BCED, EFHG,
DEIH
4. DBHG, ECIH, ADEC, EACF, ABHD,
BCIE, HDFI, GDFH, BCGD, GCFH, DAIH, ABFI
5. pentagon, scalene triangle, two rectangles, two trapezoids,
two isosceles right triangles
6.
B
74°
62°
A
C
44°
44°
44°
92°
D
62° 62°
56
44°
F
62°
E
Because AJ = AB - BJ, AJ = 4.64 - 2.4 = 2.24 cm.
2. 44; LMNO is a rhombus, so MO # LN. So
mMRL = 90 and mMLR = 46; therefore
mMLR = 44.
3. 2.24 cm; because mAML = 46 =
mMLR, AJ 6 LN. So AJLN is a parallelogram, and AJ =
LN = 2.24 cm.
4. 2.4 cm; because mMLA = 44 =
mLMR, AD 6 MO. AMOD therefore is a parallelogram,
so MO = AD = 2.4 cm.
5. 44; because mLMR =
mRMN, mRMN = 44.
6. 90; because JBCK is a
square, mBJK = 90. Because mBJK + mNJM = 180,
mNJM = 90.
7. 88; mLMR + mRMN =
mLMN = 88.
8. No; the base angles of the triangle
are not congruent.
9. parallelogram
10. rhombus
11. square
12. rectangle
Enrichment 6-5
1. 64 square units
2. 65 square units
3. When you
actually cut out the shapes and reassemble them, you find that
the “diagonal” of the rectangle is not a straight line. Trapezoid
IV does not quite meet the top edge of triangle I, and similarly
there is a space between trapezoid III and triangle II. The
extra space represents the extra one square unit of area.
4. The areas differ by 4 square inches.
10 in.
6 in.
10 in.
1a. Given
1b. Definition of a regular hexagon
1c. SAS
1d. CPCTC
1e. Given any two distinct
1. 2.24 cm; JBCK is a square, so BC = BJ = 2.4 cm.
6 in.
Enrichment 6-3
Enrichment 6-4
10 in.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Enrichment 6-1
endpoints.
1f. Definition of a diagonal
1g. Definition
of a regular hexagon
1h. Reflexive Property of
Congruence
1i. SSS
1j. CPCTC
1k. If two lines
and a transversal form alternate interior angles that are
congruent, then the two lines are parallel.
1l. Definition
of a regular hexagon
1m. SAS
1n. CPCTC
1o. Definition of a regular hexagon
1p. SSS
1q. CPCTC
1r. If two lines and a transversal form
alternate interior angles that are congruent, then the two
lines are parallel.
1s. Definition of a parallelogram
2. A parallelogram can be constructed in an octagon by
drawing the diagonals as shown. Other answers are possible.
6 in.
Chapter 6 Answers
10 in. 6 in.
points, there is a unique line segment with these points as
Geometry Chapter 6
Answers
37
Chapter 6 Answers
(continued)
Enrichment 6-6
F
D
210
4
B
E
0
360
2
O
6 4 2
330
4
G
240
11. (1; 300°)
14. (4; 315°)
17. (5; 255°)
2.
2
B
4
2 A
C
6 x
6 4 2 O
2
4
D
rhombus
3.
8
6
change by a constant amount.
A
Chapter Project
B
2
C
Activity 1: Doing
6 4 2 DO
2
Check students’ work.
2
4
6
x
4
Activity 2: Analyzing
1. The effective area is rectangular.
2. The effective area is rectangular. The effective area is
Activity 3: Researching
6 x
6
Enrichment 6-7
larger in Figure 2 than in Figure 1 because the diagonal is
longer than the face.
3. You should tie the string to a vertical stick of the kite.
4. If the faces of the kite were unchanged, one diagonal of
the rhombus is longer than the diagonals of the square, so the
effective area would increase.
4
8
12. (3.5; 225°)
15. (5; 120°)
18. (1; 180°)
1. (0, 7)
2. (0, 1)
3. (5, 1)
4. (5, 7)
5. The x-coordinate of each point decreased by 4.
6. (7, 0)
7. (-3, -5)
8. (-1.5, -2.5)
9. Add 3
to the y-coordinate.
10. The y-coordinate of each point
decreased by 3.
11. The y-coordinate decreased by 3.
12. (3, 4)
13. M(10, 6), N(2, 6)
14. rectangle
15. Either all the x-coordinates or all the y-coordinates
A
trapezoid
300
270
10. (5; 60°)
13. (6; 150°)
16. (2; 90°)
D
C
30
A
I H
C
1.
60
B
150
180
90
rectangle
4.
8
6 A
D
4
B
2
6 4 2 O
2
4
6 x
Check students’ work.
4
✔ Checkpoint Quiz 1
1. 70, 110, 70
2. 53, 53, 52
3. 96, 84, 46
4. square
5. x = 20, y = 3
6. x = 2, y = 5
7. 30.5
✔ Checkpoint Quiz 2
1. x = 66, y = 57
2. x = 35, y = 35
3. x = 3,
y=4
4. True; they are the only quadrilaterals that
possess these properties.
5. False; only two triangles
at a time are congruent.
6. (n + 1, m)
7. (k, 0)
38
Answers
All rights reserved.
120
Chapter Test, Form A
C
kite
5. 5 cm
6. 3 in.
7. 4 m
8. 20
9. x = 33;
y = 81
10. 10
11. 20
12. x = 30; y = 30
13. 12
14. D(a, 0); E(b, c); (a 12 b, 2c )
15. D(-c, 0);
E(0, b); (-2c , b2)
16. D(0, b); E(a, 0); (a2, b2)
17. 28; 28
18. 105; 75
19. 90; 48
20. 55; 90
21. 22; 68
22. 53; 37 23. The lengths of segments AB, BC, and AC are:
AB = " j2 1 k2, BC = " k2 1 l2, AC = l + j. Thus,
Geometry Chapter 6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1.–9.
Chapter 6 Answers
(continued)
the perimeter of ABC is l + j + " j2 1 k2 + " k2 1 l2.
j
-2, k2
The midpoints of segments AB, BC, and AC are: M(
N(
l k
2, 2
), O(
l 2 j
2 ,0
),
). The lengths of segments MN, NO,
and MO are: MN = 12(l + j), NO = 12 " j2 1 k2,
MO = 12 " k2 1 l2 . Thus, the perimeter of MNO is
1
2 (l +
+ " j2
j
1
k2 +
perimeter of ABC.
" k2
1
l2), which
is half the
24. parallelogram
25. kite
27. parallelogram
28. square
30. isosceles trapezoid
26. rectangle
29. rhombus
Chapter Test, Form B
All rights reserved.
1.
y
6
A
4
square
2
B
6 4 2
2
2
4
Dx
6
2.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y
2
kite
A
10
D
x = 90 (Diagonals of a kite are #.)
y = 5 (Def. of isos. trapezoid)
z = 75 (Base angles of isos. trap. are .)
4
2
C
4
6
4.
x
8 10
y
8
A
trapezoid
B
3
2
1
0
Student gives correct answers and reasons.
Student gives mostly correct answers and reasons.
Student gives mostly incorrect answers and reasons.
Student makes little or no effort.
TASK 4: Scoring Guide
a. Q = (5 - a, 5); S = (5, 5 - a)
6
4
PR
C
2
8 6 4 2
2
D
3 Student gives accurate and complete answers and diagram.
2 Student gives answers and a diagram that are mostly accurate.
1 Student gives answers or a diagram containing significant
errors.
0 Student makes little or no effort.
TASK 3: Scoring Guide
B
6
2
Rhombuses
Squares
Cx
4 6
6
2
3 Student lists all statements accurately in part a and gives the
correct answers in part b.
2 Student gives mostly correct answers but with some errors.
1 Student gives answers that fail to demonstrate
understanding of the properties of parallelograms.
0 Student makes little or no effort.
Rectangles
B
4
4
8
Samples:
a. AB CD, BC AD, AB 6 CD, BC 6 AD,
ABD BDC, ACD BAC, CBD BDA,
CAD BCA, BE = ED, AE = EC,
ABC CDA, BCD BAD,
b. C, E, F
parallelogram
2
D
6 4 2
2
3.
TASK 1: Scoring Guide
Parallelograms
C
6 y
A
Alternative Assessment, Form C
TASK 2: Scoring Guide
4
6
15. 90; 45; 45 16. 71; 71; 38 17. parallelogram 18. rhombus
19. trapezoid 20. square
2
x
4
5. 7 in. 6. 6 cm 7. 22 m 8. x = 3.5 9. x = 19; y = 123
10. x = 6 11. 78; 102 12. 90; 61 13. 64; 128 14. 90; 63; 27
Geometry Chapter 6
0
= 55 2
2 0 =
1. Slope of QS
b. Slope of
5
= 5522(5a 2
2 a) = -1.
Because the product of their slopes = -1, PR # QS.
3 Student gives correct coordinates and a valid proof.
2 Student gives answers or a proof that contains minor errors.
1 Student gives incorrect coordinates in part a or a poorly
constructed proof in part b.
0 Student makes little or no effort.
Answers
39
Chapter 6 Answers
(continued)
Cumulative Review
1. B
2. J
3. B
4. H
5. D
6. J
7. A
8. G
9. B
10. G
11. B
12. J
13. A
14. 10
15. 38, 50, 92
16. Proof: AB BC DC AD by the definition of
a rhombus. Also, AC AC. Therefore, ABC CDA
by the SSS Theorem.
A
D
C
B
17.
18.
A
B O
N
L
M
Q
D
30
C
All rights reserved.
T
R
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
19. Sample: The construction of the undercarriage of a bridge;
it is a combination of triangles, which are the strongest
geometric polygon.
40
Answers
Geometry Chapter 6
Chapter 7 Answers
Practice 7-1
Practice 7-5
1. 1 : 278
2. 18 ft by 10 ft
3. 18 ft by 14 ft
4. 18 ft by 16 ft
5. 8 ft by 8 ft
6. 3 ft by 10 ft
7. true
8. false
9. true
10. true
11. false
12. true
13. true
14. false
15. false
16. 12
17. 12
18. 33
19. 21
5
20. 48
21. 88
22. 6
23. 25
24. -25
3
25. 14 : 5
26. 3 : 2
27. 12 : 7
28. 83
29.
1. BE
6. BE
y=4
y = 198
5
7
13
All rights reserved.
Practice 7-2
1. ABC XYZ, with similarity ratio 2 : 1
2. QMN RST, with similarity ratio 5 : 3
3. Not similar; corresponding sides are not proportional.
4. Not similar; corresponding angles are not congruent.
5. ABC KMN, with similarity ratio 4 : 7
6. Not similar; corresponding sides are not proportional.
7. I
8. O
9. J
10. NO
11. LO
12. LO
13. 3.96 ft
14. 4.8 in.
15. 3.75 cm
16. 10 m
17. 32
18. 53
19. 7 12
20. 4 12
21. 53
22. 37
23. 5
Practice 7-3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. AXB RXQ because vertical angles are .
A R (Given). Therefore AXB RXQ by the
3
XM
PX
AA Postulate.
2. Because MP
LW = WA = AL = 4 ,
MPX LWA by the SSS Theorem.
3. QMP
AMB because vertical s are . Then, because
QM
2
PM
AM = BM = 1 , QMP AMB by the SAS Theorem.
4. M A (Given). Because there
are 180° in a triangle, mJ = 130, and J C. So
MJN ACB by the AA Postulate.
5. Because
AX BX
AX = BX and CX = RX, CX = RX . AXB CXR
because vertical angles are . Therefore AXB CXR
by the SAS Theorem.
6. Because AB = BC = CA and
AB BC CA
XY = YZ = ZX, XY = YZ = ZX . Then ABC XYZ
by the SSS Theorem.
7. 15
8. 55
9. 48
7
2
4
10. 85
11. 20
12. 36
13. 33 ft
3
3
Practice 7-4
1. 16
2. 8
3. "77
4. 2 "11
5. 10"2
6. 6"5
7. h
8. y
9. x
10. a
11. b
12. c
13. x = 6; y = 6"3
14. x = 8 "3; y = 4 "3
15. 92
16. x = 4 "5 ; y = "55
17. x = "3;
y = "6; z = "2
18. x = 8; y = 2 "2; z = 6 "2
19. 2"15 in.
Geometry Chapter 7
2. EH
3. BC
4. JD
7. 16
8.
4
9.
4
10.
3
15
12. 4
13. x = 6; y = 6
15. 2
16. 4
17. 10
5.
25
2
JG
JI
11. x =
14. x =
189
5 ;
25
9;
Reteaching 7-1
1. 534 apples
2. 768 gallons
3. 4627 employees
4. about 4800 picture frames
5. 90
6. 2450
7. 85
8. 560
9. 58
10. 7308
11. 12
12. 20,250
13. 2500
Reteaching 7-2
1. KLM QRS; 5 : 3
2. ABC XYZ; 10 : 7
3. not similar
4. not similar
5. LMPR QICK; 1 : 1
6. not similar
7. similar with ratio 10 : 11
8. not
similar
9. similar with ratio 4 : 9
Reteaching 7-3
1. ABC ZYX by AA 2. not similar
3. QEU SIO by SSS 4. ABC RST
by AA 5. not similar
6. BAC XQR by
SAS 7. yes; by SSS or by AA 8. No; the
vertex angles may differ in measure.
9. All congruent
triangles are similar with ratio 1 : 1. Similar triangles are not
necessarily congruent.
Reteaching 7-4
25
60
144
13 ; y = 13 ; z = 13
= 420
3. x = 4.8; y =
29
441
2. x = 400
29 ; y = 29 ;
z
8
4. x = 12; y = 16
5. x = "2; y = "3; z = "6
6. x = 2"10;
120
y = "15; z = 2"6
7. 13
1.x =
Reteaching 7-5
1. x = 4.5
2. x = 38
3. x = 7.2; y = 4.8
15
25
4. x = 8 ; y = 8
5. x = 9
6. x = 23 ; y =
16
24
7. BN = 5 ; CN = 5
8. BY = 32
5
2
Enrichment 7-1
1. 10 lb
past year.
6. $106
2. $250; costs have not changed during the
3. $240,000
4. $168,750
5. $6785
7. 360 yd
8. 40
9. 42
10. 63
Enrichment 7-2
1.–4. Check students’ work.
Answers
29
Chapter 7 Answers
(continued)
Enrichment 7-3
4. Opposite s of a
parallelogram are .
5. UTV S
5. Opposite s of a
parallelogram are .
6. QRS VUT
6. AA Similarity Post.
3. Using the distance formula, AB = "180 or 6 "5,
BC = "45 or 3 "5, CA = 15, ST = 10, TB = "80 or
4 "5, and BS = "20 or 2 "5. If the two triangles are
similar, then corresponding sides are proportional. Therefore,
6"5
3"5
BC
CA
3
3
15
3
AB
ST = 10 = 2 = TB = 4"5 = 2 = BS = 2"5 = 2.
BC
AB
Because CA
ST = TB = BS , ABC TBS by the
SSS Similarity Postulate.
Enrichment 7-4
1. The triangles have two congruent angles; AA Similarity
(DE 2 BC)AB
1 BD
= AB
3. BD =
Postulate. 2. AB DE
BC
BC
4. 800 yd
5. 4500 ft
6. 43 ft
7. 9 m
Enrichment 7-5
1. 18
2. 1
3. 19
4. 3
5. 8
6. 7
7. 22
8. 9
9. 13
10. 4
11. 5
12. 6
13. 12
14. 17
15. 2
16. 10
17. 21
18. 15
19. 20
20. 16 ARCHIMEDES OF SYRACUSE
Chapter Project
Activity 1: Doing
Check students’ work.
Activity 2: Analyzing
Stage 1: 12; 4
Stage 2: 19; 5 31
1 1
Stage 3: 192; 27
; 79
Activity 3: Doing
Check students’ work.
30
Answers
The entire snowflake at each stage will be inside the regular
hexagon. The area of the hexagon is 2 square units, so the area
of the snowflake is always less than 2 square units.
Activity 5: Modeling
Check students’ work.
✔ Checkpoint Quiz 1
6
1. 12
2. 1 : 48
3. 11
8
7. 3.6 ft
8. no; 20
2 73
4. 6
5. O
9. 780 mi
6. MO
✔ Checkpoint Quiz 2
1. yes; AA similarity postulate
postulate
3. 5
4. 13.7
6. 8 13
7. 2.5
2. yes; SAS similarity
5. x = 4.5, y = 7.5
Chapter 7 Test, Form A
24
19
2
3
5. ABC XYZ
by the AA Postulate
6. not similar
7. KPM RPQ by the SAS Theorem
8. 30 ft
9. 9
3
10. 6 "5
11. B
12. 12
13. 2
14. 6
15. 18
16.
Check
students’
work.
17.
If two parallel
7
lines are cut by a transversal, alternate interior angles are congruent. Therefore, ABC EDC by the AA Postulate.
8
18. x = 9; y = 6 "3; z = 3 "3
19. x = 10
3;y = 3
20. C
21. 60
13
1. 28
2.
3. 45
4.
Chapter 7 Test, Form B
1. 21 2. 3.5 3. 12.5 4. SLQ NTR by SSS Theorem 5. MBD FWY by AA Postulate
6. PRG KRN by SAS Theorem 7. not similar
8. 9.6 m 9. 15 10. B 11. 9 12. 6.4 13. 3 : 1 14. 2 : 3
15. x = 27 16. x = 10; y = 4.5 17. 4.8
Alternative Assessment, Form C
TASK 1: Scoring Guide
AA Similarity Postulate and SAS and SSS Similarity
Theorems; check students’ work.
3 Student gives correct answers and accurate drawings
and explanations.
2 Student gives mostly correct answers, drawings, and
explanations.
1 Student gives answers, drawings, and explanations that
contain significant errors.
0 Student makes little or no effort.
Geometry Chapter 7
All rights reserved.
1. ABCD is a trapezoid
2. AD 6 BC
3. AED CEB
4. EBC EDA
5. AED CEB
2. Statements
1. T, U, and V are mdpts
2. TU 6 RV, UV 6 TR,
and TV 6 US
3. TUVR and TUSV are
parallelograms.
4. TUV R
Reasons
1. Given
2. Def. of a trapezoid
3. Vertical angles are .
4. Alt. int. angles are .
5. AA Similarity Post.
Reasons
1. Given
2. Triangle Midsegment
Theorem
3. Def. of parallelogram
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. Statements
Activity 4: Thinking
Chapter 7 Answers
(continued)
Claim 1 is true, and Claim 2 is false. Claim 3 is partly true; the
triangles are similar, but the ratio of the areas is 1 : 4, not 1 : 2.
Because the ratio of heights is not close to the ratio of head
diameter, the baby and the adult are not similar. (The baby’s
head will grow more slowly than the rest of the body.)
3 Student gives correct answers and provides logical reasons
to support the answers.
2 Student gives mostly correct answers, but the work may
contain minor errors or omissions.
1 Student gives answers that contain significant errors.
0 Student makes little or no effort.
3 Student gives a mathematically sound analysis.
2 Student gives an explanation that contains minor errors
or omissions.
1 Student gives an explanation containing significant errors
or omissions.
0 Student makes little or no effort.
TASK 3: Scoring Guide
Cumulative Review
All rights reserved.
TASK 2: Scoring Guide
One method is to pick any
point A on land and measure
AB and AC. Then select a
fraction (say, 12), and mark
X and Y halfway from A
to B and C, respectively.
Then measure XY. Then
BC = 2 ? XY.
?
C
B
X
Y
3 Student gives a mathematically
correct procedure and
explanation.
A
2 Student gives a procedure and
explanation that may contain minor errors.
1 Student gives an invalid procedure or a valid procedure
with little or no explanation.
0 Student makes little or no effort.
1. D
7. D
13. D
2. G
3. A
4. G
5. C
6. H
8. F
9. B
10. J
11. B
12. H
14. F
15. 10
16. Congruent triangles have
congruent corresponding angles and congruent corresponding
sides, so they have a similarity ratio of 1 : 1. Similar triangles
have congruent corresponding angles but do not necessarily
have congruent corresponding sides.
17. Sample:
18. If cows do not have six legs, then pigs cannot fly.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
TASK 4: Scoring Guide
Comparing the adult to the baby, we have:
• ratio of heights 3.5 : 1
• ratio of head diameter 1.7 : 1
Geometry Chapter 7
Answers
31
Chapter 8 Answers
Practice 8-1
Practice 8-6
1. " 51
2. 4 " 7
3. 2 " 65
4. " 7
5. 2 " 21
6. 18 " 2
7. 46 in.
8. 78 ft
9. 279 cm
10. 19 m
11. acute
12. right
13. obtuse
14. right
15. obtuse
16. acute
Practice 8-2
1. x = 2; y = " 3
3.
6.
All rights reserved.
14. w =
b=3
6
4
2. a = 4.5; b = 4.5" 3
3
25"3
c = 50"
4. 8 " 2
5.
3 ;d = 3
28"3
7. 2
8. x = 15; y = 15 " 3
3
10. 42 cm
1. 46.0, 46.0
2. -37.1, -19.7
3. 89.2, -80.3
4. 38.6 mi/h; 31.2º north of east
5. 134.5 m; 42.0º south
of west
6. 1.3 m/sec; 38.7º south of east
7. 55º north
of east
8. 20º west of south
9. 33º west of north
y
10a. 1, 5
10b.
11. 5.9 in.
12. 10.4 ft, 12 ft
10"3
3 ;x =
14 " 2
9. 3 " 2
13. 170 in.2
3
5; y = 5 " 2 ; z = 5"
15. a = 4;
3
16. p = 4 " 3; q = 4 " 3; r = 8; s = 4 " 6
2 4 6x
642
2
4
6
11a. 1, -1
y
11b.
6
4
Practice 8-3
1. tan E
3. tan E
7. 7.1
12. 71.6
17. 39
= 34 ; tan F = 34
= 52; tan F = 52
2. tan E = 1; tan F = 1
4. 12.4
5. 31.0
6. 14.1
8. 2.3
9. 6.4
10. 78.7
11. 26.6
13. 39
14. 72
15. 27
16. 68
18. 54
12a. -7, -1
y
12b.
6
4
2
Practice 8-4
1. sin P = 2"710 ; cos P =
5
12
13 ; cos P = 13
"2
"2
= 2 ; cos P = 2
3. sin P =
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
5. sin P
7. 64
13. 6.6
3
7
8. 11.0
3
5
42
4. sin P = "611 ; cos P =
9. 7.0
14. 37
2. sin P = 54; cos P =
15. 11.0
6. sin P =
10. 42
16. 56
15
17 ; cos
P
11. 7.8
4 6x
642
2
4
6
5
6
8
= 17
2 4 6x
4
6
12. 53
17. 11.5
13. Sample:
N
18. 9.8
Practice 8-5
1a. angle of depression from the birds to the ship
1b. angle of elevation from the ship to the birds
1c. angle of depression from the ship to the submarine
1d. angle of elevation from the submarine to the ship
2a. angle of depression from the plane to the person
2b. angle of elevation from the person to the plane
2c. angle of depression from the person to the sailboat
2d. angle of elevation from the sailboat to the person
3. 116.6 ft
4. 84.8 ft
5. 46.7 ft
6. 31.2 yd
7. 127.8 m
8. 323.6 m
9a.
W
48
E
S
14. Sample:
N
30
W
E
S
35
30 ft
9b. 26 ft
Geometry Chapter 8
Answers
33
(continued)
Reteaching 8-1
Enrichment 8-1
1.–2. Sample:
Friar Tuck is the winner. By using the Pythagorean Theorem,
Friar Tuck’s string length is 4 " 2 + 2 " 5 or 10.1 in. Little
John’s string length is 4 " 5 + 2 " 2 or 11.8 in.
Enrichment 8-2
1
Reteaching 8-2
1
1.–2. Sample:
1
1
7
1
8
1
9
or
3
10
11
1
1
1
3. Sample: 3.5 and 7
4. Sample: 3.5 and 7
5. Sample: They are approximately equal.
6. Check students’ work.
7. d = 2 " 3; e = 4
4"3
8"3
8. f = 3 ; g = 3
9. x = 3 " 3; y = 3
10. u = 6 " 3; v = 12
1. 3
2. –2
3. 45°
4. 135°; definition of
supplementary angles
5. 45°; vertical angles are
congruent.
6. 135°; vertical angles are congruent.
7.
y
x
1. DF = 2.7, EF = 7.5
2. XZ = 8.7, YZ = 10
3. RS = 8.4, ST = 13.1
4. 5.8
5. 30; 60
6. 16.6;
7. 23.6; 66.4
8. 2.1; 3.4
9a. 7.1
9b. "51
73.4
Reteaching 8-5
3. 2.2°
8. -2, 12
9. 90°, 90°
10. No; because the slopes of the
lines are opposite reciprocals, the lines must be perpendicular.
11. 70.6°, 109.4°
4. 42.0 ft
Reteaching 8-6
1. 12.8 mi/h at 51.3° off directly across
2a. 10 mi
2b. 36.9° west of due north
3. The bird’s path shifts 31.0°
east of due south.
4. 4 mi/h
Answers
2y
4
Enrichment 8-4
1. 5.8 cm
2. 3.4 cm
3. 7.5 cm
4. 47
5. 25.5
6. 107.5
7. 0.731
8. 0.431
9. 0.951
10. 7.9
11. 7.9
12. 7.9
13. They are equal.
14. The sides
of a triangle are proportional to the sides of their opposite
angles.
15. 70
16. 15.55 km
17. 11.83 km
18. Lighthouse Q; it is closer.
Geometry Chapter 8
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
x
Reteaching 8-4
34
1
Enrichment 8-3
1. 26.6
2. 76.0
3. 80.5
4. 26.6
5. 8.1
6. 0.8
7. mA = 53.1; mB = 36.9
8. mA = 67.4; mB = 22.6
9. mA = 70.5;
mB = 19.5
10. mA = 63.4; mB = 26.6
11. 26.6
12. 63.4
13. 49.4
2. 1113.8 ft
6. 19.1 ft
1
1
y 2x 3
Reteaching 8-3
1. 42.9 ft
5. 76.1 ft
4. "5
14 15
sum of the squares of the two legs equals the square of the
6. 4 " 5
7. 7 " 2
8. 8
9. 6
hypotenuse.
1
4
or
5
2
6
13
5. The
1
1
2
4. They are equal.
3. " 4 or 2
1
3. Sample: 9, 16, and 25
2. " 3
3
2
1. " 2
5.
All rights reserved.
Chapter 8 Answers
Chapter 8 Answers
(continued)
Chapter Test, Form A
Enrichment 8-5
1. the top of the tower
3.
P
1. right
2. PM
2. acute
6. x 5 7Í 2; y 5 7Í 2
P
8. sin A
All rights reserved.
B
x
x
9. sin A
60
y M
45
A y 100 M
11. 20.9
13. 26.6
4. the angle of elevation from B to the top of the tower;
5. the angle of elevation
30°-60°-90°; scalene right triangle
from A to the top of the tower; 45°-45°-90°; isosceles right
x
x
6. tan 60 = y
7. tan 45 = y 1 100
triangle
8. x = tan 60(y); x = tan 45(y + 100)
9. 136.6
10. 236.6; the height of the tower
11. yes
1. (0.5, 0.5)
2. (-0.25, 1.80)
4. 2.9 km; 38.7° south of east
3. (-2.25, 1.80)
Chapter Project
67
elevation from the top of the tree to the birds
y
as the side adjacent to Y. The side adjacent Z
to X is the same as the side opposite Y.
Y
adjacent /Y
So, tan x° = adjacent /X = opposite /Y = tan1 y8 .
18. 92.9, 83.6
19. 120.4, -258.3
20. 211.9 mi;
21. 11, 0
22. 4, 3
19° south of west
23. Sample:
y
4
2
4 2
2
line of sight
2
4 x
4
1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
7. x 5 4; y 5 4Í 3
14d. angle of depression from the birds to the x top of the tree
15. 102 ft
16. 143 ft
17. Sample: The side opposite X is the same
1 and 2 are complementary, and 2 and 3 are
complementary, so m1 = m3.
horizon
line
2
3
Chapter Test, Form B
edge
of
protractor
1. right
5. sin T
Activity 2: Measuring
6. sin T
8. 15.3
Check students’ work.
Check students’ work.
Activity 4: Researching
Check students’ work.
✔ Checkpoint Quiz 1
2. right
= 54, cos
A
3. acute
= 53; tan
B
= 34, sin
4. tan A = 34,
B
= 35, cos B = 45
5. tan A = 2, sin A = 0.8955, cos A = 0.4478; tan B = 0.5,
sin B = 0.4478, cos B = 0.8955
6. tan A = 0.8833,
sin A = 0.6625, cos A = 0.75; tan B = 1.1321, sin B = 0.75,
cos B = 0.6625
7. 11.6
8. 41°
9. 16.5
✔ Checkpoint Quiz 2
1. (71, 79)
2. (16, -9)
3. (-60, 148)
north of east
5. 16.1, 29.7° south of west
48.4° north of west
7. 22.5 ft
Geometry Chapter 8
2. obtuse
3. 2Í 85
4. x 5 4Í 3; y 5 8Í 3
6 "2
6 "2
= 19 ; cos T = 17
19 ; tan T = 17
93
14
= 17
; cos T = "1793; tan T = 14"
93
7. 60
12a. angle of
9. 63
10. 36.9
11. 78.7
depression from the cloud to the person in the castle
12b. angle of elevation from the person in the castle to the
12c. angle of depression from the person in the
cloud
12d. angle of elevation
castle to the person on the ground
13. 1710 ft
from the person on the ground to the castle
Activity 3: Comparing
sin A
5. 3
12. 76.0
14a. angle of elevation from the person to the
top of the tree
14b. angle of depression from the top of
the tree to the person
14c. angle of
X
Activity 1: Building
1. obtuse
4. 2Í 41
2"42
42
19
= 23 ; cos A = 23
; tan A = 2"
19
3 "7
"7
= 43 ; cos A = 4 ; tan A = 7
10.
opposite /X
Enrichment 8-6
string
3. obtuse
4. 31.6, 71.6°
6. 120.4,
14. sin x° =
opposite
hypotenuse
adjacent
and cos x° = hypotenuse
opposite
adjacent
sin x8
cos x8 = hypotenuse hypotenuse =
X
x
opposite
hypotenuse
hypotenuse ? adjacent =
opposite
adjacent =
15. -92.3, 59.9
17. 1, 10
tan x°
y
Z
Y
16. 18.0 mi; 33.7° north of east
Answers
35
Chapter 8 Answers
(continued)
18. Sample:
TASK 4 Scoring Guide:
a. a 12, -3; b 8, 8
b. a 12, 4; b 11, 3
c. a 14° south of east; b 45° north of east
W
W
y
4
4
2
2
4 x
4
3
2
1
0
W
W
W
W
Student gives correct answers.
Student gives answers that are mostly correct.
Student gives mostly incorrect answers.
Student makes little or no effort.
x = 57.3
1. A
2. H
3. D
4. G
5. D
6. G
7. C
8. J
9. D
10. F
11. B
12. J
13. D
14. 10
15. Sample: You may have used the branches to
climb into the tree.
16. Sample: Suppose that an
3 Student gives correct answer and shows a valid method.
2 Student gives generally correct work that contains
minor errors.
1 Student gives an incorrect answer, and the method is
not correct.
0 Student makes little or no effort.
equilateral triangle has an obtuse angle. Then one angle in the
triangle has measure greater than 90. This contradicts the fact
that, in an equilateral triangle, the measure of each angle is 60.
17. Sample: The side opposite the 90° angle is also the
hypotenuse, and there are two adjacent sides.
18. Sample:
The sum of the three angles of a triangle is 180°. On a sphere,
the sum of the three angles will be greater than 180°.
Alternative Assessment, Form C
TASK 1 Scoring Guide:
All rights reserved.
Cumulative Review
TASK 2 Scoring Guide:
111 ft
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3 Student gives a correct answer and shows a valid method.
2 Student gives generally correct work that contains
minor errors.
1 Student gives an incorrect answer, and the method is
not correct.
0 Student makes little or no effort.
TASK 3 Scoring Guide:
x = 1.53 mi; y = 0.52 mi
3 Student gives correct answer and shows a valid method.
2 Student gives generally correct work that contains minor
errors.
1 Student gives an incorrect answer, and the method is not
correct.
0 Student makes little or no effort.
36
Answers
Geometry Chapter 8
Chapter 9 Answers
9.
Practice 9-1
1. No; the triangles are not the same size.
All rights reserved.
2. Yes; the
hexagons are the same shape and size.
3. Yes; the ovals
are the same shape and size.
4a. C and F
4b. CD and CD, DE and DE, EF and EF, CF and
CF
5a. M and N
5b. MN and MN, NO
and NO, MO and MO
6. (x, y) S (x - 2, y - 4)
7. (x, y) S (x - 2, y - 2)
8. (x, y) S (x - 3, y - 1)
9. (x, y) S (x + 4, y - 2)
10. (x, y) S (x - 5, y + 1)
11. (x, y) S (x + 2, y + 2)
12. W(-2, 2), X(-1, 4),
Y(3, 3), Z(2, 1)
13. J(-5, 0), K(-3, 4), L(-3, -2)
14. M(3, -2), N(6, -2), P(7, -7), Q(4, -6)
15. (x, y) S (x + 4.2, y + 11.2)
16. (x, y) S (x + 13, y - 13)
17. (x, y) S (x,y)
18. (x, y) S (x + 3, y + 3)
19a. P(-3, -1)
19b. P(0, 8), N(-5, 2), Q(2, 3)
A
Practice 9-2
1. (-3, -2)
2. (-2, -3)
3. (-1, -4)
4. (4, -2)
5. (4, -1)
6. (3, -4)
y
7a.
y
T
B
4
2
6 4 2
2
2
4 B T
A
10.
T
y
T
4
6 4 2
A
11.
6 x
4
2 4
4
B B
A
y
T
6 x
T
4
4
L
2
6 4 2 O
2 I
4
6 x
K
J
B
A 4
A
y
B
12.
y
7b.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
8 6 4 2
A 2
6
8 x
B
I
J
O
2
6
L
2
4
6
6 x
2
6 4 2
2
y
4
Z
A
2
Y
6 4 2
2 4
W
4
8b.
6x
X
13. (-6, 4)
1. I
7.
2. I
4
6x
14. (-8, 0)
3. I
15. (0, -12)
4. GH
5. G
6. ST
U
T
P
4
Y
2
2
4 B
Practice 9-3
y
X
T, T
4
4
8a.
6 4
4
10
4
K
2
W
Z
4
Geometry Chapter 9
4
6x
S
T
U
S
Answers
35
Chapter 9 Answers
(continued)
8.
Practice 9-4
1. The helmet has reflectional symmetry.
2. The teapot
has reflectional symmetry.
3. The hat has both rotational
and reflectional symmetry.
4. The hairbrush has
P
reflectional symmetry.
5.
O
9.
P P
N
Q O
6.
Q
O
P
N
Q
M
N
L L
Q
O
L
M
N
L
M
N
M
L
O
Q
Q
P
line symmetry and 72° rotational
symmetry
P
9.
N
O
11. L
12.
E
7. This figure has no lines of symmetry.
8.
P
Q
M
P
O
line symmetry
Q'
10.
P'
S
R'
P
line symmetry and 90° rotational
symmetry
Q
11.
R
13.
P'
S
Q'
line symmetry and 45° rotational
symmetry
P
R'
12.
Q
X
O
X
R
180° rotational symmetry
13.
line symmetry
36
Answers
Geometry Chapter 9
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
10.
All rights reserved.
N
Chapter 9 Answers
(continued)
14.
12.
R
R
line symmetry and 45° rotational
symmetry
O
15.
A
180° rotational symmetry
A
All rights reserved.
16.
T
13. P(-12, -12), Q(-6, 0), R(0, -6)
14. P(-12, 41),
3
1
1
Q(1 4, -4), R(1 4 , 2)
15. P(-21, 6), Q(3, 24),
R(-6, 6)
16. P(-2, 1), Q(-1, 0), R(0, 1)
Practice 9-6
line symmetry
17.
18.
1. I. D II. C III. B IV. A
3.
COOK
2. I. B II. A III. C IV. D
B B B
20.
H
O
A
X
m
m
4.
C C
C
19.
21.
m
5.
1. L(-2, -2), M(-1, 0), N(2, -1), O(0, -1)
2. L(-30, -30), M(-15, 0), N(30, -15), O(0, -15)
3. L(-12, -12), M(-6, 0), N(12, -6), O(0, -6)
4. 35
5. 12
6. 2
7. yes
8. no
9. no
R
A
R
11.
R
O
T
A
m
J
Practice 9-5
10.
J
R
Geometry Chapter 9
A
m
y
6.
4
E
S
2
T
A
J
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
T
T
T
O
2
2
4
6 x
B
Answers
37
Chapter 9 Answers
y
y
13.
10
B
T
4
B
8
6
4 2 O
2
4
E
S
2
4 2 O
4 x
2
4
2
4
8 x
6
B
T
translational symmetry
y
T
2.
S
4
2
2 O
B 2
16. glide reflection
1.
4
line symmetry,
rotational symmetry,
translational symmetry,
glide reflectional symmetry
E
2
6 x
4
3.
y
10.
4
B
T
2
translational symmetry
6 4 2 O
2
E
4
4 x
2
4.
line symmetry,
rotational symmetry,
translational symmetry,
glide reflectional symmetry
S
y
11.
B
2
5.
4 x
6 E 2 O
2
line symmetry,
rotational symmetry,
translational symmetry,
glide reflectional symmetry
4
S
T
6.
y
12.
S
4
E
2
2 O
2
T
38
2
4
4
rotational symmetry,
translational symmetry
6 x
7.
8.
B
Answers
Geometry Chapter 9
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
9.
S
Practice 9-7
2
2 O
2
6 x
4
15. rotation
14. reflection
17. translation
E
S
2
4 E
y
8.
T
2
All rights reserved.
7.
(continued)
Chapter 9 Answers
(continued)
9.–11. Samples:
9.
10.
8.
Z
Y
X
11.
Z
X
All rights reserved.
Reteaching 9-4
12. yes
17. no
13. yes
14. no
15. yes
16. no
Reteaching 9-1
1.–5. Check students’ work.
6. A(0, -3), B(1, 1),
C(4, -1), D(5, -4)
7. A(4, -2), B(5, 2), C(8, 0),
D(9, -3)
8. A(3, 5), B(4, 9), C(7, 7), D(8, 4)
1. two lines of symmetry (vertical and horizontal), 180°
rotational symmetry (point symmetry)
2. one line of
symmetry (horizontal)
3. one line of symmetry (vertical)
4. two lines of symmetry (vertical and horizontal), 180°
rotational symmetry (point symmetry)
5. one line of
symmetry (vertical)
6. one line of symmetry (vertical)
Reteaching 9-5
1. Check students’ work.
y
2.
4
Reteaching 9-2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1.–5. Check students’ work.
6. reflection across x-axis:
F(-1, -3), G(-5, -1), H(-3, -5); reflection across
y-axis: F(1, 3), G(5, 1), H(3, 5)
7. reflection across
x-axis: C(2, -4), D(5, -2), E(6, -3); reflection across
y-axis: C(-2, 4), D(-5, 2), E(-6, 3)
8. reflection
across x-axis: J(-1, 5), K(-2, 3), L(-4, 6); reflection
across y-axis: J(1, -5), K(2, -3), L(4, -6)
2
4 2 O
2
2
4 x
4
y
3.
4
Reteaching 9-3
2
Y
1.–5.
4
Z
Y
X
X
2
4 x
4
Z
y
4.
T
Y
6.
O
2
4
2
Z
X
X
Y
T
Z
12
8 6 4 2 O
2
2
4
6
8
12 x
4
7. Sample:
Y
Z
S
X
X
Y
Z
Geometry Chapter 9
Answers
39
Chapter 9 Answers
12
10
y
5.
line symmetry across the
dashed lines, rotational
symmetry around points,
translational symmetry, glide
reflectional symmetry
6
4
2
8 6
2
2 4 6 8 x
Enrichment 9-1
4
6
8
10
1.
3.
5.
8.
(x, y) S (x - 7, y - 2)
2. (x, y) S (x + 4, y - 7)
(x, y) S (x - 10, y + 5)
4. (x, y) S (x - 3, y - 4)
Foster
6. yes
7. Wilson
(x, y) S (x - 7, - 3)
9. C
12
Enrichment 9-2
1. (-3, -1)
Reteaching 9-6
1. translation
2. reflection
3. rotation
4. glide reflection
5. rotation
6. glide reflection
7. reflection
8. translation
2. (-1, 0)
3. (5, 3)
4. the midpoint formula: M =
y
5.
x1 1 x2 y1 1 y2
(
,
2
2
)
6
L
K
4
Reteaching 9-7
2
1.
L'
J
6 4 2
J'
All rights reserved.
5.
(continued)
0
2
4
6 x
2
4
K'
2. Sample:
6. y = 21 x +
7.
1
2
T'
line symmetry across the dashed
lines, rotational symmetry around
points, translational symmetry,
glide reflectional symmetry
6 4 2
Q'
4
2
0
6x
R'
R
6
Q
8
line symmetry across the dashed
lines, rotational symmetry around
points, translational symmetry,
glide reflectional symmetry
2 S'4
2
4
3.
y
S
T
y = -4
4.
rotational symmetry around
points, translational symmetry
40
Answers
Geometry Chapter 9
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
6
Chapter 9 Answers
8.
6
y
(continued)
12.
E'
6
D'
4
4
D
2
H' 2
F'
0
6 4 2
2
4
8 6 4 2 0
H 2
6
I
G
4
2 J' 4
6x
J
6
y = -x
K
9.
6
All rights reserved.
I'
K'
6x
2
F
4
E
y
y
4
A
D
A'
6 4
0
B
2
C
4
2
C'
Enrichment 9-4
x=1
10.
M
6
y
N
M'
2
P
0
8 6 4 2
N'
Q
2
Q'
4
P'
6x
4
6
y = -2x - 2
11.
1
1. (0, -2, 3)
2. (-2, -2, 3)
3. (-2, -2, 0)
4. (0, -2, 0)
5. (0, -6, 0)
6. (0, -6, 3)
7. (-2, -6, 3)
8. (-2, -6, 0)
9. (2, 0, 3)
10. (2, -2, 3)
11. (2, -2, 0)
12. (2, 0, 0)
13. (6, 0, 0)
14. (6, 0, 3)
15. (6, -2, 3)
16. (6, -2, 0)
x
6
B'
= -12 x -
Enrichment 9-3
D'
2
6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y
G'
U'
V'
6
y
U
4
Enrichment 9-5
2
W'
T'
8 6 4 2
0
2
4
6
y = 3x + 2
Geometry Chapter 9
1. yes; rotational and point symmetry
2. no
3. yes;
rotational and point symmetry
4. yes; rotational and point
symmetry
5. no
6. no
7. diamonds
8. 12
9. Seven of diamonds; this does not have symmetry because
the diamond in the middle is toward either the bottom or top
of the card, and when you rotate the card 180°, the position
will be reversed.
10. All the face cards have symmetry.
11. yes; 2, 4, 10
12. No; Sample: When you look at the
card one way, three of the points of the hearts are pointing
down, and five are pointing up. When you rotate the card 180º,
five of the points of the hearts are pointing down, and three
are pointing up.
13. No; because the number and suit of
each card are placed in opposite corners, none of the cards
have line symmetry.
14. You can add a backward 3 with
the small club below it to the two empty corners to create line
symmetry, or you can remove the 3 with the small club below
it from each of the two corners.
V
6x
4
T
W
1a. (2, 0, 2)
1e. (2, 0, 0)
2a. (4, 0, 4)
2e. (4, 0, 0)
3. (0, 0, 0)
1b. (0, 0, 2)
1c. (0, 2, 2)
1d. (2, 2, 2)
1f. (0, 0, 0)
1g. (0, 2, 0)
1h. (2, 2, 0)
2b. (0, 0, 4)
2c. (0, 4, 4)
2d. (4, 4, 4)
2f. (0, 0, 0)
2g. (0, 4, 0)
2h. (4, 4, 0)
4. 2
5. Each edge in the image is double
that in the preimage.
6. Samples: Three faces are
coplanar for image and preimage; three faces are parallel;
each face in the image has two times the perimeter and four
times the area of the corresponding face in the preimage.
7. Surface area of image = 96 sq. units; surface area of
preimage = 24 sq. units; the ratio of the surface areas is 4 : 1.
Answers
41
Chapter 9 Answers
(continued)
8. Volume of image = 64 cubic units; volume of preimage =
8 cubic units; the ratio of the volumes is 8 : 1.
9. Dilations
in three dimensions are proportional to dilations in two
dimensions.
1.–9. Check students’ work.
10. 12-3; rotation; twice
11. 130
12. 100
13. 160
7.
8.
C
C'
L'
B
B'
B
D'
A
B'
D
C C'
A, A'
9.
L
C'
C
A
D'
D
12
R Y
O
T
A
T
I
O
N
10. T( 8, -5), Q(2, -3) and R(4, 0)
✔ Checkpoint Quiz 2
1. line symmetry;
rotational; 180°
H
2. line symmetry;
rotational; 180°
3. line symmetry
Activity 2: Modeling
a. Ancient Egyptian ornament
b. Oriental design
4. rotational; 180°
5. translational; no turns
6. rotational; preimage turns
7. translational
8. rotational, reflectional, translational
9. translational
Chapter Test, Form A
c. Greek vase design
42
Answers
All rights reserved.
11 units up
1.
2.
3.
4.
5.
P(13, 0), Q(10, -2), R(11, -4), S(13, -6)
P(-1, -2), Q(2, -4), R(1, -6), S(-1, -8)
P(0, -5), Q(2, -2), R(4, -3), S(6, -5)
P(-2.5, 0), Q(-1, -1), R(-1.5, -2), S(-2.5, -3)
P(5, 2), Q(2, 0), R(3, -2), S(5, -4)
6. P(-11, 3),
Q(-8, 1), R(-9, -1), S(-11, -3)
7. glide reflection
8. translation
9. translation
10. rotation
Geometry Chapter 9
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Activity 1: Investigating
a. vertical
b. vertical
c. both
Check students’ work.
1. No; the figures are not the same size.
2. yes
3. yes
4. (x, y) S (x - 3, y - 2)
5. 6 units right and
4 units down
6. a resultant translation of 4 units right and
2
C
3
O
R
5
4
M
P
E
G
P
R
F
L
O
E
L
I
6
I
S
E
D
I
M
I
E
C
S
7
R O T A T I O N A L
G
I
I
E
M
E
O
O
F
E
N
N
L
T
10
A
E
R
R
11
C
L
S Y M M E T
13
P O I N T
F
I
L
14
15
E
T R A N S F O R M A T I O N
N
C
I
A
T
L
16
I
T E S S E L L A T I O N
N
O
G
N
Chapter Project
Activity 5: Creating
✔ Checkpoint Quiz 1
Enrichment 9-7
T
R
A
N
S
L
A
T
8
I M A G
O
N
9
L I N
a, b
Activity 4: Classifying
a. mg
b. 12
Enrichment 9-6
1
Activity 3: Investigating
Chapter 9 Answers
(continued)
11.
16.
B'
C'
A'
P
A
C
All rights reserved.
12. line symmetry
13. rotational symmetry
14a. Any two of the following: ABCDEHIKMOTUVWXY
14b. Any two of the following: HINOSXZ
15. Translations can be performed on the pieces because they
can slide. Rotations can be performed on the pieces because
each piece can be turned. Reflections and glide reflections
cannot be performed on the pieces because the back of a
puzzle piece is useless in solving the puzzle. Dilations are
impossible because the pieces cannot change size.
16. A
17. Sample: glide reflectional, rotational, line,
and translational
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
18. L(-3, 6), M(6, 6), N(3, -9)
19. L(12, -34),
3
5
1
M(4, 1), N(4, -2)
20. L(-18, -18), M(-18, 18),
N(0, 0)
21. L(3, -2), M(0, -53 ), N(73 , -2)
22. YZ = 15 cm; XZ = 24 cm; scale factor = 73
23. XY = 25 in.; YZ = 40 in.; scale factor = 53
24. (-1, 3)
25. (5, 4)
26. (-2, -7)
27. (2, -7)
28. (1, -1)
29. (-3, 1)
B
17. 180° rotational symmetry
18. line symmetry
19. line symmetry; 180° rotational symmetry
20. (-7, 0) 21. (0, -3) 22. (-6, 1) 23. (-9, -4)
24. (2, -2) 25. (-5, 11)
Alternative Assessment, Form C
TASK 1: Scoring Guide
20; 200 cm; all distances are magnified 20 times.
Chapter Test, Form B
1. Yes; the figures are the same shape and size. 2. No;
the figures are not the same size. 3. Yes; the figures are the
same shape and size. 4. (x, y) S (x - 6, y + 1)
5. (x, y) S (x + 3, y - 7) 6. A9(0, -3), B9(-4, 0),
C9(-6, -4), D9(-3, -5) 7. A9(6, 3), B9(10, 0), C9(12, 4),
D9(9, 5) 8. A9(3, 8), B9(-1, 5), C9(-3, 9), D9(0, 10)
9. A9(5, 2), B9(1, -1), C9(-1, 3), D9(2, 4) 10. A9(-3, 0),
B9(0, -4), C9(-4, -6), D9(-5, -3) 11. A9(0, 3), B9(4, 6),
C9(6, 2), D9(3, 1) 12. (x, y) S (x + 4, y - 3)
13. (x, y) S (x - 3, y + 6) 14. rhombus and two triangles;
3 Student gives correct answers and a clear and accurate
explanation.
2 Student gives answers and an explanation that are largely
correct.
1 Student gives answers or an explanation that may contain
serious errors.
0 Student makes little or no effort.
TASK 2: Scoring Guide
Sample:
translation
15. A'
B'
P C'
A
C
B
Geometry Chapter 9
3 Student draws a figure that accurately reflects the stated
conditions.
2 Student draws a figure that is mostly correct but falls just
short of satisfying all the stated conditions.
1 Student draws a figure that has significant flaws relative to
the stated conditions.
0 Student makes little or no effort.
Answers
43
Chapter 9 Answers
(continued)
d. glide reflection:
TASK 3: Scoring Guide
Sample:
The car can undergo translations and rotations as it moves
and turns. But it can undergo neither a dilation, which would
change its size, nor a reflection, which would, for example,
put the steering wheel on the passenger side.
3 Student gives a correct and thorough explanation.
2 Student gives an explanation that demonstrates
understanding of transformations but may contain minor
errors or omissions.
1 Student gives an explanation containing major errors or
omissions.
0 Student makes little or no effort.
4 y
The glide reflection is equivalent to
a translation 2, 0 followed by a
reflection across the line y = -2.
2
4 2
2
2
4x
4
e. rotation:
4 y
4x
4 2
4 y
4
2
3 Student gives accurate graphs and correctly identifies the
details the problem calls for.
2 Student gives generally correct graphs and details, although
the work may contain some errors.
1 Student gives graphs and accompanying details that contain
significant errors or omissions.
0 Student makes little or no effort.
4x
4
4
a. translation:
4 y
2
translation vector: 1, 0
4 2
2
Cumulative Review
1. D
8. H
14. J
2. G
3. B
4. H
5. A
6. J
7. B
9. C
10. J
11. B
12. H
13. A
15. B
16. 49 cm3
17. A, H, I, M, O, T, U, V,
W, X, Y
18. Rotational; it is turning around a fixed point,
4x
4
the corner.
b. dilation:
19.
4
4 y
2
T
scale factor: 2
y
U
G
U
G
2
T
2
2
x
2
4
2
4 x
20. Equilateral triangle, square, and regular hexagon;
the measure of each interior angle is a factor of 360.
21. If a triangle is not a right triangle, then the Pythagorean
Theorem cannot be applied.
c. reflection:
y
2
4 2
2
2
line of reflection: y = 1
2
y
22.
4
4x
2
U
4
V
V
U
4
2
2
4
W
6 x
W
4
23. 21.21
44
Answers
Geometry Chapter 9
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Sample:
All rights reserved.
degrees of rotation: 90
TASK 4: Scoring Guide
Chapter 10 Answers
Practice 10-1
1. 8
2. 8 3. 60
4. 240
5. 2.635
6. 1.92
7. 0.8125
8. 9
9. 1500
10. 8.75
11. 3.44
12. 7
13. 110
14. 30
15. 16
16. 119
17. 12
18. 48
19. 40
20. 56
Practice 10-2
All rights reserved.
1. 48 cm2
2. 784 in.2
3. 11.4 ft2
4. 90 m2
2
2
2
5. 2400 in.
6. 374 ft
7. 160 cm
8. 176.25 in.2
2
9. 54 ft
10. 42.5 square units
11. 96 " 3 square units
12. 36 " 3 square units
13. 45 cm2
14. 226.2 in.2
2
15. 49,500 m
Practice 10-3
8"3
8"3
= 3 ;d = 3
1. x = 7; y = 3.5 " 3
2. a = 60; c
3. p = 4 " 3; q = 8; x = 30
4. m1 = 45; m2 = 45;
m3 = 90; m4 = 90
5. m5 = 60; m6 = 30;
m7 = 120
6. m8 = 60; m9 = 30; m10 = 60
7. 12 " 3
8. 36.75 " 3
9. 75 " 3
10. 120 in.2
2
2
11. 137 in.
12. 97 in.
1. 4 : 5; 16 : 25
2. 5 : 3; 25 : 9
3. 3 : 4; 9 : 16
2
5. 9 : 5
6. 1 : 6
7. 8 : 3
8. 576
9.
5 in.
1152
2
10. 25 in.
11. 2 : 3
12. 1 : 2
13. 2 : 5
14. 108 ft2
4. 1 : 2
2700
7
cm2
17.
p
2
18. 3
Practice 10-8
7
3
1. 8.7%
2. 10.9%
3. 15.3%
4. 10
5. 10
1
1
2
6. 5
7. 10
8. 1
9. 5
10. 22.2%
11. 4.0%
12. 37.5%
13. 20%
14. 50%
15. 40%
16. 33 13 %
Reteaching 10-1
1. Samples:
4
4
6
4
6
6
2. Samples:
4
3
6
cm2
3. 20
4. 55
8. Samples:
Practice 10-4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
15. 63
16. 25
8p
3p
19. 2
20. 10
3
8
in.2
2
8
5. 6 m
6. 8 ft
7. 4 in.
4
12
3
6
8
9. 5 in.
Practice 10-5
10. 23 cm
11. 9 ft
1. 174.8 cm2
2. 578 ft2
3. 1250.5 mm2
2
2
4. 192.6 m
5. 1131.4 in.
6. 419.2 cm2
7. 324.9 in.2
8. 162 cm2
9. 37.6 m2
10. 30.1 mi2
11. 55.4 km2
12. 357.6 in.2
13. 9.7 mm2
14. 384.0 in.2
15. 54.5 cm2
16. 9.1 ft2
2
2
17. 80.9 m
18. 83.1 m
19. 65.0 ft2
20. 119.3 ft2
2
2
21. 8 m
22. 55.4 ft
Reteaching 10-2
Practice 10-6
2. Sample: bases = 6 cm and 10 cm, height = 6 cm
3. Sample: 48 cm2
4. Check students’ work.
5. Sample:
1 1
1. 32p
2. 16p
3. 7.8p
4. Samples: DFA , ABF
00
1 1
5. Samples: FE , FD
6. Samples: FEB , EDA
0 0
7. Samples: FE , ED
8. ACB
9. FCD
10. 32
11. 43
12. 54
13. 39
14. 71
15. 90
16. 50
17. 220
18. 140
19. 220
20. 130
21. 6p in.
22. 16p cm
23. 49 p m
Practice 10-7
1. 49p
2. 21.2
6. 81
7.
or
9
18
10. 5 p
11. 8 p
p
4
3. 49
6p
p
1
16 p or 16
12. 23 p
Geometry Chapter 10
4. 49
6p p
8. 16 - 81
13. 15
2p
21.2
5.
9. 4p
14. 6p
1
4p
1. Sample:
6. Sample: diagonals = 4 cm and 9 cm
8. Check students’ work.
9. 144 ft2
2
2
11. 75 m
12. 98.4 cm
7. Sample: 18 cm2
10. 112 in.2
Answers
39
(continued)
Reteaching 10-3
Enrichment 10-1
1.–2. Sample:
1. c; each segment is the hypotenuse of a right triangle with
2. No; the angles must be shown to be
legs a and b.
3. They are congruent; SSS
4. 180
right angles.
5. 90; acute angles of a right triangle are complementary.
6. CPCTC
7. mSPQ = 180 - (mAPS + mPSA)
= 180 - 90 = 90
8. Repeat steps similar to those in
Exercises 4–7 to show that PQR, QRS, and RSP are
9. a + b; (a + b)2 = a2 + 2ab + b2
right angles.
2
10. c; c
11. a, b; 21 ab
12. The areas of congruent
13. a2 + 2ab + b2 = 4(12 ab) + c2
triangles are equal.
or a2 + 2ab + b2 = 2ab + c2
14. By subtracting 2ab
from both sides, you obtain a2 + b2 = c2 (The Pythagorean
Theorem).
radius
apothem
3. Sample: apothem = 2.5, radius = 3.5, side = 5
4. yes
5. They are approximately equal.
6. 100 in.2
7. 200 ft2
8. 72 cm2
9. 8 m2
Reteaching 10-4
1. 4 : 3; 16 : 9
2. 8 : 5; 64 : 25
4. 4 : 3
5. 4 : 5
6. 3 : 5
9. 9 : 25
3. 9 : 5; 81 : 25
7. 16 : 9
8. 16 : 25
Enrichment 10-2
Reteaching 10-5
1. 19.3 in.2
5. 363.3 in.2
2. 123.1 cm2
6. 126.3 in.2
3. 86.6 in.2
7a. 81 in.2
4. 172.0 in.2
7b. area of
square = s2
Reteaching 10-6
1. Sample:
1. Choose an interior point X; construct PX 6 BC, XQ 6 AC ,
and XR 6 AB.
2. A triangle is a trapezoid with one
side where b1 = 0. Hence A = 21 h(b1 + b2) = 12 hb2 .
3. Area of BDE = Area ABEC - Area ABDC
= [ 21 h(y + x + b)] - [ 21 h(y + x )]
= 12 hy + 12 hx + 21 hb - (12 hy + 21 hx)
= 12 hb
All rights reserved.
Chapter 10 Answers
Enrichment 10-3
2. Sample: 135° 3. Sample: 2 cm
4. Sample: 3p
2 cm 4.7 cm
5. Sample: 4.5 cm
6. They are approximately
10p
equal.
7. 5p
8.
9.
5p
10. 20p
11. 5p
3
3
2
6
15p
25p
12. 6
13. 2
2. 47 square units
3. 73 square
Enrichment 10-4
1. 14
2. 21
3. 43
4. 19
Enrichment 10-5
1. 36 cm2
2. 4 "3 cm or about 6.93 cm
4. A = 6 sin u, 0° u 180°
Reteaching 10-7
1.–2. Sample:
3. 6 cm
Enrichment 10-6
4 cm
1.–5.
Object
3. Sample: 50
cm2
cm2
4. Sample: 50.3
5. They
are approximately equal.
6. Check students’ work.
7. 25p ft2
8. 4p in.2
9. 64p m2
10. 20.25p ft2
2
2
11. 324p cm 12. 64p in.
13. 14 cm
14. 615 cm2
1. Quarter
2. CD
3. Circle with a
radius of 4 in.
4.
5.
Radius
(r)
1.15 cm
238 in.
4 in.
Diameter Circumference
(d)
(C)
2.3 cm
4 34 in.
8 in.
7.222 cm
14.92 in.
8p
Check student’s work.
Check student’s work.
6. They are approximately equal.
8.
C
d
3.14
3.14
p
3.14
3.14
7. p 3.14
Reteaching 10-8
1. Check students’ work.
2. 21%
3. Check
students’ work.
4. Sample: They are approximately equal.
5. Sample: The ratio would tend to be close to 21%.
6. 43%
40
Answers
Geometry Chapter 10
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. 19.5 square units
units
Chapter 10 Answers
(continued)
Enrichment 10-7
3
3
1. p
2. " 3
3. 3"
4. p - 3"
4
4
6. p4
7. 4
8. 1
9. p
10. " 2
12. p - 2
Enrichment 10-8
1. 0.06 or
1
18
2. 0.41
3.
Alternative Assessment, Form C
5. 21
11. 2
0.63 or "3
1 1 "3
b
h
36 in.2; 30.25 in.2; the seams in the finished quilt block account
for the difference.
b
A parallelogram with base b and height h has the same area
as two triangles with base b and height h. Thus, the area of the
parallelogram is: A = 2 (12 bh) = bh.
b.
b1
Activity 2: Creating
h
Check students’ work.
b2
Activity 3: Researching
Check students’ work.
✔ Checkpoint Quiz 1
1. 37.2 m2
5. 200 ft2
2. 20 ft2
6. 648 m2
h
4. 0.21
Activity 1: Modeling
All rights reserved.
Sample:
a.
Chapter Project
3. 63 m2
4. 96 in.2
7. 37; 949
✔ Checkpoint Quiz 2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
TASK 1: Scoring Guide
1. 119.3 ft2
2. 293.9 m2
3. 523.1 in.2
4. 6.9 in.
2
5. 20.9 ft
6. 10p mi; 25p mi
7. 18p cm; 81p cm2
p or about 21%
8. 4 2
9. 12 or 50%
4
A trapezoid with bases b1 and b2 and height h has the same
area as two triangles. The bases of the two triangles are b1
and b2, and the heights of both are h. Thus, the area of the
trapezoid is: A = 21 b1h + 12 b2h = 12 h(b1 + b2).
3 Student gives accurate explanations accompanied by clear
and correct diagrams.
2 Student gives generally correct explanations, although
there may be some errors in either the explanations or
the diagrams.
1 Student gives inaccurate or incomplete explanations and
diagrams.
0 Student makes little or no attempt.
Chapter 10 Test, Form A
TASK 2: Scoring Guide
1. 15.6 ft2
2. 41.6 cm2
3. 192 ft2
4. 80 square
units
5. 120 square units
6. 50 square units
7. 64
8. 91
9. D
10. 18.8 cm2
11. 19.1 in.2
9
12. 268.4 ft2
13. 166.3 in.2
14. 172.0 cm2
15. Sample: Draw an altitude to form a 30°-60°-90° triangle.
16. 13.5 cm
The altitude is 4" 3, and the area is 32 " 3.
17. 20 square units
18. 12p
19. 36p
20. 45 p
21. 54 p
22. p5
23. 4.44 square units
24. 21.5%
25. 9.1%
Sample: C(2, 8), D(-2, 5). The slopes of AB and CD are
equal, and the slopes of BC and AD are equal, so ABCD
is a parallelogram. The slopes of AB and BC are opposite
reciprocals, so AB # BC, and ABCD is a rectangle. AB = 5
and BC = 5, so the area of ABCD is 25 square units.
Chapter 10 Test, Form B
TASK 3: Scoring Guide
1. 110.9 ft2
2. 649.5 cm2 ; 649.6 cm2 ; or 649.7 cm2
3. 40.2 ft2
4. 60 ft2
5. 92 cm2
6. 72 in.2
16
1
2
7. 81
8. 4
9. A
10. 72.7 in.
11. 42.2 cm2
12. 21.6 cm
13. 385 cm2 ; 386 cm2 ; or 387 cm2
14. 36p
15. 324p
16. 4p
17. 18p
18. 75.4 cm2
19. 144p cm2
20. 4p in.2
Geometry Chapter 10
3 Student gives correct answers and provides complete and
accurate reasoning.
2 Student gives generally correct answers and reasoning.
1 Student gives generally incorrect answers and reasoning.
0 Student makes little or no attempt.
Q
Sample:
1
a. QXY is a 45°-45°-90° triangle,
so the apothem QY = XY = 1 ,
or "22.
"2
X
Y
The perimeter is 4(2XY) = 4(2"22) = 4 " 2 .
The area is (2XY)2 = 4 ? ("22)2 = 4 (24) = 2.
Answers
41
Chapter 10 Answers
(continued)
b. PAB forms a 30°-60°-90°
triangle, so AB = 1 , or "33.
P
"3
1
The perimeter is 6(2AB) =
12"33 = 4 " 3.
The area is
A
1
1
2 ap = 2 (1)(4"3) =
B
2 " 3.
All rights reserved.
3 Student gives clear and correct diagrams and explanations.
2 Student gives diagrams and explanations that may contain
some minor errors.
1 Student gives diagrams or explanations that contain major
errors or omissions.
0 Student makes little or no effort.
TASK 4: Scoring Guide
Sample:
a. a circle
b. a and r are nearly equal; p and C are nearly equal.
c. If the number of sides is great enough that the polygon
becomes indistinguishable from a circle, the two formulas
are equivalent, and one can be derived from the other.
For example:
A = 12 ap
A = 12 rC
A = 12 r(2pr)
(r a and p C)
(C = 2pr)
= pr2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3 Student gives correct answers to parts a and b and gives an
adequate explanation relating the two formulas in part c.
2 Student gives correct answers to parts a and b but exhibits
difficulty with the logic of part c.
1 Student gives answers and an explanation that may contain
significant errors.
0 Student makes little or no effort.
Cumulative Review
1. B
8. H
14. J
2. G
3. D
4. J
5. C
6. J
7. C
9. A
10. G
11. B
12. J
13. A
15. A
16. J
17. Sample: You may have
used the branches to climb into the tree.
18. Sample:
Suppose that an equilateral triangle has an obtuse angle. Then
one angle in the triangle has measure greater than 90. This
contradicts the fact that, in an equilateral triangle, the measure
of each angle is 60.
19. Sample:
8 cm
5 cm
2 cm
5 cm
20. 16p - 24 " 3
42
21. 61.5
Answers
m2
22. Euclid
Geometry Chapter 10
Chapter 11 Answers
Practice 11-1
1. 8
2. 5
3. 12
4. C
5. D
6. B
8. triangle
9. rectangle
10. rectangle
11. rectangle
Reteaching 11-1
7. A
1. Sample:
5 in.
8 in.
8 in.
3 in.
4 in.
5 in.
12. rectangle
3 in.
4 in.
4 in.
4 in. 4 in.
3 in.
5 in.
3 in.
3 in.
8 in.
3 in.
5 in.
All rights reserved.
Practice 11-2
1. 62.8 cm2
2. 552.9 ft2
3. 226.2 cm2
4. 113.1 m2
2
2
2
5. 44.0 ft
6. 84.8 cm
7a. 320 m
7b. 440 m2
8a. 576 in.2
8b. 684 in.2
9a. 216 mm2
9b. 264 mm2
10a. 48 cm2
10b. 60 cm2
2
2
11a. 1500 m
11b. 1800 m
12a. 2000 ft2
2
2
12b. 2120 ft
13. 8p m
14. 94.5p cm2
2
15. 290p ft
3 in. 3 in.
4 in.
5 in.
5 in.
2. Check students’ work.
3. vertices = 6; faces = 5;
?
edges = 9
4. E = V + F - 2; 9 =
6 + 5 - 2; 9 = 9
5.
6.
7.
8.
Practice 11-3
1. 64 m2
2. 620 cm2
3. 188 ft2
4. 36p cm2
2
2
5. 800p m
6. 300p in.
7. 109.6 m2
8. 160.0 cm2
2
2
2
9. 387.7 ft
10. 147.6 m
11. 118.4 cm
12. 668.2 ft2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 11-4
1. 1131.0 m3
2. 94,247.8 cm3
3. 7.1 in.3
3
3
4. 1781.3 in.
5. 785.4 cm
6. 603.2 cm3
3
3
7. 120 in.
8. 35 ft
9. 42 m3
10. 1728 ft3
3
3
11. 784 cm
12. 265 in.
13. 76 ft3
14. 943 in.3
3
15. 1152 m
1.–2. Check students’ work.
3. L.A. = 300 cm2;
S.A. = 350 cm2
4. 722.6 cm2
5. 864 in.2
6. 153.3 m2
Practice 11-5
Reteaching 11-3
1. 34,992 cm3
2. 400 in.3
3. 10,240 in.3
3
3
4. 4800 yd
5. 150 m
6. 48 cm3
7. 628.3 cm3
8. 314.2 in.3
9. 4955.3 m3
10. 1415.8 in.3
11. 1005.3 m3
12. 18.8 ft3
13. 20
14. 2
15. 6
1. Sample:
Reteaching 11-2
Practice 11-6
1. 2463.0 in.2
2. 6,157,521.6 m2
3. 12.6 cm2
2
2
4. 1256.6 m
5. 50.3 ft
6. 153.9 m2
7. 1436.8 mi3
3
3
8. 268,082.6 cm
9. 7238.2 m
10. 14.1 cm3
3
11. 2,572,354.6 cm
12. 904,810.7 yd3
13. 546 ft2
2
2
3
14. 399 m
15. 1318 cm
16. 233 cm
17. 7124 cm3
18. 5547 cm3
19. 42 cm3
Practice 11-7
1. 7 : 9
2. 5 : 8
3. yes; 4 : 7
4. yes; 4 : 3
5. not similar
6. yes; 4 : 5
7. 125 cm3
8. 256 in.3
3
2
2
9. 432 ft
10. 25 ft
11. 90 m
12. 600 cm2
13. 54,000 lb
14. 16 lb
Geometry Chapter 11
2. Check students’ work.
3. Sample: base = 3 cm,
slant height = 4 cm
4. Sample: 33 cm2
5. 896 cm2
6. 314.2 m2
7. 416.2 in.2
8. 349.3 ft2
Reteaching 11-4
1. 490 cm3
5. 756 ft3
2. 242.5 cm3
6. 2261.9 cm3
3. 735 m3
4. 2544.7 in.3
Reteaching 11-5
1. B
2. 301.6 cm3
5. 211.6 m3
3. 1296 ft3
4. 48 in.3
Answers
35
Chapter 11 Answers
(continued)
6. 25 cubic units
S.A. = 221.7 mm2
z
1. 2 : 1
2. 3 : 2
3. 64 : 125
4. 512 : 125
5. 250 "2 : 27
6. 16 : 9
7. 49 : 25
Enrichment 11-1
1. 8
2. 5
3. 6
4. 11
5. 4
6. 18
7. 7
8. 0
9. 10
10. 15
11. 3
12. 12
13. 2
14. 14
15. 9
16. 25
17. 34
18. 20
19. 22
20. 30
Enrichment 11-2
in.2
1. 96
2. 8
3. 88
4. 4
5. 22 in.2
6. 11 in.
7. 96 in.2
8. (2s2) in.2
9. (96 - 2s2) in.2
10. 4
11. (24 – 0.5s2) in.2
12. s in.
13. ( 24
14. 96 in.2
s - 0.5s) in.
2
15. 96 in. ; no matter what the size of the base is, the surface
1. on a line perpendicular to the base at the center of the
square
2., 4.
3. isosceles triangle
4. the height of the pyramid
5. Pythagorean Theorem
6. h = 12"4l2 2 s 2
2
8. s6 "4l 2 2 s 2
V
9., 13.
M
1. a2
2. b2
3. trapezoid
4. 12 l(a + b)
2
2
5. 2l(a + b)
6. a + b + 2l(a + b)
7. πR2
2
2
2
8. pr
9. pRl - prl
10. p(R + r + Rl + rl)
11. 152 cm2
12. 722.6 in.2
13. 506.6 ft2
2
14. 1952 m
Enrichment 11-4
1. 48 cubic units
4. 15 cubic units
5. 90 cubic units
2. 30 cubic units
z (3, 5, 6)
(0, 0, 6)
(0, 5, 6)
(3, 0, 6)
(3, 0, 0)
(0, 5, 0)
y
(3, 5, 0) (0, 0, 0)
7. s2
h
a
C
10. slant height
11. apothem
13. h = "l 2 2 a2
14. A =
12. height
2
2
15. nas
2 "l 2 a
nas
2
Enrichment 11-6
1. 43 pr 3
2. 34 pR3
3. 43 p(r 3 - R3)
3
4. 266,000,000,000 mi
5. 34 p(r - R)(r 2 + rR + R2)
4
6. 3 pd(3R3) = 4pdR3
7.
R
d
Exact Volume
of Shell
Approximate
Volume of Shell
10
0.1
126.93
125.66
50
1
32,048.44
31,415.93
100
3
388,413.95
376,991.12
8. 0.421 in.3
9. 1.34 in.3
Enrichment 11-7
1. 180 ft; 120 ft
2. $118.13
3. 6.53 in.
4. 5.52 in.
5. $5.40
6. 8.14 in.
7. 13.58 in.
8. 16.98 in.
9. 3.43 in.
10. 19.48 in.
x
36
3. 64 cubic units
h
s
2
area of the rectangular prism will always be 96 in.2 as long as
the entire piece of paper is used.
Enrichment 11-3
y
Enrichment 11-5
LEONHARD EULER WAS A FAMOUS MATHEMATICIAN
KNOWN FOR HIS CONTRIBUTIONS TO GEOMETRY,
CALCULUS, AND NUMERICAL ANALYSIS.
in.2
(0, 0, 0)
(5, 0, 0)
(5, 0, 2)
x
Reteaching 11-7
in.2
(0, 0, 2)
(0, 5, 2)
(0, 5, 0)
Answers
All rights reserved.
1.–4. Check students’ work.
5. four circles
6. They
are the same.
7. V = 113.1 in.3; S.A. = 113.1 in.2
8. V = 523.6 cm3; S.A. = 314.2 cm2
9. V = 7238.2 m3;
2
3
S.A. = 1809.6 m
10. V = 8181.2 ft ; S.A. = 1963.5 ft2
3
11. V = 1047.4 in. ; S.A. = 498.8 in.2
12. V = 310.3 mm3;
Geometry Chapter 11
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Reteaching 11-6
Chapter 11 Answers
(continued)
✔ Checkpoint Quiz 2
Chapter Project
1. 201.1 m2 ; 268.1 m3
2. 452.4 ft2 ; 402.1 ft3
3. 158.7 in.2 ; 84 in.3
4. 534.1 yd2 ; 942.5 yd3
5. 384 m2 ; 384 m3
6. 113.1 cm2 ; 50.3 cm3
Activity 1: Measuring
Check students’ work.
Activity 2: Analyzing
Sample:
l
w
d
V
S.A
V : S.A.
6
6
6
216
216
1:1
2
3
36
216
372
18 : 31
3
3
24
216
306
36 : 51
4
6
9
216
228
18 : 19
7. 21.2 in.3
Chapter Test, Form A
1. Sample:
2. Sample:
6 ft
6 ft
All rights reserved.
3m
1. 6 by 6 by 6; 2 by 3 by 36; the lower the ratio, the more
material is being used to enclose the same volume.
2. to minimize the cost of packaging
3. Sample: A cereal box is used for promoting the product,
so it needs a large surface area on the front.
Activity 3: Investigating
Check students’ work.
Activity 4: Creating
Check students’ work.
✔ Checkpoint Quiz 1
1.
6 in.
4 in.
4 in.
4 in.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2 in.
4 in.
3. The lateral area of a paint roller determines the amount of
paint it can spread on a wall in one revolution. Because the
lateral area of roller A is 24p in.2 and the lateral area of roller
B is 24p in.2, they both spread the same amount of paint in
one revolution.
4. V = 400.0 cm3; S.A. = 360.0 cm2
3
5. V = 254.5 in. ; S.A. = 226.2 in.2
6. V = 1050.0 cm3;
2
3
S.A. = 734.0 cm
7. V = 1005.3 ft ; S.A. = 628.3 ft2
3
8. V = 113.1 m ; S.A. = 113.1 m2
9. V = 119 cm3;
2
S.A. = 174 cm
10a. a cylinder with radius 2 units and
height 3 units
10b. 12p cubic units
10c. 12p square
units
11a. half of a sphere and a cone
11b. 251.3 cm2
12. Two regular cans of soup each have a volume of
423.3 cm3 for a total of 846.7 cm3. The single family-size can
with its volume of 863.9 cm3 is larger than the total volume
of the two regular cans.
13. 729 : 64
14. 9 : 4
15. 137.2p cm3
16. S.A. = 603.2 m2; V = 1206.4 m3
17. S.A. = 3327.7 cm2; V = 12,250.6 cm3
18. 18 cm
Chapter Test, Form B
2 in.
2.
10 m
7 ft
1. Sample:
4m
2. Sample:
2 cm
10 cm
6m
22 in.
4m
4 cm
3. 88
in.2
4. 80p
m2
6.
8. rectangle
Geometry Chapter 11
8 in.
5. 1814.4p
7.
in.2
2 cm
3. The lateral area of a paint roller determines the amount of
paint it can spread on a wall in one revolution. Because the
lateral area of roller A is 22p in.2 and the lateral area of roller
B is 21p in.2, roller A spreads more paint on a wall in one
revolution.
4. V = 523.6 m3; S.A. = 314.2 m2
5. V = 1583.4 cm3; S.A. = 754.0 cm2
6. V = 67.0 ft3;
S.A. = 138.2 ft2
7. V = 880.0 in.3; S.A. = 696.0 in.2
8a. half of a sphere and a cone
8b. 301.6 cm2
9. Two
regular cans of soup each have a volume of 461.8 cm3 for a
total of 923.6 cm3. The single family-size can with its volume of
942.5 cm3 is larger than the total volume of the two regular
cans.
10. D
11. S.A. = 1357.2 m2; V = 4071.5 m3
12. S.A. = 3296.8 cm2; V = 12,113.3 cm3
Answers
37
Chapter 11 Answers
(continued)
Alternative Assessment, Form C
TASK 3: Scoring Guide
TASK 1: Scoring Guide
a.
Sample:
Sample:
2 cm
4
8
8
4 cm
4
3 cm
4
8
4 cm
Prism: S.A. = 60 cm2; V = 24 cm3
Cylinder: S.A. = 10p cm2 or about 31.4 cm2; V = 4p cm3 or
about 12.6 cm3
3 Student gives accurate drawings and calculations.
2 Student gives drawings and calculations that may contain
minor errors.
1 Student gives drawings and calculations that contain
significant errors or omissions.
0 Student makes little or no attempt.
3 Student draws an accurate net and gives correct answers.
2 Student draws a net and gives answers that may contain
minor errors.
1 Student draws a net and does calculations incorrectly.
0 Student makes little or no attempt.
TASK 4: Scoring Guide
Foundation drawing:
1
All rights reserved.
b. V 390 m3; S.A. 306 m2
4 cm
1
Sample:
Pyramids have polygonal sides, whereas cones have rounded
sides. Each of these figures has a vertex, and each has volume
1
3 Bh. The base of a pyramid is a polygon, and the base of a
cone is a circle.
s
a pyramid with square base,
height h, slant height s
A
4( 12 xs)
2xs
V = 13 x2 h
a cone with radius r,
height h, and slant height s
S.A.
V
= pr 2 +
= 31 pr 2h
prs
3 Student gives accurate drawings and formulas.
2 Student gives drawings and formulas that may contain
minor errors.
1 Student gives drawings and formulas that contain significant
errors.
0 Student makes little or no attempt.
38
2
Orthographic drawing:
Front
x
S.A.
2
Answers
1
r
x
= x2 +
= x2 +
2
s
h
h
2
Top
Right
3 Student draws completely accurate figures.
2 Student draws generally accurate figures, but these may
contain a minor error.
1 Student draws figures containing significant errors.
0 Student makes little or no attempt.
Cumulative Review
1. B
2. J
3. D
4. J
5. C
6. G
7. D
8. H
9. B
10. G
11. B
12. F
13. A
14. J
15. between 4 yd and 12 yd
16. 2 cylinders
17. 14p in.2
18. Sample: A sharpened pencil is a
composite of a cone and a cylinder. The cone is the top of
the pencil with the point, and the trunk of the pencil is the
cylinder.
19. Check students’ work.
20. Both shapes
are triangles, and the sides are proportional in a 2 : 3 ratio.
Geometry Chapter 11
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
TASK 2: Scoring Guide
Chapter 12 Answers
Practice 12-1
1. 32
2. 50
3. 72
4. 15
7. "634
8. "901
9. 4 "3
11. inscribed
12. circumscribed
14. 28 in.
15. 52 ft
5. "91
6. 6
10. circumscribed
13. 24 cm
y
(0, 5)
17.
(0, 0)
(5, 0)
x
5
Practice 12-2
0
1. r = 13; mAB 134.8
3. r =
0
#41
2 ; mAB
102.7
2. r = 3 "5; mAB 53.1
0
4. 3
5. 4.5
7. 20.8
8. 13.7
9. 8.5
10. Q T;
0 0
PR SU
11. A J; BC KL
12. Construct
radii OD and OB . FD = EB because a diameter perpendicular to a chord bisects it, and chords CD and AB are congruent
(given). Then, by HL, OEB OFD, and by CPCTC,
OE = OF.
13. Because congruent arcs have congruent
chords, AB = BC = CA. Then, because an equilateral
triangle is equiangular, mABC = mBCA = mCAB.
All rights reserved.
6. 3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 12-3
1. A and D; B and C
2. ADB and CDB
3. ADB and CAD
4. 55
5. x = 45; y = 50;
z = 85
6. x = 90; y = 70
7. 180
8. 70
9. x = 120; y = 60; z = 60
10. x = 120; y = 100;
z = 140
11. x = 63; y = 63; z = 540
12. x = 50; y = 80; z = 80
13a. mAE = 170
13b. mC = 85
13c. mBEC = 10
13d. mD = 85
14a. mA = 90
14b. mB = 80
14c. mC = 90
14d. mD = 100
Practice 12-4
y
18.
3
5
4
3
2
1
(3, 5)
y
19.
2
2
2
x
4 6
20.
(1, 7)
y
6
4
(1, 1)
4 2
Practice 12-5
Geometry Chapter 12
4
2
(2, 4)
4
1. 87
2. 35
3. 45
4. 120
5. 72
6. 186
7. x = 58; y = 59; z = 63
8. x = 30; y = 66
9. x = 30; y = 30; z = 120
10. x = 16; y = 52
11. x = 138; y = 111; z = 111
12. x = 30; y = 60
13. 10
14. 14.8
15. 4.7
16. 4
17. 3.2
18. 6
1. C(0, 0), r = 6
2. C(2, 7), r = 7
3. C(-1, -6), r = 4
4. C(-3, 11), r = 2"3
5. x2 + y2 = 49
6. (x - 4)2 + (y - 3)2 = 64 7. (x - 5)2 + (y - 3)2 = 4
8. (x + 5)2 + (y - 4)2 = 14
9. (x + 2)2 + (y + 5)2 = 2
10. (x + 1)2 + (y - 6)2 = 5
11. x2 + y2 = 4
2
2
12. (x + 3) + (y - 3) = 1
13. x2 + (y - 3)2 = 16
2
2
14. (x - 7) + (y + 2) = 4
15. x2 + (y + 20)2 = 100
2
2
16. (x + 4) + (y + 6) = 25
x
1 2 3 4 5
6
2
(5, 1)
2
4
x
2
4
21.
23.
24.
25.
26.
x2
(x
(x
(x
(x
+
+
+
-
y2 = 25
4)2 + (y
7)2 + (y
4)2 + (y
4)2 + (y
22. (x - 5)2 + (y - 9)2 = 9
+
+
+
-
3)2
2)2
3)2
1)2
=
=
=
=
61
80
4
16
Answers
33
Chapter 12 Answers
(continued)
11.
Practice 12-6
1.
2 cm
Q
T
1.5 cm
12.
0.75 in.
2.
1 in.
Q
S
0.75 in.
T
U
13.
0.5 cm
3.
14.
A
B
5 mm
6 mm
4.
All rights reserved.
P
R
0.75 in.
0.5 in.
R 0.5 in.
Reteaching 12-1
S
0.75 in.
1. "157
2. "741
3. "2
4. 7.5
Reteaching 12-2
5.
B
A
E
C
7.
H
F
Reteaching 12-3
1. 87
6. 120
2. 40
7. 40
3. 60
8. 20
4. 55
9. 70
5. x = 94, y = 80
Reteaching 12-4
1. 93
2. 156
3. 42
7. x = 36; y = 60; z = 48
9. x = 46; y = 90; z = 44
4. 35
5. 60
6. 55
8. x = 64; y = 64; z = 52
Reteaching 12-5
P
J
G
8. a third line parallel to the two given lines and midway
between them
9. a sphere whose center is the given
point and whose radius is the given distance
10. the
points within, but not on, a circle of radius 1 in. centered
at the given point
1. (x - 3)2 + (y 3. (x - 6)2 + (y +
5. (x + 2)2 + (y +
6. (x - 3)2 + (y 7. (x - 5)2 + (y 8. (x - 4)2 + (y 9. (x + 2)2 + (y 11. C(0, 0); r = 0.2
13. C(3, 5); r = 4
11)2 = 4
2. (x + 5)2 + y2 = 225
2
6) = 7
4. x2 + y2 = 20
2
2) = 4
1)2 = 50
2)2 = 36
2)2 = 25
3)2 = 25
10. C(-3, -5); r = 5
12. C(4, 0); r = "6
Reteaching 12-6
1.–5. Check students’ work.
6. the line perpendicular to
AB at its midpoint
7. one, the midpoint of AB
34
Answers
Geometry Chapter 12
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
6.
1. 5 "5
2. 12 "2
3. 2 "14
4. 11.83 cm
5. 2.54 cm
6. 8.66 in.
7. 2.78 in.
Y
X
All rights reserved.
Chapter 12 Answers
(continued)
Enrichment 12-1
Enrichment 12-6
1. Given
2. Two points determine a line segment.
3. Two tangents drawn to a circle from an external point
are congruent.
4. Radii of a circle are congruent.
5. A radius and a tangent drawn to the same point of contact
form a right angle.
6. Definition of a square
7. Sides
of a square are congruent.
8. Addition Property
9. Segment Addition Postulate
10. Substitution Property
11. Addition Property
12. Substitution Property
13. Subtraction Property
1.–9. Check students’ drawings.
1. P with a radius
of 4 in.
2. two lines parallel to line l, one 2 in. above l
and the other 2 in. below l
3. 4 points
4. two lines
5. 2 points
Enrichment 12-2
Activity 1: Doing
1a. 1
1b. 3
1c. 6
1d. 10
1e. 15
N(N 2 1)
2. 1, 3, 6, 10, 15
3. triangular numbers
4.
2
5. From each point, (N - 1) chords may be drawn. So, a total
Activity 2: Exploring
of N(N - 1) chords may be drawn for N points. Because each
chord has been drawn twice, divide by 2.
6. N(N - 1)
Enrichment 12-3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. Given
2. Two points determine a line segment.
3. The measures of exterior angles of a triangle equal the
sum of the measures of the remote interior angles.
4. Radii of a circle are congruent.
5. Definition of
isosceles triangle
6. Base angles of an isosceles triangle
are congruent.
7. Definition of congruent angles
8. Base angles of an isosceles triangle are congruent.
9. Definition of congruent angles
10. The measures of
exterior angles of a triangle equal the sum of the measures
of the remote interior angles.
11. Substitution Property
12. Division Property
13. drawing diameter DE passing
through A so that radius OB AB
Enrichment 12-4
1. 2x + 2x + 8 + x + x - 32 = 360; 6x - 24 = 360;
6x = 384
2. x = 64
3. 128
4. 136
5. 64
6. 32
7. chords
AB
,
BD
;
16
8.
secants
EB
, EC; 52
)
9. tangent FB and chord AB; 64
10. chords
BD
, AC; 84
)
11. chords AC, BC; 64
12. tangent FB and secant FD;
36
13. chords AC, BD; 96
14.) AF, AE; 100
15. DE, DA; 48
16. tangent FB and chord BC; 68
Enrichment 12-5
parallel to line l, one 4 in. above l and the other 4 in. below l
6. two lines parallel to line l, one 6 in. above l
and the other 6 in. below l
7. empty set
8. 6 points
9. empty set
Chapter Project
Check students’ work.
Both figures are made from 9 circles with the same center. In
Figure A, the smallest circle and alternate rings are black. In
Figure B, vertical and horizontal tangents to the smallest circle
are drawn. Then the tangents at 45° angles to the first ones are
drawn. Alternating sections of the 6 outer rings are black, as is
the smallest ring.
Check students’ work.
Activity 3: Constructing
Check students’ work.
✔ Checkpoint Quiz 1
1. 52 in.
6. 11.5
z = 79
Geometry Chapter 12
4. 6
5. 12
8. x = 116, y = 88,
✔ Checkpoint Quiz 2
1. x = 140
2. x = 75, y = 105
3. x = 20, y = 70
4. x2 + (y - 1)2 = 6.76
5. (x + 3)2 + (y - 2)2 = 100
6. 12
7. 51
8. 20
7
Chapter Test, Form A
1.
3.
5.
7.
9.
C(0, 0); r = 10
2. C(11, -6); r = 9
C(-1, -4); r = "7
4. (x + 3)2 + (y - 2)2 = 1
2
2
(x - 2) + (y - 1) = 16
6. x2 + (y - 2)2 = 9
C = 31.4; A = 78.5
8. (x + 3)2 + (y - 4)2 = 41
y
C(1, 2)
r2
1a. the edge of the fountain pool
1b. plaza
1c. fountain pool
2. a circle with center (-3, 4) and a
radius of 3
3. all the interior points of a circle with center
(4, 0) and a radius of "12
4. all points outside a circle
with center (0, 0) and a radius of "18
5. a circle and all
its interior points with center (0, -2) and a radius of 5
6. x 2 + y 2 = 100
7. x2 + y2 36
8. The sergeant
can draw a circle with center (0, 0) and a radius of 2. The team
will search in the circle.
2. 44 cm
3. 40 m
7. x = 37, y = 100
9. x = 120
x
10. 90
Answers
35
Chapter 12 Answers
11.
(continued)
y
Chapter Test, Form B
x
0
0
1. inscribed 2. circumscribed 3. R; QS 4. M; LN
5. center (0, 0); r = 5 6. center (3, 5); r = 10
7. x2 + y2 = 9 8. (x - 4)2 + (y + 1)2 = 4
9. 37.7 units; 113.1 square units 10. x2 + (y + 1)2 = 25
11. center (-3, 2); r = 3
y
6
4
2
y
6 4 2
2
2
4
x
6
4
6
x
12. 90 13. 106 14. 105 15. 90 16. 45 17. 90 18. 180
19. 270 20. 47 21. 10 22. 8 23. 22 24. 126 25. 55
All rights reserved.
12.
Alternative Assessment, Form C
x 5
TASK 1: Scoring Guide
y
(a) center (2, 5), radius = 4 (b) The distance from point
(x, y) on a circle to the center (2, 5) is given by the formula
D = "(x 2 2) 2 1 (y 2 5) 2. In any circle, the distance, or
radius, is the same for all points on the circle. Therefore the
distance formula, when applied to the circle in this problem,
becomes 4 = "(x 2 2) 2 1 (y 2 5) 2 for all points (x, y) on
the circle. Squaring both sides of this equation yields the
equation of the circle, 16 = (x - 2)2 + (y - 5)2.
x
14.
y
y4
3 Student gives accurate answers and a correct explanation.
2 Student gives answers or an explanation that may contain
minor errors.
1 Student gives wrong answers or an incomplete or inaccurate
explanation.
0 Student makes little or no effort.
x
TASK 2: Scoring Guide
y 6
15. 120
20. 115
25. 35
16. 105
17. 118.0
18. 40
19. 70
21. 6
22. 11.8
23. 5.5
24. 45
26. 80
27. 5.8
28. 5.7
29. x = 63;
y = 71
30. x = 87; y = 95; w = 85; z = 93
31. x = 120; y = 64; z = 176
32. 44
36
Answers
(a) Construct the perpendicular bisector of two chords. The
point at which they meet is the center of the circle. (b)
Because the pentagon is regular, all chords are congruent.
Therefore the corresponding arcs are congruent. Because
there are 50
congruent arcs in the circle, each arc, and in
particular AB , has measure 72. To find
AB, call the center O,
0
and consider triangle AOB. Because AB has measure 72, so
does angle AOB. Because OA = OB, AOB is isosceles.
Therefore, the bisector of angle AOB is perpendicular to
(and bisects) chord AB, forming a 36°-54°-90° triangle. Then
1
AB
AB , yielding AB = 7.05.
=
sin 36 = 2
12
OA
Geometry Chapter 12
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
13.
Chapter 12 Answers
(continued)
3 Student devises a correct method and gives a correct answer
and a valid explanation.
2 Student devises a method, gives an answer, and gives an
explanation that may contain some errors.
1 Student gives a method, an answer, and an explanation that
may contain major errors or omissions.
0 Student makes little or no effort.
All rights reserved.
TASK 3: Scoring Guide
(a) Angles A and C are inscribed angles. Each is
inscribed in a different arc, but together the arc in which
they are inscribed comprise the entire circle. Therefore,
mA + mC = 12(360) = 180. And, because ABCD is a
parallelogram, angles A and C are congruent. Finally, if angles
are congruent and supplementary, then they are right. (b) The
diagonals of ABCD bisect each other because ABCD is a
parallelogram. Suppose that they intersect at point X. Then
AX = CX and BX = DX. Also, because ABCD is
inscribed in a circle, the diagonals are chords, and therefore
AX ? CX = BX ? DX. Substituting yields AX2 = BX2,
and therefore AX = BX = CX = DX. Then AC = BD.
(c) Either (a) or (b) allows you to conclude that ABCD must
be a rectangle.
Cumulative Review
1. B
8. F
14. H
2. J
3. A
4. F
5. D
6. G
7. A
9. D
10. G
11. C
12. H
13. D
15. B
16. "89
17. Check students’
work.
18. A tangent intersects a circle at exactly one
point. A secant intersects a circle at two points.
19. Check
students’ work.
20. This follows from the Pythagorean
Theorem (c2 = a2 + b2) or the fact that the longest side of
a triangle is opposite the largest angle.
21. If a triangle
is not a right triangle, then it is acute.
22. 135
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3 Student gives valid and accurate arguments and
explanations.
2 Student gives arguments that, although basically valid,
may contain minor flaws.
1 Student gives arguments containing major flaws.
0 Student makes little or no attempt.
TASK 4: Scoring Guide
(a) Because the circumscribing lines are tangent to the circle,
AB = AX, CB = CY, and DX = DY. Then, because it is
given that AB = BC, we have AX = CY. Adding segments
together yields DA0
= DC, showing that1
ACD is isosceles.
Secondly, because XY has measure 100, XBY has measure
260. Then mD = 21 (260 - 100) = 80, and because
ACD is isosceles, mA = mC = 50. Finally, AC = 20
AB
because AB = BC = 10, and, by trigonometry, cos 50 = AD
,
10
AB
AD = cos 50 = 0.6428 = 15.56. And again, AD = CD.
(b) Because angle A has measure 50 and line AB is tangent
at B, triangle AOB is a 25°-65°-90° triangle. Therefore, by
OB
trigonometry, tan 25 = AB
, OB = 10 ? tan 25 = 4.66.
3 Student gives accurate answers and explanations.
2 Student gives answers and explanations that may contain
minor errors.
1 Student gives answers and explanations that contain
significant errors.
0 Student makes little or no attempt.
Geometry Chapter 12
Answers
37
Geometry: All-In-One Answers Version A
2
Practice 1-1
Front
6.
Right
1
Guided Problem Solving 1-1
1
1. The pattern is easier to visualize. 2. The graph will go up.
3. Use Years for the horizontal axis. 4. Use Number of
Stations for the vertical axis. 5. increasing 6. greater
7. Yes. Since the number of stations increases steadily from
3
1950 to 2000, we can be confident that the number of stations
in 2010 will be greater than in 2000. 8. Patterns are necessary
to reach a conclusion through inductive reasoning. 9. (any
list of numbers without a pattern would apply) 2,435; 16,439;
16,454; 3,765; 210,564
Top
Right
ratio gets to a number that is approximately 0.618.
20. 0, 1, 1, 2, 3, 5, 8, 13
1
Front
1. 47, 53 2. 1.00001, 1.00001 3. 42, 54 4. -64, 128
5. 22, 29 6. 63.5, 63.75 7. Sample: 2 or 3 8. 51 or 49
9. 6 or 8 10. A or AA 11. D or G 12. Y or A
13. any hexagon 14. circle with 8 equally spaced diameters
15. a 168.75° angle 16. 21 hand-shakes 17. h = n(n 22 1)
18. 34 19. Sample: The farther out you go, the closer the
All rights reserved.
2
Right
5.
Chapter 1
Front
Front
Top
Right
7. A, C, D 8. C 9. D 10. B 11. A
Guided Problem Solving 1-2
1. They represent three-dimensional objects on a twodimensional surface. 2. nine 3. See the figure in 4, below.
4.
Practice 1-2
Fro
nt
Rig
ht
2.
Fro
5. Yes. It is similar to the foundation drawing, except there are
no numbers. 6. no
nt
Rig
7.
ht
3.
Fro
Top
nt
Rig
4.
1
1
1
1
1
1
ht
Front
8. Yes.
9.
Right
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1.
4
2
3
1
Right
Front
Practice
* ) 1-3
Front
Top
All-In-One Answers Version A
Right
1. AC 2. any two of the following: ABD, DBC, CBE,
ABE, ECD, ADE, ACE, ACD 3. Points E, B, and D are
collinear. 4. yes 5. yes 6. no 7. no 8. yes 9. no
Geometry
63
Geometry: All-In-One Answers Version A
yes 11.* yes) 12. yes
* ) 13. no 14. yes 15. yes* )
G 17. LM 18. PN 19. the empty set 20. KP
*M ) 22. Sample: plane ABD 23. Sample: plane ABC
AB 25. yes 26. no 27. yes 28. the empty set
no 30. yes 31. yes
5. Answers may vary. Sample: (0, 0) (Answers will be
coordinates (x, y), where y = 32x, x < 2.)
y
6
Guided Problem Solving 1-3
4
1. Collinear points lie on the same line. 2. Answers may vary.
(Some people might note that the y-coordinate of two of the
points is the same so that the third point must have the same
y-coordinate to be collinear. Since it does not, the points are
not collinear.) 3. horizontal 4. No. 5. No. 6. All points
must have the same y-coordinate, -3. 7. No. 8. a1, 21 b
2
A(2,3)
2
x
2
2
2
C(0,0)
4
6
6. yes 7. L(4, 2)
1. true 2. false 3. true 4. false 5. false 6. false
* )
7. JK, HG 8. EH 9. any three of the following pairs: EF
* )*
)
* )*
)
* )* )
and
)
*and JH
) * ; )EF and
* )GK
* ;) HG* and) JE
* ; )HG *and) FK
* ); JK *KG
EH ; JK and FG ; EJ and FG ; EH and FK ; JE and
;
* )
* ) *KG ) *JH ) * ) *JH ) GE
EH and
;
and KF ;
and
10. planes A and
B 11. planes A and C; planes B and C 12. planes A and C
)
)
13. planes B and C 14. Sample: EG 15. 6 16. EF and
)
)
)
) )
) )
ED or EG and ED 17. FE , FD 18. GF , GD 19. yes
20. Sample:
Practice 1-5
1. 4 2. 12 3. 20 4. 6 5. 22 6. -10 or 6 7. -1 or 1
8. 3; 4; no 9. 6; 6; yes 10. C or -2 11. 15 12. 31
13. 14 14. x = 11 32 ; AB = 31; BC = 31 15. x = 35 23 ;
AB = 103; BC = 103
Guided Problem Solving 1-5
T
1. AD DC 2. AD = DC 3. Segment Addition Postulate.
4. Since AD = DC, AC = 2(AD). 5. AC = 2(12) = 24
6. y = 15 7. DC = AD = 12 8. Answers may vary.
9. ED = 11, DB = 11, EB = 22
U
Practice 1-6
S
1. any three of the following: &O, &MOP, &POM, &1
2. &AOB 3. &EOC 4. &DOC 5. 51 6. 90
7. 17 8. 107 9. 141 10. 68 11. &ABD, &DBE, &EBF,
&DBF, &FBC 12. &ABF, &DBC 13. &ABE, &EBC
14. x = 5; 8; 21; 13 15. x = 9; 85; 35; 120
21. Sample:
R
Q
W
P
Guided Problem Solving 1-6
1. Angle Addition Postulate 2. supplementary angles
3. m&RQS + m&TQS = 180 4. (2x + 4) + (6x + 20) = 180
5. x = 19.5 6. m&RQS = 43; m&TQS = 137 7. The sum
of the angle measures should be 180; m&RQS + m&TQS =
43 + 137 = 180. 8a. x = 11 8b. m&AOB = 17;
m&COB = 73
Guided Problem Solving 1-4
1. Opposite rays are two collinear rays with the same
endpoint. 2. a line 3–4. See graph in Exercise 5 answer.
Practice 1-7
1.
X
64
Geometry
All rights reserved.
Practice 1-4
* )* )
B(4,6)
Y A
B
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
10.
16.
21.
24.
29.
(continued)
Geometry: All-In-One Answers Version A (continued)
2.
9.
1 2
1
X
2
B
Y
10.
2 ? 2
2
3.
C
All rights reserved.
11.
X
Y
X
X
4.
12.
2 ? X
Z
W
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
5.
13.
X
45°
Z
90°
14. true 15. false 16. false 17. true 18. true
19. false 20. true
6.
MN
OP
Guided Problem Solving 1-7
B
A
7.
1. &DBC > &ABC 2. complementary angles
3. &CBD 4. m&CBD = m&CBA = 41
5. m&ABD = m&CBA + m&CBD = 41 + 41 = 82
6. m&ABE + m&CBA = 90
OP
MN
K
L
m&ABE + 41 = 90
m&ABE = 49
7. m&DBF = m&ABE = 49 8. Answers may vary. Sample:
The sum of the measures of the complementary angles should
be 90 and the sum of the measures of the supplementary
angles should be 180. 9. m&CBD= 21, m&FBD = 69,
m&CBA =21, and m&EBA = 69
8.
2
1
A
1
All-In-One Answers Version A
2
Geometry
65
Geometry: All-In-One Answers Version A
Practice 1-8
Guided Problem Solving 1-9
1.–5.
1. six 2. It is a two-dimensional pattern you can fold to form
a three-dimensional object. 3. rectangles
y
6
A
C
4.
4
6
4
2
6
D
4
(continued)
2
E
2
4
4
4
6
6 x
4
8
2
8
B
4
4
6
Q
4
2
P
R
2
4
6
8x
4
4
6
6
5. 208 in.2 6. They are equal. 7. 208 in.2 8. Answers will
vary. Sample: 2(4 6) + 2(4 8) + 2(6 8); the results are the
same, 208 in.2 9. 6(72) = 294 in.2
1A: Graphic Organizer
1. Tools of Geometry 2. Answers may vary. Sample: patterns
and inductive reasoning; measuring segments and angles; basic
constructions; and the coordinate plane 3. Check students’
work.
1B: Reading Comprehension
* ) * )*
) *
)
1. Answer may vary. Sample: AB 6 CD , EF 6 GH ,
JK > LM, JL > KM, m/AJF + m/FJK = 180°,
) * )
/HKB > /KMD,DN ' CD 2. Points A, M, and S are
* ) * )
* )
collinear. 3. AB , HI , and LN intersect at point M. 4. a
24. 24.7 25. (3.5, 3)
1C: Reading/Writing Math Symbols
Guided Problem Solving 1-8
1. Distance Formula 2. The distance d between two points
A(x1,y1) to B(x2,y2) is d 5 Í (x2 2 x1) 2 1 (y 2 2 y1) 2.
3. No; the differences are opposites but the squares of the
1. Line BC is parallel to line MN. 2. Line CD
3. Line segment GH 4. Ray AB 5. The length of
segment XY is greater than the length of segment ST.
6. MN = XY 7. GH = 2(KL) 8. ST ' UV
9. plane ABC || plane XYZ 10. AB 6 DE
differences are the same.
4. XY 5 Í (5 2 (26)) 2 1 (22 2 9) 2 5. To the nearest
tenth, XY = 15.6 units. 6. To the nearest tenth, XZ =
12.0 units. 7. Z is closer to X. 8. The results are the same;
e.g., XY 5 Í (26 2 5) 2 1 (9 2 (22)) 2 5 Í 242 , or
about 15.6 units, as before.
9. YZ 5 Í (17 2 (26)) 2 1 (23 2 9) 2 5 Í 673 ; to the
nearest tenth, YZ = 25.9 units. To the nearest tenth,
XY+YZ+XZ = 53.5 units.
Practice 1-9
1. 792 in.2 2. 3240 in.2 3. 2.4 m2 4. 32p 5. 16p
6. 7.8p 7. 26 cm; 42 cm2 8. 46 cm; 42 cm2 9. 29 in.;
42 in.2 10. 40 ft; 51 ft2 11. 40 m; 99 m2 12. 40 m; 91 m2
13. 68; 285 14. 26;3 22 15. 30; 44 16. 156.25p
17. 10,000p 18. p16 19. 48; 28 20. 36 21. 26; 13
66
Geometry
All rights reserved.
6. 5"2 7.1 7. 2"17 8.2 8. 12 9. 8 10. 12
11. "26 5.1 12. (5, 5) 13. (21 , 1) 14. (10 21 , -5)
15. (-2 12 , 6) 16. (-0.3, 3.4) 17. (2 78 , -58 ) 18. (5, -2)
19. (4, 10) 20. (-3, 4) 21. yes; AB = BC = CD =
DA = 6 22. "401 20.025
23.
y
S
4
1D: Visual Vocabulary Practice
1. parallel planes 2. Segment Addition Postulate
3. supplementary angles 4. opposite rays 5. isometric
drawing 6. perpendicular lines 7. foundation drawing
8. right angle 9. congruent sides
1E: Vocabulary Check
Net: A two-dimensional pattern that you can fold to form a
three-dimensional figure.
Conjecture: A conclusion reached using inductive reasoning.
Collinear points: Points that lie on the same line.
Midpoint: A point that divides a line segment into two
congruent segments.
Postulate: An accepted statement of fact.
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
6
Geometry: All-In-One Answers Version A
(continued)
All rights reserved.
1F: Vocabulary Review Puzzle
C
O
L
L
I
N
E
A
R
P
P
D
C
L
Z
J
P
S
Y
J
E
Z
C
C
C
E
O
R
H
L
O
E
G
E
L
K
C
A
F
R
T
S
V
I
L
A
B
B
N
L
I
L
A
E
A
O
R
E
U
N
N
X
W
L
T
Z
S
L
B
G
N
W
I
N
N
Y
L
T
I
Y
Y
U
R
G
T
A
C
N
E
L
F
Q
G
G
D
P
C
O
S
C
S
B
R
R
K
A
C
I
E
N
I
L
R
D
M
E
P
I
D
O
U
A
X
T
F
N
R
V
P
V
E
U
A
A
J
D
V
K
C
P
Y
H
U
E
D
I
C
V
D
N
E
H
X
N
I
I
T
X
S
G
M
S
L
X
S
E
G
M
E
N
T
E
O
M
I
A
V
I
X
K
F
L
P
L
S
D
K
V
T
P
R
C
O
F
J
R
N
E
B
L
E
V
D
V
J
S
S
R
R
E
N
L
W
F
Q
H
G
N
I
N
O
S
A
E
R
E
V
I
T
C
U
D
N
I
R
O
T
C
E
S
I
B
L
P
J
A
E
N
R
U
K
R
S
T
R
A
I
G
H
T
A
N
G
L
E
W
U
Q
A
P
E
L
G
N
A
E
T
U
C
A
W
K
M
F
B
O
Z
N
T
U
B
A
R
N
K
I
W
T
K
Q
R
U
I
V
C
F
Chapter 2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 2-1
1. Sample: It is 12:00 noon on a rainy day. 2. Sample: The
car will not start because of a dead battery. 3. Sample: 6
4. If you are strong, then you drink milk. 5. If a rectangle is a
square, then it has four sides the same length. 6. If you are
tired, then you did not sleep. 7. If x = 26, then x - 4 = 22;
true. 8. If x 0, then ∆x∆ 0; true. 9. If m is positive,
then m2 is positive; true. 10. If 2y - 1 = 5, then y = 3; true.
11. If x 0, then point A is in the first quadrant; false;
(4, -5) is in the fourth quadrant. 12. If lines are parallel, then
their slopes are equal; true. 13. If you have a sibling, then you
are a twin; false; a brother or sister born on a different day.
14.
siblings
twin
15. Hypothesis: If you like to shop; conclusion: Visit Pigeon
Forge outlets in Tennessee. 16. Pigeon Forge outlets are
good for shopping. 17. If you visit Pigeon Forge outlets, then
you like to shop. 18. It is not necessarily true. People may go
to Pigeon Forge outlets because the people they are with want
to go there. 19. Drinking Sustain makes you train harder and
run faster. 20. If you drink Sustain, then you will train harder
and run faster. 21. If you train harder and run faster, then
you drink Sustain.
Guided Problem Solving 2-1
1. Hypothesis: x is an integer divisible by 3. 2. Conclusion:
x2 is an integer divisible by 3. 3. Yes, it is true. Since 3 is a factor
All-In-One Answers Version A
of x, it must be a factor of x ? x = x2. 4. If x2 is an integer
divisible by 3 then x is an integer divisible by 3. 5. The converse
is false. Counterexamples may vary. Let x2 = 3. Then x 5 Í 3 ,
which is not an integer and is not divisible by 3. 6. No. The
conditional is true, so there is no such counterexample. 7. No.
By definition, a general statement is false if a counterexample
can be provided. 8. If 5x + 3 = 23, then x = 4. The original
statement and the converse are both true.
Practice 2-2
1. Two angles have the same measure if and only if they are
congruent. 2. 2x - 5 = 11 if and only if x = 8.
3. The converse, “If ∆n∆ = 17, then n = 17,” is not true.
4. A figure has eight sides if and only if it is an octagon.
5. If a whole number is a multiple of 5, then its last digit is
either 0 or 5. If a whole number has a last digit of 0 or 5, then
it is a multiple of 5. 6. If two lines are perpendicular, then
the lines form four right angles. If two lines form four right
angles, then the lines are perpendicular. 7. If you live
in Texas, then you live in the largest state in the contiguous
United States. If you live in the largest state in the contiguous
United States, then you live in Texas. 8. Sample: Other
vehicles, such as trucks, fit this description. 9. Sample: Other
objects, such as spheres, are round. 10. Sample: This is not
specific enough; many numbers in a set could fit this
description. 11. Sample: Baseball also fits this definition.
12. Sample: Pleasing, smooth, and rigid all are too vague.
13. yes 14. no 15. yes
Guided Problem Solving 2-2
1. A good definition is clearly understood, precise, and
reversible. 2. &3 and &4, &5 and &6 3. No
Geometry
67
Geometry: All-In-One Answers Version A
5.
y
B(0,3)
pairs: &1 and &3, &1 and &4, &2 and &3, &2 and &4
A(1,3)
Practice 2-3
Guided Problem Solving 2-3
1. conditional; hypothesis 2. Yes 3. Beth will go.
4. Anita, Beth, Aisha, Ramon 5. No; only two students went.
6. Beth, Aisha, Ramon; no—only two went. 7. Aisha, Ramon
8. The answer is reasonable. It is not possible for another pair
to go to the concert. 9. Ramon
Practice 2-4
1. UT = MN 2. m&QWR = 30 3. SB = MN
4. y = 51 5. JL 6. Given; Addition Property of Equality;
Division Property of Equality 7. Given; Addition; Subtraction
Property of Equality; Multiplication Property of Equality;
Division Property of Equality 8. Substitution 9. Substitution
10. Transitive Property of Equality 11. Symmetric Property
of Congruence 12. Transitive Property of Congruence
13. Definition of Complementary Angles; 90, Substitution; 3x,
Simplify; 3x, 84, Subtraction Property of Equality; 28, Division
Property of Equality 14. Given; (2x - 4), Substitution;
6x - 12, Distributive Property; -x, Subtraction Property of
Equality; 12, Multiplication Property of Equality
C(3,1)
6. They are adjacent complementary angles. 7. Answers may
vary. C can be any point on the line y 5 21 x, x . 0.
3
Sample: C(3, -1). 8. a right angle 9. Yes; their sum
corresponds to the right angle formed by the positive x-axis
and the positive y-axis. 10. Answers may vary. D can be any
point on the negative x-axis, sample: D(-4, 0)
2A: Graphic Organizer
1. Reasoning and Proof 2. Answers may vary. Sample:
conditional statements; writing biconditionals; converses; and
using the Law of Detachment 3. Check students’ work.
2B: Reading Comprehension
1. 42 degrees 2. 38 degrees 3. b
2C: Reading/Writing Math Symbols
1. Segment MN is congruent to segment PQ. 2. If p, then q.
3. The length of MN is equal to the length of PQ.
4. Angle XQV is congruent to angle RDC. 5. If q, then p.
6. The measure of angle XQV is equal to the measure of
angle RDC. 7. p if and only if q. 8. a S b 9. AB = MN
10. m/XYZ = m/RPS 11. b S a 12. AB > MN
13. a 4 b 14. /XYZ /RPS
Guided Problem Solving 2-4
2D: Visual Vocabulary Practice
1. Angle Addition Postulate 2. Substitution Property of
1. Law of Detachment 2. hypothesis 3. Distributive
Property 4. Reflexive Property 5. Law of Syllogism
6. biconditional 7. conclusion 8. good definition
9. Symmetric Property
Equality; Simplify; Addition Property of Equality; Division
Property of Equality 3. 40 4. yes; yes 5. 13; 13
Practice 2-5
1. 30 2. 15 3. 20 4. 6 5. 16 6. 9 7. m&A = 135;
m&B = 45 8. m&A = 36; m&B = 144 9. m&A = 75,
m&B = 15 10. m&A = 10; m&B = 80
11. m&PMO = 55; m&PMQ = 125; m&QMN = 55
12. m&BOD = m&COE = 90; m&BOC =
2E: Vocabulary Check
m&COD = 45; m&AOB = m&DOE = 45
13. m&BWC = m&CWD; m&AWB + m&BWC = 180;
m&CWD + m&DWA = 180; m&AWB = m&AWD
and its converse; it contains the words “if and only if.”
Conclusion: The part of an if-then statement that follows
then.
Conditional: An if-then statement.
Truth value: “True” or “false” according to whether the
statement is true or false, respectively
Hypothesis: The part that follows if in an if-then statement.
Biconditional: The combination of a conditional statement
Guided Problem Solving 2-5
1. 90 2. See graph in Exercise 5 answer. 3. on the positive
y-axis 4. Answers may vary. B can be any point on the
positive y-axis. Sample: B(0, 3).
68
Geometry
All rights reserved.
1. &A and &B are complementary. 2. Football practice is
canceled for Monday. 3. DEF is a right triangle. 4. If
you liked the movie, then you enjoyed yourself. 5. If two
lines are not parallel, then they intersect at a point. 6. If you
vacation at the beach, then you like Florida. 7. not possible
8. Tamika lives in Nebraska. 9. not possible 10. It is not
freezing outside. 11. Shannon lives in the smallest state in
the United States. 12. On Thursday, the track team warms
up by jogging 2 miles.
X(4,0)
x
O(0,0)
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4. They are not supplementary. 5. A linear pair has a common
vertex, shares a common side, and is supplementary. 6. yes
7. yes; yes 8. linear pairs: &1 and &2, &3 and &4; not linear
(continued)
Geometry: All-In-One Answers Version A (continued)
2F: Vocabulary Review Puzzle
1
3
P
O
S
T
U
L
A
T
2
V
4
E
P
R
5
7
C
O
G
O
N
S
All rights reserved.
C
T
D
I
J
C
C
L
P
A
R
C
O
L
I
N
A
A
J
T
N
E
E
U
T
G
C
E
A
L
T
M
N
E
U
I
G
S
R
C
T
I
O
11
N
T
D
12
H
Y
P
L
O
T
H
O
E
A
L
L
E
L
N
E
E
U
T
I
N
R
N
D
8
A
O
I
O
6
T
A
9
N
10
A
C
O
R
E
S
I
S
S
I
N
T
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Chapter 3
Practice 3-1
1. corresponding angles 2. alternate interior angles
3. same-side interior angles 4. alternate interior angles
5. same-side interior angles 6. corresponding angles 7. 1
and 5, 2 and 6, 3 and 8, 4 and 7 8. 4 and 6,
3 and 5 9. 4 and 5, 3 and 6 10. m1 = 100,
alternate interior angles; m2 = 100, corresponding angles or
vertical angles 11. m1 = 75, alternate interior angles; m2
= 75, vertical angles or corresponding angles 12. m1 = 135,
corresponding angles; m2 = 135, vertical angles 13. x =
103; 77°, 103° 14. x = 24; 12°, 168° 15. x = 30; 85°, 85°
16a. Alternate Interior Angles Theorem 16b. Vertical angles
are congruent. 16c. Transitive Property of Congruence
Guided Problem Solving 3-1
1. The top and bottom sides are parallel, and the left and right
sides are parallel. 2. The two diagonals are transversals, and
also each side of the parallelogram is a transversal for the two
sides adjacent to it. 3. Corresponding angles, interior and
exterior angles are formed. 4. v, w and x; By the Alternate
Interior Angles Theorem, v = 42, w = 25 and x = 76.
5. Answers may vary. Possible answer: By the Same-Side
Interior Angles Theorem, (w + 42) + (y + 76) = 180. Since
w = 25, y = 37. (The two y’s are equal by Theorem 3-1.)
6. w = 25, y = 37, v = 42, x = 76; yes 7. v = 42, w = 35,
x = 57, y = 46
All-In-One Answers Version A
Practice 3-2
* )
* )
1a. same-side interior 1b. QR 1c. TS 1d. same-side
* )
interior 1e. Same-Side Interior Angles 1f. TS 1g. 3-5
2. l and m, Converse of Same-Side Interior Angles Theorem
3. none 4. BC and AD , Converse of Same-Side Interior
Angles Theorem 5. RT and HU , Converse of Corresponding
Angles Postulate 6. BH and CI, Converse of Corresponding
Angles Postulate 7. a and b, Converse of Same-Side Interior
Angles Theorem 8. 43 9. 90 10. 38 11. 100 12. 70
13. 48
Guided Problem Solving 3-2
1. and m 2. transversals 3. x 4. the angles measuring
19xº and 17xº 5. 180º 6. 17xº 7. 180 - 19x = 17x or
19x + 17x = 180 8. x = 5 9. With x = 5, 19x = 95 and
17x = 85. 10. x = 6
Practice 3-3
1. True. Every avenue will be parallel to Founders Avenue, and
therefore every avenue will be perpendicular to Center City
Boulevard, and therefore every avenue will be perpendicular
to any street that is parallel to Center City Boulevard.
2. Not necessarily true. No information has been given about
the spacing of the streets. 3. True. The fact that one
intersection is perpendicular, plus the fact that every street
belongs to one of two groups of parallel streets, is enough to
guarantee that all intersections are perpendicular. 4. True.
Opposite sides of each block must be of the same type (avenue
Geometry
69
Geometry: All-In-One Answers Version A
l1
1. A theater stage, consisting of a large platform surrounding
a smaller platform. The shapes in the bottom part of the figure
may represent a ramp for actors to enter and exit. 2. The
measures of angles 1 and 2 3. 8; octagon 4. (8 - 2)180 =
1080 degrees 5. 135 6. 45 7. Yes, angle 1 is an obtuse angle
and angle 2 is an acute angle. 8. trapezoids 9. 360
Practice 3-6
1. y = 31 x - 7 2. y = -2x + 12 3. y = 7x - 18
1
4. y = -2 x - 3 5. y = 16 x - 23 6. y = 45 x - 2
7. y = 4x - 13 8. y = -x + 6
9.
2 4 6 x
2 4 6 x
–6–4–2
–2
–4
–6 y 1 x 5
4
–6
Guided Problem Solving 3-3
1. supplementary angles 2. right angle 3. Any one of the
following: Postulate 3-1, or Theorem 3-1, 3-2, 3-3 or 3-4 4. 90
5. It is congruent; Postulate 3-1 6. 90 7. a ⊥ c 8. It is true
for any line parallel to b. 9. Yes. The point is that a
11.
Practice 3-4
13.
1. three 2. 180 3. right triangle 4. z = 90; Because it is
given in the figure that BD ⊥ AC. 5. Theorem 3-12, the
Triangle Angle-Sum Theorem 6. x = 38 7. y = 36
8. #ABD is a 36-54-90 right triangle. #BCD is a 38-52-90
right triangle. 9. 74 10. #ABC is a 52-54-74 acute triangle.
11. Yes, all three are acute angles, with &ABC visibly larger
than &A and &C. 12. &BCD
Practice 3-5
1. x = 120; y = 60 2. n = 5137 3. a = 108; b = 72
4. 109 5. 133 6. 129 7. 129 8. 47 9. 127 10. 30
11. 150 12. 6 13. 5 14. 8 15. BEDC 16. FAE
17. FAE and BAE 18. ABCDE 19. 20
70
Geometry
x
4
2
–6–4 –2
–2
–4
–6
17.
y
6 y 2x 3
4
2
–6–4–2
–2
y
16.
y 3x 5
2 4 6 x
y
2
6 y x2
3
4
4
x
2 4 6
–6
2 4 6 x
–2
–2
–4
–6
18.
y
–4
–6
14.
2 4 6 x
6
4 y 5x 4
2
–6–4–2
2 4 6 x
–6–4–2
–4
–6
y
15.
6
4
2
2 4 6
6 y 2x 7
4
2
–6–4 –2
–2
–4
–6
y yx1
12.
y
6
4
2
–6–4–2
–2
–4
–6
1. 125 2. 69 3. 143 4. 129 5. 140 6. 136 7. x = 35;
y = 145; z = 25 8. a = 55; b = 97; c = 83 9. v = 118;
w = 37; t = 62 10. 50 11. 88 12. m3 = 22; m4 =
22; m5 = 88 13. 57.1 14. 136 15. m1 = 33; m2
= 52 16. isosceles 17. obtuse scalene 18. right scalene
19. obtuse isosceles 20. equiangular equilateral
Guided Problem Solving 3-4
y 1 x 2
3
transversal cannot be perpendicular to just one of two parallel
lines. It has to be perpendicular to both, or else to neither.
y
6
4
2
y 4x 1
–6–4–2
–2
l3
10.
y
6
4
2
All rights reserved.
l2
Guided Problem Solving 3-5
2 4 6 x
y
6
4
2
2
6x
–6–4–2
–2
1
–4 y 2 x 3
–6
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
or boulevard), and adjacent sides must be of opposite type.
5. Not necessarily true. If there are more than three avenues
and more than three boulevards, there will be some blocks
bordered by neither Center City Boulevard nor Founders
Avenue. 6. a ⊥ e 7. a || e 8. a || e 9. a ⊥ e 10. a ⊥ e
11. a || e 12. If the number of ⊥ statements is even, then
1 || n. If it is odd, then 1 ⊥ n. 13. The proof can instead
use alternate interior angles or alternate exterior angles (if the
angles are congruent, the lines are parallel) or same-side
interior or same-side exterior angles (if the angles are
supplementary, the lines are parallel).
14. It is possible.
(continued)
Geometry: All-In-One Answers Version A (continued)
19.
20.
y
6
4
2
x 2
–6–4
y 2x
2 4 6 x
–2
–4
–6
43.
y
2 4 6 x
–6–4–2
–2
–4
–6
22.
y
All rights reserved.
–6–4–2
–2
–4
–6
6
4
2
2 4 6 x
y
6
4
2
–
y 3x 2 2
–6–4–2
–2
–4
–6
25. y = –3x + 13
28. y = -21 x - 12
31. y = -15 x - 65
y = -11 34. x =
36. x = -1; y = 8
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
37.
–4
–6
24.
2 4 6 x
38.
45a. m = $0.10 45b. the amount of money the worker is
paid for each box loaded onto the truck 45c. b = $3.90
45d. the base amount the worker is paid per hour
46. y = -25 x + 8
–6 –4 –2 2 4 6 x
–2
–4
–6 (0, –1)
y
6
4
1
(– , 0) 2 (0, 2)
2
2 4 6 x
–6–4–2
(6, 0)
–4
–6
42.
6
4
2 (4, 0)
2 4 6 x
–6–4–2
–2
–4 (0, –4)
–6
All-In-One Answers Version A
y
6
4
2 (0, 1) (8, 0)
–6–4–2
–2
–4
–6
Guided Problem Solving 3-6
1.
y
B
2
O A
2
x
C
2
y 2y
4
2
2 4 6 x
y
–2
–2 2 4 6 x
–4
–6
2. m 5 x22 2 x11 3. y - y1 = m(x-x1) 4. Slope of
(–2, 0) 6
6 (0, 6)
4
2
41.
2 4 6 8
–4
–6
2
y
40.
y
(9, 0)
x
2 4 6 x
–6–4–2
–2
–4
–6
–6–4–2 2
6 x
–2
–4
–6
–8
–10
–12 (0, 12)
–6–4–2
–2
–4
–6
x 2.5
26. y = x + 4 27. y = 21 x - 3
29. y = 2x + 4 30. y = 13 x + 4
32. y = -6x + 45 33. x = 2;
0; y = 2 35. x = -4; y = -4
6
4 (4, 0)
2
(0, 6)
y
6
4
2
y
39.
2 4 6 x
–6–4–2
y 5
23.
yx
–6 –4–2
–2
6
4
(–3, 0) 2
y
6
4
2
y
6
4
2
12 (0, 12)
10
–6
21.
44.
y
6
4
2 4 6 8 x
å
å
AB = 5 ; slope of BC = - 5 . The absolute values of the
2
2
slopes are the same, but one slope is positive and the other
is negative. 5. Point-slope form: y - 0 = 5 (x - 0);
2
5
slope-intercept form: y = x 6. Point-slope form:
2
y - 5 = - 5 (x - 2) or y - 0 = - 5 (x - 4);
2
2
å
5
slope-intercept form: y = - x + 10 7. Of line AB : (0, 0);
2
å
of line BC : (0, 10) 8. #ABC appears to be an isosceles
triangle, which is consistent with a horizontal base and two
remiaining sides having slopes of equal magnitude and opposite
sign. 9. Slope = 0; y = 0; y-intercept = (0, 0) just as for line
å
AB (they intersect on the y-axis).
Practice 3-7
1. neither; 3 2 13, 3 ? 13 2 -1 2. perpendicular;
2
1
2
2 ? -2 = -1 3. parallel; -3 = -3 4. parallel;
-1 = -1 5. perpendicular; y = 2 is a horizontal line,
x = 0 is a vertical line 6. parallel; -21 = -12
7. neither; 1 2 18, 1 ? 81 2 -1 8. parallel; -32 = -23
9. perpendicular; -1 ? 1 = -1 10. neither; 21 2
7
7
-53, 21 ? –53 2 -1 11. neither; -23 2 -12
, -32 ? -12
2 -1
Geometry
71
Geometry: All-In-One Answers Version A
neither; 6 2 -51, 6 ? -51 2 -1
neither; 29 2 4, 29 ? 4 2 -1
parallel; 21 = 21 15. y = 23 x
y = -43 x + 24
y
y = -x - 3
6
4
y = 35 x + 6
2
y=0
4 6 x
–6–4 L
y = 2x - 4
–4
y = 2x
10. Sample:
b
a
11. Sample:
Guided Problem Solving 3-7
1.
b
y
c
K
H
4
x
G
4
All rights reserved.
12.
13.
14.
16.
17.
18.
19.
20.
21.
(continued)
O
12.
4
4
b
2. a right angle 3. m1 m2 = -1 4. sides GH and GK
b
5. Slope of GH = 35 ; slope of GK = - 38
b
6. Product = - 85 -1. Sides GH and GK are not
perpendicular. 7. #GHK has no pair of perpendicular sides.
It is not a right triangle. 8. No 9. &HGK; approximately 80
10. Slope of LM = 72 and slope of LN = - 27 . The product
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
of the slopes is -1, so LM and LN are perpendicular.
b
13.
Practice 3-8
1.–3.
4.–6.
a
Q
T
c
14. Sample:
7.–9.
K
b
72
Geometry
c
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
8. Line AB is perpendicular to line DF. 9. Angle ABC and
angle ABD are complementary. 10. Angle 2 is a right angle,
or the measure of angle 2 is 90°. 11. Sample answer:
å å
CB GD, m/BAF = m/GFA
15.
a
3D: Visual Vocabulary Practice/High-Use
Academic Words
1. property 2. conclusion 3. describe 4. formula
5. measure 6. approximate 7. compare 8. contradiction
9. pattern
a
Guided Problem Solving 3-8
All rights reserved.
1. a line segment of length c 2. Construct a quadrilateral
with one pair of parallel sides of length c, and then examine
the other pair. 3. The procedure is given on p. 181 of the text.
3E: Vocabulary Check
Transversal: A line that intersects two coplanar lines in two
points.
Alternate interior angles: Nonadjacent interior angles that
lie on opposite sides of the transversal.
Same-side interior angles: Interior angles that lie on the
m
l
4. Adjust the compass to exactly span line segment c, end to
end. Then tighten down the compass adjustment as necessary.
same side of a transversal between two lines.
Corresponding angles: Angles that lie on the same side of a
transversal between two lines, in corresponding positions.
Flow proof: A convincing argument that uses deductive
reasoning, in which arrows show the logical connections
between the statements.
5.
m
3F: Vocabulary Review
l
1. C 2. E 3. D 4. B 5. A 6. F 7. K 8. H 9. L 10. G
11. I 12. J
Chapter 4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
6.
m
Practice 4-1
l
7. They appear to be both congruent and parallel. 8. The
answers to Step 7 are confirmed. 9. yes; a parallelogram
3A: Graphic Organizer
1. Parallel and Perpendicular Lines 2. Answers may vary.
Sample: properties of parallel lines; finding the measures of
angles in triangles; classifying polygons; and graphing lines
3. Check students’ work.
3B: Reading Comprehension
1. 14 spaces 2. 6 spaces 3. 60° 4. corresponding
5. $7000; $480 6. the width of the stalls, 10 ft 7. b
3C: Reading/Writing Math Symbols
å å
1. m # n 2. m/1 + m/2 = 180 3. AB CD
4. m/MNP + m/MNQ = 90 5. /3 /EFD
6. Line 1 is parallel to line 2. 7. The measure of angle ABC
is equal to the measure of angle XYZ.
All-In-One Answers Version A
1. m1 = 110; m2 = 120 2. m3 = 90; m4 = 135
3. m5 = 140; m6 = 90; m7 = 40; m8 = 90
4. CA JS , AT SD , CT JD 5. C J,
A S, T D 6. WZ JM , WX JK ,
XY KL , ZY ML 7. W J, X K,
Y L, Z M 8. Yes; GHJ IHJ by
Theorem 4-1 and by the Reflexive Property of . Therefore,
GHJ IHJ by the definition of triangles.
9. No; QSR TSV because vertical angles are
congruent, and QRS TVS by Theorem 4-1, but none
of the sides are necessarily congruent. 10a. Given
10b. Vertical angles are . 10c. Theorem 4-1 10d. Given
10e. Definition of triangles
Guided Problem Solving 4-1
1. right triangles 2. m&A = 45, m&B = m&L = 90,
and AB = 4 in. 3. #ABC is congruent to #KLM means
corresponding sides and angles are congruent. 4. x and t
5. 45 6. m&K = m&M = 45 7. 3x = 45 8. x = 15 9. 4
10. 2t = 4 11. t = 2 12. The angle measures indicate that
the two triangles are isosceles right triangles. This matches the
appearance of the figure. 13. m&M = 60
Geometry
73
Geometry: All-In-One Answers Version A
(continued)
Practice 4-2
Practice 4-4
1. ADB CDB by SAS 2. not possible 3. not
possible 4. TUS XWV by SSS 5. not possible
6. DEC GHF by SAS 7. MKL KMJ by
SAS 8. PRN PRQ by SSS 9. not possible
10. C 11. AB and BC 12. A and B
13. AC 14a. Given 14b. Reflexive Property of
Congruence 14c. SAS Postulate
15. Statements
Reasons
1. EF FG , DF FH
1. Given
1. BD is a common side, so ADB CDB by SAS, and
A C by CPCTC. 2. FH is a common side, so FHE
Guided Problem Solving 4-2
1. IS > OS and SP bisects &ISO. 2. Prove whatever
additional facts can be proven about #ISP and #OSP, based
on the given information. 3. IS > OS 4. &ISP &PSO
5. SP 6. #ISP #OSP by Postulate 4-2, the Side-AngleSide (SAS) Postulate 7. It does not matter. The Side-AngleSide Postulate applies whether or not they are collinear.
8. It does follow, because #ISP #OSP and because IP and
PQ are corresponding parts.
Practice 4-3
1. not possible 2. ASA Postulate 3. AAS Theorem
4. AAS Theorem 5. not possible 6. not possible 7. ASA
Postulate 8. not possible 9. AAS Theorem
10. Statements
Reasons
1. K M, KL ML
2. JLK PLM
3. JKL PML
11.
1. Given
2. Vertical s are .
3. ASA Postulate
Q S
Given
TRS RTQ
QRT STR
Given
AAS Theorem
RT TR
Reflexive Property of 12. BC EF 13. KHJ HKG or
KJH HGK 14. NLM NPQ
Guided Problem Solving 4-3
1. corresponding angles and alternate interior angles
2. &EAB and &DBC 3. &EBA and &DCB. 4. &EAB
and &DBC 5. &EAB > &DBC, AE > BD, and
&E > &D 6. #AEB > #BDC by Postulate 4-3, the AngleSide-Angle (ASA) Postulate 7. Yes; Theorem 4-2, the
Angle-Angle-Side (AAS) Theorem; &EBA > &DCB
8. No, because now there is no way to demonstrate a second
Guided Problem Solving 4-4
1. A compass with a fixed setting was used to draw two
circular arcs, both centered at point P but crossing in
different locations, which were labeled A and B. The compass
was used again, with a wider setting, to draw two intersecting
circular arcs, one centered at A and one at B. The point at
å
which the new arcs intersected was labeled C. Finally, line CP
was drawn. 2. Find equal lengths or distances and explain
å
why CP is perpendicular to . 3. #ACP and #BCP
4. AP = PB and AC = BC. 5. #APC > #BPC, by
Postulate 4-1, the Side-Side-Side (SSS) Postulate
6. &APC > &BPC by CPCTC 7. Since &APC > &BPC,
m&APC = m&BPC and m&APC + m&BPC = 180, it
follows that m&APC = m&BPC = 90. 8. from the
definition of perpendicular and the fact that m&APC =
m&BPC = 90 9. The distances do not matter, so long as
AP = BP and AC = BC. That is what is required in order that
#APC > #BPC. 10. Draw a line, and use the construction
technique of the problem to construct a second line
perpendicular to the first. Then do the same thing again to
construct a third line perpendicular to the second line. The
first and third lines will be parallel, by Theorem 3-10.
Practice 4-5
1. x = 35; y = 35 2. x = 80; y = 90 3. t = 150
4. r = 45; s = 45 5. x = 55; y = 70; z = 125
6. a = 132; b = 36; c = 60 7. x = 6 8. a = 30; b = 30;
c = 75 9. z = 120 10. AD ; D F 11. GA ;
ACG AGC 12. KJ ; KIJ KJI 13. DC ;
CDE CED 14. BA ; ABJ AJB 15. CB ;
BCH BHC 16. 130 17. 65 18. 130 19. 90
20. x = 70; y = 55 21. x = 70; y = 20
22. x = 45; y = 45
pair of congruent sides, nor a second pair of congruent angles.
74
Geometry
All-In-One Answers Version A
All rights reserved.
2. Vertical s are .
3. SAS Postulate
UVX WVX by SSS, and U W by CPCTC.
6. ZAY and CAB are vertical angles, so ABC AYZ by ASA. ZA AC by CPCTC. 7. EG is a common side, so DEG FEG by SAS, and FG DG by
CPCTC. 8. JKH and LKM are vertical angles, so
HJK MLK by AAS, and JK KL by CPCTC.
9. PR is a common side, so PNR RQP by SSS, and
N Q by CPCTC. 10. First, show that ACB and
ECD are vertical angles. Then, show ABC EDC by
ASA. Last, show A E by CPCTC. 11. First, show
FH as a common side. Then, show JFH GHF by
SAS. Last, show FG JH by CPCTC.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2. DFE HFG
3. DFE HFG
HFG by ASA, and HE FG by CPCTC.
3. KLJ PMN by ASA, so K P by CPCTC.
4. QS is a common side, so QTS SRQ by AAS.
QST SQR by CPCTC. 5. VX is a common side, so
Geometry: All-In-One Answers Version A (continued)
Guided Problem Solving 4-5
GI JI
4.
1. One angle is obtuse. The other two angles are acute and
congruent. 2. Highlight an obtuse isosceles angle and find its
Given
angle measures, then find all the other angle measures
represented in the figure.
3. Possible answer:
HI HI
IHG IHJ
Reflexive Property of HL Theorem
GHI JHI
GHI and JHI
are right s.
Given
Theorem 2-5
mGHI + mJHI = 180
Angle Addition Postulate
All rights reserved.
4. 60 5. 30, because the measure of each base angle is half
the measure of an angle of the equilateral triangle. 6. 120°
because the sum of the angles of the highlighted triangle must
equal 180°. 7. The other measures are 90° and 150°.
Examples:
5. RS VW 6. none 7. mC and mF = 90 8.
GH JH 9. LN PR 10. ST UV or SV UT 11. mA and mX = 90 12. mF and mD =
90 13. GI # JH
Guided Problem Solving 4-6
1. Two congruent right triangles. Each one has a leg and a
hypotenuse labeled with a variable expression. 2. the values
of x and y for which the triangles are congruent by HL
3. The two shorter legs are congruent. 4. x = y + 1 5. The
hypotenuses are congruent. 6. x + 3 = 3y 7. x = 3; y = 2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
8.
hypotenuse
shorter leg =
2; yes, this matches the figure. 9. The
8. Yes; 3 120 = 360. 9. Answers may vary.
solution remains the same: x = 3 and y = 2. The reason is
that one is still solving the same two equations, x = y + 1 and
x + 3 = 3y.
Practice 4-6
Practice 4-7
1. Statements
1. AB # BC , ED # FE
2. B, E are right s.
3. AC FD , AB ED
4. ABC DEF
2. Statements
1. P, R are right s.
2. PS QR
3. SQ QS
4. PQS RSQ
3.
Reasons
1. Given
2. Perpendicular lines
form right s.
3. Given
4. HL Theorem
Reasons
1. Given
2. Given
3. Reflexive
Property of 4. HL Theorem
MJN and MJK
are right s.
1. ZWX YXW; SAS 2. ABC DCB; ASA
3. EJG FKH; ASA 4. LNP LMO; SAS
5. ADF BGE; SAS 6. UVY VUX; ASA
7.
B
A
C
B
C
common side: BC
F
common side: FG
D
8. F
G
E
G
H
9.
common angle: L
L
Perpendicular lines
form right s.
L
N
K
MN MK
MJN MJK
Given
HL Theorem
M
J
MJ MJ
Reflexive Property of All-In-One Answers Version A
Geometry
75
Geometry: All-In-One Answers Version A
Reasons
1. Given
2. Given
3. Perpendicular lines
form right s.
4. Reflexive Property
of 5. BYA CXA
5. ASA Postulate
11. Sample: Because FH GE, HFG EGF,
and FG GF, then FGE GFH by SAS. Thus, FE
GH by CPCTC and EH HE, then GEH FHE
by SSS.
Guided Problem Solving 4-7
4E: Vocabulary Check
Angle: Formed by two rays with the same endpoint.
Congruent angles: Angles that have the same measure.
Congruent segments: Segments that have the same length.
Corresponding polygons: Polygons that have
corresponding sides congruent and corresponding angles
congruent.
CPCTC: An abbreviation for “corresponding parts of
congruent triangles are congruent.”
4F: Vocabulary Review Puzzle
1. postulate 2. hypotenuse 3. angle 4. vertex 5. side
6. leg 7. perpendicular 8. polygon 9. supplementary
10. parallel 11. corresponding
1. The figure, a list of parallel and perpendicular pairs of sides,
and one known angle measure, namely m&A = 56
2. nine 3. They are congruent and have equal measures.
4. m&A = m&1 = m&2 = 56 5. m&4 = 90
6. m&3 = 34 7. m&DCE = 56 8. m&5 = 22
9. m&FCG = 90 10. m&6 = 34 11. m&7 = 34,
m&8 = 68, and m&9 = 112 12. m&9 = 56 + 56 = 112
13. m&FIC = 180 -(m&2 + m&3) = 90;
m&DHC = m&4 = 90; m&FJC = 180 - m&9 = 68;
m&BIG = m&FIC = 90
Chapter 5
4A: Graphic Organizer
Guided Problem Solving 5-1
1. Congruent Triangles 2. Answers may vary. Sample:
congruent figures; triangle congruence by SSS, SAS, ASA,
and AAS; proving parts of triangles congruent; the Isosceles
Triangle Theorem 3. Check students’ work.
4B: Reading Comprehension
1. Yes. Using the Isosceles Triangle Theorem, /W /Y.
It is given that WX YX and WU YV. Therefore
nWUX nYVX by SAS. 2. There is not enough
information. You need to know if AC EC, if /A /E, or
if /B /D. 3. a
4C: Reading/Writing Math Symbols
Practice 5-1
1a. 8 cm 1b. 16 cm 1c. 14 cm 2a. 22.5 in. 2b. 15.5 in.
2c. 15.5 in. 3a. 9.5 cm 3b. 17.5 cm 3c. 14.5 cm
4. 17 5. 20.5 6. 7 7. 128
19 8. 42 9. 16.5 10a. 18
10b. 61 11. GH 6 AC, HI 6 BA, GI 6 BC
12. PR 6 YZ, PQ 6 XZ, XY 6 RQ
1. 30 units 2. The three sides of the large triangle are each
bisected by intersections with the two line segments lying in
the interior of the large triangle. 3. the value of x 4. They
are called midsegments. 5. They are parallel, and the side
labeled 30 is half the length of the side labeled x. 6. x = 60
7. Yes; the side labeled x appears to be about twice as long as
the side labeled 30. 8. No. Those lengths are not fixed by the
given information. (The triangle could be vertically stretched
or shrunk without changing the lengths of the labeled sides.)
All one can say is that the midsegment is half as long as the
side it is parallel to.
Practice 5-2
1. Angle-Angle-Side 2. triangle XYZ 3. angle PQR
4. line segment BD 5. line ST 6. ray WX 7. hypotenuseleg 8. line 3 9. angle 6 10. Angle-Side-Angle
1. WY is the perpendicular bisector of XZ. 2. 4 3. 7.5
4. 9 5. right triangle 6. 5 7. 17 8. 17 9. equidistant
10. isosceles triangle
) 11. 3.5 12. 21 13. 21 14. right
triangle 15. JP is the bisector of LJN. 16. 9) 17. 45
18. 45 19. 14 20. Sample: Point M lies on JP . 21. right
4D: Visual Vocabulary Practice
isosceles triangle
1. theorem 2. congruent polygons 3. base angle of an
isosceles triangle 4. CPCTC 5. postulate 6. vertex angle of
an isosceles triangle 7. corollary 8. base of an isosceles
triangle 9. legs of an isosceles triangle
76
Geometry
All rights reserved.
Statements
1. AX AY
2. CX # AB, BY # AC
3. mCXA and
mBYA = 90
4. A A
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
10. Sample:
(continued)
Geometry: All-In-One Answers Version A (continued)
Guided Problem Solving 5-2
1.
<
y
m
A
C
4
B x
4
O
All rights reserved.
2. See answer to Step 1, above. 3. See answer to Step 1,
above. 4. Plot a point and explain why it lies on the bisector
of the angle at the origin. 5. line : y 5 23 x 1 25 ; line m:
4
2
x = 10 6. C(10, 5) 7. CA = CB = 5; yes 8. Theorem 5-5,
the Converse of the Angle Bisector Theorem 9. m&AOC =
m&BOC < 27 10. Draw , m, and C, then draw OC. Since
OA = OB = 10, it follows that #OAC > #OBC, by HL.
Then CA > CB and &AOC > &BOC by CPCTC.
Practice 5-3
1. (-2, 2) 2. (4, 0) 3. (2, 1) 4. Check students’ work. The
final result of the construction is shown.
B
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
C
5. altitude 6. median 7. none of these 8. perpendicular
bisector 9. angle bisector 10. altitude 11a. (2, 0)
11b. (-2, -2) 12a. (0, 0) 12b. (3, -4) 13a. (0, 0)
13b. (0, 3)
Guided Problem Solving 5-3
1. the figure and a proof with some parts left blank 2. Fill in
the blanks. 3. AB 4. Theorem 5-2, the Perpendicular
Bisector Theorem 5. BC; XC 6. the Transitive Property of
Equality 7. Perpendicular Bisector. (This converse is
Theorem 5-3.) 8. The point of the proof is to demonstrate
that n runs through point X. It would not be appropriate to
show that fact as already given in the figure. 9. Nothing
essential would change. Point X would lie outside #ABC
å
(below BC ), but the proof woud run just the same.
Practice 5-4
1. I and III 2. I and II 3. The angle measure is not 65.
4. Tina does not have her driver’s license. 5. The figure does
not have eight sides. 6. The restaurant is open on Sunday.
7. ABC is congruent to XYZ. 8. mY 50 9a. If
Toronto; true. 11. Assume that mA 2 mB. 12. Assume
that TUVW is not a trapezoid. 13. Assume that LM does not
intersect NO. 14. Assume that FGH is not equilateral.
15. Assume that it is not sunny outside. 16. Assume that D
is obtuse. 17. Assume that mA 90. This means that
mA + mC 180. This, in turn, means that the sum of the
angles of ABC exceeds 180, which contradicts the Triangle
Angle-Sum Theorem. So the assumption that mA 90 must
be incorrect. Therefore, mA 90.
Guided Problem Solving 5-4
1. Ice is forming on the sidewalk in front of Toni’s house.
2. Use indirect reasoning to show that the temperature of the
sidewalk surface must be 32°F or lower. 3. The temperature
of the sidewalk in front of Toni’s house is greater than 32°F.
4. Water is liquid (ice does not form) above 32°F. 5. There is
no ice forming on the sidewalk in front of Toni’s house.
6. The result from Step 5 contradicts the information
identified as given in Step 1. 7. The temperature of the
sidewalk in front of Toni’s house is less than or equal to 32°F.
8. If the temperature is above 32°F, water remains liquid. This
is reliably true. Converse: If water remains liquid, the
temperature is above 32°F. This is not reliably true. Adding salt
will cause water to remain liquid even below 32°F.
9. Suppose two people are each the world’s tallest person. Call
them person A and person B. Then person A would be taller
than everyone else, including B, but by the same token B
would be taller than A. It is a contradiction for two people
each to be taller than the other. So it is impossible for two
people each to be the World’s Tallest Person.
Practice 5-5
1. M, N 2. C, D 3. S, Q
4. R, P 5. A, T 6. S, A
7. yes; 4 + 7 8, 7 + 8 4, 8 + 4 7
8. no; 6 + 10 17 9. yes; 4 + 4 4
10. yes; 1 + 9 9, 9 + 9 1, 9 + 1 9
11. yes; 11 + 12 13, 12 + 13 11, 13 + 11 12
12. no; 18 + 20 40 13. no; 1.2 + 2.6 4.9
14. no; 8 21 + 9 14 18 15. no; 2. 5 + 3.5 6
16. BC, AB, AC 17. BO, BL, LO 18. RS, ST, RT
19. D, S, A 20. N, S, J 21. R, O, P
22. 3 x 11 23. 8 x 26 24. 0 x 10
25. 9 x 31 26. 2 x 14 27. 13 x 61
Guided Problem Solving 5-5
1.
two triangles are not congruent, then their corresponding angles
are not congruent; false. 9b. If corresponding angles are not
congruent, then the triangles are not congruent; true. 10a. If
you do not live in Toronto, then you do not live in Canada; false.
10b. If you do not live in Canada, then you do not live in
All-In-One Answers Version A
Geometry
77
Geometry: All-In-One Answers Version A
2. The side opposite the larger included angle is greater than
the side opposite the smaller included angle. 3. The angle
opposite the larger side is greater than the angle opposite the
smaller side 4. The opposite sides each have a length of
nearly the sum of the other two side lengths. 5. The opposite
sides are the same length. They are corresponding parts of
triangles that are congruent by SAS.
(continued)
isosceles trapezoid, quadrilateral 8. square, rectangle,
parallelogram, rhombus, quadrilateral 9. x = 7;
AB = BD = DC = CA = 11 10. m = 9; s = 42;
ON = LM = 26; OL = MN = 43 11. f = 5; g = 11;
FG = GH = HI = IF = 17 12. parallelogram
13. rectangle 14. kite 15. parallelogram
Guided Problem Solving 6-1
Sample: midsegments of triangles; bisectors in triangles;
concurrent lines, medians, and altitudes; and inverses,
contrapositives, and indirect reasoning 3. Check students’
work.
5B: Reading Comprehension
1. The width of the tar pit is 10 meters. 2. b
1. a labeled figure, which shows an isosceles trapezoid
2. The nonparallel sides are congruent. 3. the measures of the
angles and the lengths of the sides 4. m&G = c
5. c + (4c - 20) = 180 6. 40 7. m&D = m&G = 40,
m&E = m&F = 140 8. a - 4 = 11 9. 15 10. DE = FG =
11, EF = 15, DG = 32 11. 40 + 40 + 140 + 140 = 360 =
(4 - 2)180 12. m&D = m&G = 39, m&E = m&F = 141
Practice 6-2
1. L 2. F 3. O 4. G 5. A 6. I 7. M 8. H 9. E
10. K 11. B 12. D 13. N 14. J 15. C
1. 15 2. 32 3. 7 4. 8 5. 12 6. 9 7. 8 8. 3 58 9. 54
10. 34 11. 54 12. 34 13. 100 14. 40; 140; 40 15. 70;
110; 70 16. 113; 45; 22 17. 115; 15; 50 18. 55; 105; 55
19. 61; 72; 108; 32 20. 32; 98; 50 21. 16 22. 35 23. 28
24. 4
5D: Visual Vocabulary Practice
Guided Problem Solving 6-2
1. median 2. negation 3. circumcenter 4. contrapositive
5. centroid 6. equivalent statements 7. incenter 8. inverse
9. altitude
1. the ratio of two different angle measures in a parallelogram
2. The consecutive angles are supplementary.
3.
9x
5C: Reading/Writing Math Symbols
5E: Vocabulary Check
Midpoint: A point that divides a line segment into two
congruent segments.
Midsegment of a triangle: The segment that joins the
midpoints of two sides of a triangle.
Proof: A convincing argument that uses deductive reasoning.
Coordinate proof: A proof in which a figure is drawn on a
coordinate plane and the formulas for slope, midpoint, and
distance are used to prove properties of the figure.
Distance from a point to a line: The length of the
perpendicular segment from the point to the line.
5F: Vocabulary Review
1. altitude 2. line 3. median 4. negation 5. contrapositive
6. incenter 7. orthocenter 8. slope-intercept
9. exterior 10. obtuse 11. alternate interior 12. centroid
13. equivalent 14. right 15. parallel
Chapter 6
Practice 6-1
1. parallelogram 2. rectangle 3. quadrilateral
4. parallelogram, quadrilateral 5. kite, quadrilateral
6. rectangle, parallelogram, quadrilateral 7. trapezoid,
78
Geometry
All rights reserved.
1. Relationships Within Triangles 2. Answers may vary.
x
4. the measures of the angles 5. The angles are
supplementary angles, because they are consecutive.
6. x + 9x = 180 7. 18 and 162 8. No; the lengths of the
sides are irrelevant in this problem. 9. 30 and 150
Practice 6-3
1. no 2. yes 3. yes 4. no 5. yes 6. yes 7. x = 2;
y = 3 8. x = 6; y = 3 9. x = 64; y = 10 10. x = 8;
the figure is a because both pairs of opposite sides are
congruent. 11. x = 40; the figure is not a because one
pair of opposite angles is not congruent. 12. x = 25; the
figure is a because the congruent opposite sides are 6 by
the Converse of the Alternate Interior Angles Theorem.
13. Yes; the diagonals bisect each other. 14. No; the
congruent opposite sides do not have to be 6. 15. No; the
figure could be a trapezoid. 16. Yes; both pairs of opposite
sides are congruent. 17. Yes; both pairs of opposite sides
are 6 by the converse of the Alternate Interior Angles
Theorem. 18. No; only one pair of opposite angles is
congruent. 19. Yes; one pair of opposite sides is both
congruent and 6. 20. No; only one pair of opposite sides is
congruent.
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
5A: Graphic Organizer
Geometry: All-In-One Answers Version A (continued)
Guided Problem Solving 6-3
Guided Problem Solving 6-5
1. a labeled figure, which shows a quadrilateral that appears to
be a parallelogram 2. The consecutive angles are
supplementary. 3. find values for x and y which make the
quadrilateral a parallelogram 4. m&A + m&D = 180, so
1. Isosceles trapezoid ABCD with AB > DC 2. &B > &C
and &BAD > &D 3. AB > DC is Given. DC > AE
because opposite sides of a parallelogram are congruent
(Theorem 6-1). AB > AE is from the Transitive Property of
Congruency. 4. Isosceles; >; because base angles of an
isosceles triangle are congruent. 5. &1 > &C because
corresponding angles on a transversal of two parallel lines are
congruent. 6. &B > &C by the Transitive Property of
Congruency. 7. &BAD is a same-side interior angle with &B,
and &D is a same-side interior angle with &C. 8. This is not
a problem, because for AD > BC there is a similar proof with
a line segment drawn from B to a point E lying on AD.
9. The two drawn segments can be shown to be congruent,
and then one has two congruent right triangles by the HL
Theorem. &B > &C follows by CPCTC and &BAD > &D
because they are supplements of congruent angles.
that &A and &D meet the requirements for same-side
interior angles on the transversal of two parallel lines
(Theorems 3-2 and 3-6). 5. &B > &D, by Theorem 6-2.
6. 3x + 10 + 5y = 180; 8x + 5 = 5y 7. x = 15, y = 25
8. m&A = m&C = 55 and m&B = m&D = 125, which
matches the appearance of the figure. 9. (3x + 10) +
(8x + 5) = 180; yes
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
Practice 6-4
1a. rhombus 1b. 72; 54; 54; 72 2a. rectangle
2b. 72; 36; 18; 144 3a. rectangle 3b. 37; 53; 106; 74
4a. rhombus 4b. 59; 90; 90; 59 5a. rectangle
5b. 60; 30; 60; 30 6a. rhombus 6b. 22; 68; 68; 90
7. Yes; the parallelogram is a rhombus. 8. Possible; opposite
angles are congruent in a parallelogram. 9. Impossible; if the
diagonals are perpendicular, then the parallelogram should be
a rhombus, but the sides are not of equal length.
10. x = 7; HJ = 7; IK = 7 11. x = 7; HJ = 26;
IK = 26 12. x = 6; HJ = 25; IK = 25 13. x = -3;
HJ = 13; IK = 13 14a. 90; 90; 29; 29 14b. 288 cm2
15a. 70; 90; 70; 70 15b. 88 in.2 16a. 38; 90; 90; 38
16b. 260 m2 17. possible 18. Impossible; because opposite
angles are congruent and supplementary, for the figure to be a
parallelogram they must measure 90, the figure therefore must
be a rectangle.
Guided Problem Solving 6-4
1. A labeled figure, which shows a parallelogram. One angle is a
right angle, and two adjacent sides are congruent. Algebraic
expressions are given for the lengths of three line segments.
2. diagonals 3. Find the values of x and y. 4. It is a square.
Theorem 6-1 and the fact that AB > AD imply that all four
sides are congruent. Theorems 3-11 and 6-2 plus the fact that
m&B = 90 imply that all four angles are right angles.
5. congruent; bisect 6. 4x - y + 1 = (2x - 1) + (3y + 5);
2x - 1 = 3y + 5 7. x = 7 1 ; y = 3 8. It was not necessary to
2
know AB > AD, but it was necessary to know m&B = 90.
The key fact, which enables the use of Theorem 6-11 in addition
to Theorem 6-3, is that ABCD is a rectangle. It does not matter
whether all four sides are congruent. 9. 40
Practice 6-5
1. 118; 62 2. 99; 81 3. 59; 121 4. 96; 84 5. 101; 79
6. 67; 113 7. x = 4 8. x = 16; y = 116 9. x = 1
10. 105.5; 105.5 11. 90; 25 12. 118; 118 13. 90; 63; 63
14. 107; 107 15. 90; 51; 39 16. x = 8 17. x = 7
18. x = 28; y = 32
All-In-One Answers Version A
Practice 6-6
1. (1.5a, 2b); a 2. (1.5a, b); " a2 1 4b2 3. (0.5a, 0); a
4. (0.5a, b); " a2 1 4b2 5. 0 6. 1 7. -12 8. 2 9. 2b
3a
2b
2b
10. -2b
11.
12.
13.
E(
a,
3
b
);
I(4
a,
0)
3a
3a
3a
14. O(3a, 2b); M(3a, -2b); E(-3a, -2b) 15. D(4a, b);
I(3a, 0) 16. T(0, 2b); A(a, 4b); L(2a, 2b) 17. (-4a, b)
18. (-b, 0)
Guided Problem Solving 6-6
1. a rhombus with coordinates given for two vertices
2. A rhombus is a parallelogram with four congruent sides.
3. the coordinates of the other two vertices 4. They are the
diagonals. 5. They bisect each other. 6. W(-2r, 0),
Z(0, -2t) 7. No; neither Theorem 6-3 nor any other theorem
or result would apply. 8. Slope of WX = 0 and slope of YZ
is undefined. This confirms Theorem 6-10, which says that the
diagonals of a rhombus are perpendicular.
Practice 6-7
p
p2
p
q 1b. y = mx + b; q = q (p) + b; b = q - q ;
p
p
p2
y = q x + q - q 1c. x = r + p 1d. y = q (r + p) +
2
2
2
rp
rp
p
p
p
q - q ; y = q + q + q - q ; y = q + q; intersection
rp
at (r + p, q + q) 1e. qr 1f. (r, q) 1g. y = mx + b;
2
2
q = qr (r) + b; b = q - rq ; y = rqx + q - rq
2
2
2
rp
1h. y = qr (r + p) + q - rq ; y = rq + q + q - rq ;
rp
rp
y = q + q; intersection at (r + p, q + q)
rp
1i. (r + p, q + q) 2a. (-2a, 0) 2b. (-a, b)
b
3
b
2c. (-3a
2 , 2 ) 2d. a 3a. (-4a, 0) 3b. (-2a, 3a) 3c. 2
3d. (2a, -a) 3e. 12 4. The coordinates for D are (0, 2b).
1a.
The coordinates for C are (2a, 0). Given these coordinates, the
lengths of DC and HP can be determined:
Geometry
79
Geometry: All-In-One Answers Version A
DC = "(2a 2 0) 2 1 (0 2 2b) 2 = " 4a2 1 4b2;
HP = "(0 2 2a) 2 1 (0 2 2b) 2 = " 4a2 1 4b2;
DC = HP, so DC HP.
(continued)
Guided Problem Solving 6-7
the given information, only one side and one angle of the two
triangles can be proven to be congruent. Another side or angle
is needed. If it were given that QUSV is a parallelogram, then
the proof could be made. 3. All four sides are congruent.
4. Yes. Since EG > EG by the Reflexive Property,
nEFG > nEHG by SSS. 5. b
1. kite DEFG with DE = EF with the midpoint of each side
identified 2. A kite is a quadrilateral with two pairs of
6C: Reading/Writing Math Symbols
adjacent sides congruent and no opposite sides congruent.
3. The midpoints are the vertices of a rectangle.
4. D(-2b, 2c), G(0, 0) 5. L(b, a + c), M(b, c), N(-b, c),
K(-b, a + c) 6. Slope of KL = slope of NM = 0, slopes
of KN and LM are undefined 7. Opposite sides are
parallel; it is a rectangle. 8. Adjacent sides are perpendicular.
9. right angles 10. Answers will vary. Example: a = 3, b = 2,
c = 2 yields the points D(-4, 4), E(0, 6), F(4, 4), G(0, 0) with
midpoints at (-2, 2), (-2, 5), (2, 5), and (2, 2). Connecting
these midpoints forms a rectangle. 11. Construct DF and
EG. Slope of DF = 0, so DF is horizontal. Slope of EG is
undefined, so EG is vertical.
1. AH, CH, or BH 2. DG, FG, or EG 3. G
4. DE 5. rhombus 6. rectangle 7. square 8. isosceles
6D: Visual Vocabulary Practice/High-Use
Adademic Words
1. solve 2. deduce 3. equivalent 4. indirect 5. equal
6. analysis 7. identify 8. convert 9. common
6E: Visual Vocabulary Check
Consecutive angles: Angles of a polygon that share a
common side.
Kite: A quadrilateral with two pairs of congruent adjacent
sides and no opposite sides congruent.
Parallelogram: A quadrilateral with two pairs of parallel
sides.
Rhombus: A parallelogram with four congruent sides.
Trapezoid: A quadrilateral with exactly one pair of parallel
sides.
6A: Graphic Organizer
1. Quadrilaterals 2. Answers may vary. Sample: classifying
quadrilaterals; properties of parallelograms; proving that a
quadrilateral is a parallelogram; and special parallelograms
3. Check students’ work.
6B: Reading Comprehension
1. QT > SR, QR > ST, QT 6 RS, QR 6 TV
2. No, it cannot be proven that nQTV > nSRU because with
1
2
P
A
R
A
L
4
T
L
5
R
E
L
A
P
O
E
G
Z
R
A
O
I
H
7
O
M
8
B
U
S
9
C
3
I
M
N
I
C
D
E
S
N
E
N
T
6
R
O
E
R
A
M
R
C
E
U
E
N
M
A
T
C
G
Geometry
I
G
Q
U
A
R
N
80
C
B
S
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
6F: Vocabulary Review Puzzle
D
E
10
H
Y
P
O
T
E
N
L
T
E
E
S
R
U
All rights reserved.
trapezoid
S
E
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Guided Problem Solving 7-3
Chapter 7
Practice 7-1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
1. 1 : 278 2. 18 ft by 10 ft 3. 18 ft by 14 ft 4. 18 ft by 16 ft
5. 8 ft by 8 ft 6. 3 ft by 10 ft 7. true 8. false 9. true
10. true 11. false 12. true 13. true 14. false 15. false
88
16. 12 17. 12 18. 33 19. 21
5 20. 48 21. 3 22. 6
23. 25 24. -52 25. 14 : 5 26. 3 : 2 27. 12 : 7 28. 83
7
29. 13
1. no; N/A 2. yes; WT, RS 3. It is a trapezoid. 4. They
are congruent. 5. They are parallel. 6. They are congruent.
7. #RSZ and #TWZ 8. AA~ or Angle-Angle Similarity
Postulate 9. No; there is only one pair of congruent angles.
10. yes; parallelogram, rhombus, rectangle, and square
Practice 7-4
Guided Problem Solving 7-1
1. 16 2. 8 3. "77 4. 2 "11 5. 10"2 6. 6"5 7. h
8. y 9. x 10. a 11. b 12. c 13. x = 6; y = 6"3
14. x = 8 "3; y = 4 "3 15. 92 16. x = 4 "5 ; y = "55
7. 0.348 8. yes 9. 21.912
Guided Problem Solving 7-4
Practice 7-2
1. #ABC, #ACD, #BCD 2. 1 3. 1; 1 4. 1 5. 1
6. 2; 1 7. 2 8. "2 9. "2 10. yes 11. no (This would
42
1
or
1. ratios 2. 42,000,000
3. the denominator
1,000,000
x 5
1
4. 29,000
5. Cross-Product Property 6. 0.029
1,000,000
17. x = "3; y = "6; z = "2 18. x = 8; y = 2 "2;
z = 6 "2 19. 2"15 in.
require the Pythagorean Theorem.)
1. ABC XYZ, with similarity ratio 2 : 1
2. QMN RST, with similarity ratio 5 : 3
3. Not similar; corresponding sides are not proportional.
4. Not similar; corresponding angles are not congruent.
5. ABC KMN, with similarity ratio 4 : 7
6. Not similar; corresponding sides are not proportional.
7. I 8. O 9. J 10. NO 11. LO 12. LO
13. 3.96 ft 14. 4.8 in. 15. 3.75 cm 16. 10 m 17. 32
18. 53 19. 7 21 20. 4 12 21. 53 22. 37 23. 5
Practice 7-5
16
1. BE 2. EH 3. BC 4. JD 5. JG
JI 6. BE 7. 3
15
25
8. 4 9. 4 10. 25
2 11. x = 9 ; y = 4 12. 4 13. x = 6;
198
189
y = 6 14. x = 5 ; y = 5 15. 2 16. 4 17. 10
Guided Problem Solving 7-5
Guided Problem Solving 7-2
1. parallel 2. CE; BD 3. 6; 15 4. 90 5. yes 6. yes 7. The
sides would not be parallel. 8. They are similar triangles.
1. equal 2. 6.14
3. 2.61; 6.14 4. 19.3662; 19.2182 5. no
2.61
6. no 7. 2.3706; 2.3525 8. Since the quotients are not equal,
7A: Graphic Organizer
1. Similarity 2. Answers may vary. Sample: ratios and
proportions; similar polygons; proving triangles similar; and
similarity in right triangles 3. Check students’ work.
the ratios are not equal, and the bills are not similar
rectangles. 9. 4.045
Practice 7-3
1. AXB RXQ because vertical angles are . A R (Given). Therefore AXB RXQ by the AA 3
XM
PX
Postulate. 2. Because MP
LW = WA = AL = 4 , MPX LWA by the SSS Theorem. 3. QMP AMB
QM
2
because vertical s are . Then, because AM = PM
BM = 1 ,
QMP AMB by the SAS Theorem. 4. M A
(Given). Because there are 180° in a triangle, mJ = 130, and
J C. So MJN ACB by the AA Postulate.
BX
5. Because AX = BX and CX = RX, AX
CX = RX . AXB
CXR because vertical angles are . Therefore AXB CXR by the SAS Theorem. 6. Because AB = BC =
AB BC CA
CA and XY = YZ = ZX, XY
= YZ = ZX . Then ABC XYZ by the SSS Theorem. 7.
11. 20
3 12. 36 13. 33 ft
15
2
8.
55
4
9.
48
7
10.
85
3
7B: Reading Comprehension
1. DHB ACB 2. AA similarity postulate
The triangles have two similar angles. 3. 1 : 2 4. 1 : 2
5 HB 6. 250 ft 7. a
5. DB
AB
CB
7C: Reading/Writing Math Symbols
1. no 2. no 3. yes 4. no 5. yes 6. no 7. yes, AAS
8. yes, Hypotenuse-Leg Theorem 9. yes, SAS or ASA or
AAS 10. not possible
7D: Visual Vocabulary Practice
1. Angle-Angle Similarity Postulate 2. golden ratio
3. Side-Side-Side Similarity Theorem 4. geometric mean
5. scale 6. Cross-Product Property 7. golden rectangle
8. simplest radical form 9. Side-Angle-Side Similarity
Theorem
All-In-One Answers Version A
Geometry
81
Geometry: All-In-One Answers Version A
(continued)
7E: Vocabulary Check
Practice 8-3
Similarity ratio: The ratio of lengths of corresponding sides
1. tan E = 34 ; tan F = 34 2.
3. tan E = 52; tan F = 52 4.
8. 2.3 9. 6.4 10. 78.7 11.
15. 27 16. 68 17. 39 18.
proportion is equal to the product of the means.
Ratio: A comparison of two quantities by division.
Golden rectangle: A rectangle that can be divided into a
square and a rectangle that is similar to the original rectangle.
Scale: The ratio of any length in a scale drawing to the
corresponding actual length.
Guided Problem Solving 8-3
1.
A
B
7F: Vocabulary Review
1. L 2. E 3. I 4. A 5. J 6. K 7. C 8. D 9. G 10. O
11. N 12. M 13. H 14. B 15. F
Chapter 8
1. " 51 2. 4 " 7 3. 2 " 65 4. " 7 5. 2 " 21
6. 18 " 2 7. 46 in. 8. 78 ft 9. 279 cm 10. 19 m
11. acute 12. right 13. obtuse 14. right 15. obtuse
16. acute
Guided Problem Solving 8-1
1. the sum of the lengths of the sides 2. Pythagorean
Theorem 3. 7 cm 4. 4 cm 3 cm 5. c2 = 42 + 32 6. 5
7. 12 cm 8. perimeter of rectangle = 14 cm; yes 9. Answers
will vary; example: Draw a 4 cm 3 cm grid, copy the given
figure, measure the lengths with a ruler, add them together.
10. 20 cm
Practice 8-2
1. x = 2; y = " 3 2. a = 4.5; b = 4.5" 3 3. c =
50"3
3 ;
3
3
d = 25"
4. 8 " 2 5. 14 " 2 6. 28"
7. 2
3
3
8. x = 15; y = 15 " 3 9. 3 " 2 10. 42 cm 11. 5.9 in.
12. 10.4 ft, 12 ft 13. 170 in.2 14. w =
y = 5" 2 ; z
80 cm
2. 180 3. m&A = 2m&X 4. 90 5. base: 40 cm, height:
10 cm 6. 4 7. 4 8. 76 9. 152 10. 28 11. yes 12. 46
Practice 8-4
Practice 8-1
5"3
= 3
20 cm
10"3
3 ;x =
5;
15. a = 4; b = 3 16. p = 4 " 3;
q = 4 " 3; r = 8; s = 4 " 6
1. sin P = 2"710 ; cos P =
3
7
3
5
= 56
8
= 17
2. sin P = 45; cos P =
"11
5
12
13 ; cos P = 13 4. sin P = 6 ; cos P
"2
"2
15
= 2 ; cos P = 2 6. sin P = 17
; cos P
3. sin P =
5. sin P
7. 64 8. 11.0 9. 7.0 10. 42 11. 7.8 12. 53
13. 6.6 14. 37 15. 11.0 16. 56 17. 11.5 18. 9.8
Guided Problem Solving 8-4
1. The sides are parallel. 2. sine 3. sin 30° 5 w6 4. 3.0
5. yes 6. cosine 7. cos x° 5 34 8. 41 9. Answers may
vary. Sample: cos 60° 36; sin 49° 34 10. 5.2, 2.6
Practice 8-5
1a. angle of depression from the birds to the ship
1b. angle of elevation from the ship to the birds
1c. angle of depression from the ship to the submarine
1d. angle of elevation from the submarine to the ship
2a. angle of depression from the plane to the person
2b. angle of elevation from the person to the plane
2c. angle of depression from the person to the sailboat
2d. angle of elevation from the sailboat to the person
3. 116.6 ft 4. 84.8 ft 5. 46.7 ft 6. 31.2 yd 7. 127.8 m
8. 323.6 m
9a.
Guided Problem Solving 8-2
1. 30°-60°-90° triangle 2. l 3. h 4. Í 3 5. 24 or 8Í 3
!3
48
6. 2 7.
or 16Í 3 8. 28 ft 9. 0.28 min 10. yes
!3
11. 34 ft
35
30 ft
9b. 26 ft
82
Geometry
All rights reserved.
Cross-Product Property: The product of the extremes of a
tan E = 1; tan F = 1
12.4 5. 31.0 6. 14.1 7. 7.1
26.6 12. 71.6 13. 39 14. 72
54
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
of similar polygons.
Geometry: All-In-One Answers Version A (continued)
Guided Problem Solving 8-5
14. Sample:
N
1. &e = 1, &d = &4 2. congruent 3. m&e = m&d
4. 7x - 5 = 4(x + 7) 5. 11 6. 72 7. 72 8. yes
9. 44, 44
30
W
Practice 8-6
1. 46.0, 46.0 2. -37.1, -19.7 3. 89.2, -80.3
4. 38.6 mi/h; 31.2º north of east 5. 134.5 m; 42.0º south
of west 6. 1.3 m/sec; 38.7º south of east 7. 55º north of east
8. 20º west of south 9. 33º west of north 10a. 1, 5
All rights reserved.
y
6
4
642
2
4
6
x ;
1. Check students’ work. 2. 100
100
3. 100 cos 30°; 100 sin 30° 4. 86.6; 50 5. 86.6, 50
6. 86.6, -50 7. 173.2, 0 8. 173; due east 9. yes
10. 100; due east
2 4 6x
8A: Graphic Organizer
11a. 1, -1 11b.
1. Right Triangles and Trigonometry 2. Answers may vary.
Sample: the Pythagorean Theorem; special right triangles; the
tangent ratio; sine and cosine ratios; angles of elevation and
depression; vectors 3. Check students’ work.
y
6
4
4 6x
642
2
4
6
8B: Reading Comprehension
1. A 2. J 3. B 4. J 5. B 6. B 7. b
8C: Reading/Writing Math Symbols
1. F 2. G 3. D 4. A 5. C 6. B 7. H 8. E
5 10. #ABC , #XYZ 11. m&A < 52°
9. sin21 A 5 12
y
12a. -7, -1 12b.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
S
Guided Problem Solving 8-6
y
10b.
E
6
4
2
42
4
6
7
12. tan Z 5 24
2 4 6x
8D: Visual Vocabulary Practice
1. 30°-60°-90° triangle 2. inverse of tangent 3. congruent
sides 4. tangent 5. Pythagorean Theorem 6. hypotenuse
7. 45°-45°-90° triangle 8. Pythagorean triple 9. obtuse
triangle
13. Sample:
8E: Vocabulary Check
N
Obtuse triangle: A triangle with one angle whose measure is
between 90 and 180.
W
48
E
Isosceles triangle: A triangle that has at least two congruent
sides.
Hypotenuse: The side opposite the right angle in a right
S
triangle.
Right triangle: A triangle that contains one right angle.
Pythagorean triple: A set of three nonzero whole numbers
a, b, and c that satisfy the equation a2 + b2 = c2.
All-In-One Answers Version A
Geometry
83
Geometry: All-In-One Answers Version A
(continued)
J
A
T
N
A
T
L
U
S
E
R
Y
G
D
H
P
G
H
S
N
L
P
U
R
F
S
I
M
E
E
M
Q
O
I
I
Y
H
G
E
A
K
U
R
R
E
L
O
F
L
S
B
T
N
T
N
L
O
N
Q
G
M
A
M
M
D
M
T
P
B
K
V
I
X
E
X
X
E
C
S
Q
E
A
L
V
Y
D
W
M
E
T
D
O
T
S
B
U
N
T
P
X
L
T
Z
I
T
O
R
N
I
F
A
D
R
Y
R
L
P
A
H
S
R
P
S
Z
S
E
O
E
E
E
J
I
I
M
T
A
H
Y
M
C
O
N
E
D
C
L
M
Z
C
R
D
E
G
H
E
D
P
I
L
U
T
I
N
E
B
M
J
C
V
O
R
F
P
S
U
T
A
B
A
F
N
V
E
C
T
O
R
Y
O
O
R
I
N
I
V
G
N
T
M
A
D
D
K
E
B
C
O
N
G
R
U
E
N
T
G
H
N
T
B
R
A
A
N
G
L
E
O
F
D
E
P
R
E
S
S
I
O
N
R
A
E
U
A
R
A
X
L
L
Y
T
N
E
J
O
B
M
C
O
O
V
Z
L
P
R
O
P
O
R
T
I
O
N
Practice 9-1
1. No; the triangles are not the same size. 2. Yes; the
hexagons are the same shape and size. 3. Yes; the ovals are
the same shape and size. 4a. C and F 4b. CD and
CD, DE and DE, EF and EF, CF and CF
5a. M and N 5b. MN and MN, NO and NO,
MO and MO 6. (x, y) S (x - 2, y - 4) 7. (x, y) S
(x - 2, y - 2) 8. (x, y) S (x - 3, y - 1) 9. (x, y) S
(x + 4, y - 2) 10. (x, y) S (x - 5, y + 1) 11. (x, y) S
(x + 2, y + 2) 12. W(-2, 2), X(-1, 4), Y(3, 3), Z(2, 1)
13. J(-5, 0), K(-3, 4), L(-3, -2) 14. M(3, -2),
N(6, -2), P(7, -7), Q(4, -6) 15. (x, y) S (x + 4.2,
y + 11.2) 16. (x, y) S (x + 13, y - 13) 17. (x, y) S
(x,y)
18. (x, y) S (x + 3, y + 3) 19a. P(-3, -1)
19b. P(0, 8), N(-5, 2), Q(2, 3)
Guided Problem Solving 9-1
1. the four vertices of a preimage and one of the vertices of
the image 2. Graph the image and preimage. 3. C(4, 2) and
C(0, 0) 4. x = 4, y = 2, x + a = 0, y + b = 0 5. a = -4;
b = -2; (x, y) → (x - 4, y - 2) 6. A(-1, 4), B(1, 3),
7.
y
A
B
A
B
D
C
D
2 O C 2
x
4
2
8. yes 9. (x, y) → (x + 1, y - 3), A9(4, 3), B9(6, 2), D9(3, 0)
Practice 9-2
1. (-3, -2) 2. (-2, -3) 3. (-1, -4)
4. (4, -2) 5. (4, -1) 6. (3, -4)
y
7a.
4
2
6 4 2 O
2 I
4
L
J
6 x
K
D(-2, 1)
84
Geometry
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Chapter 9
All rights reserved.
8F: Vocabulary Review Puzzle
Geometry: All-In-One Answers Version A (continued)
y
7b.
12.
4
K
I
J
10
O
2
6
L
y
B
A
2
4
6 x
6
4
T, T
4
2
y
8a.
4
Z
2
Y
6 4 2
All rights reserved.
6 4 2
2
2 4
W
4
8b.
A
6x
X
X
A
4 B
1. a point at the origin, and two reflection lines
2. A reflection is an isometry in which a figure and its image
have opposite orientations. 3. the image after two successive
W
reflections
6x
4
Z
4.
y
y=3
4
9.
6x
Guided Problem Solving 9-2
4
Y
2
4
13. (-6, 4) 14. (-8, 0) 15. (0, -12)
y
6 4
2
x
y
T
4
B
O(0,0)
4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2
6 4 2
2
2
6 x
4
5.
y=3
4 B T A
10.
T
y
4
6 4 2
x
O(0,0)
T
2 4
4
A
B B
y
O'(0,6)
6 x
6.
y=3
4
y
O'(0,6)
A
x
11.
y
T
4
O(0,0)
T
4
O"(0,6)
8 6 4 2
A 2
B
A 4
2
4
6
8 x
7. (0, - 6) 8. Yes. The x-coordinate remains 0 throughout.
9. O(0, 0) and O(0, 6)
B
All-In-One Answers Version A
Geometry
85
Geometry: All-In-One Answers Version A
Practice 9-3
(continued)
3.
slope of OA 5 2
5
y
1. I 2. I 3. I 4. GH 5. G 6. ST
7.
A
U
T
x
P
O
T
S
U
S
4.
y
B
8.
A
O
9.
x
O
P P
N
C
Q O
5. slope of OB 5 252 ; slope of OC 5 52 ; slope of OD 5 252 ;
N
O
P
N
Q
M
N
L L
Q
O
L
M
N
L
O
P
Q
M
P
the slopes of perpendicular line segments are negative
reciprocals. 6. square 7. yes 8. B(2, 7), C(7, -2),
D(-2, -7)
Practice 9-4
N
L
E
1. The helmet has reflectional symmetry. 2. The teapot has
reflectional symmetry. 3. The hat has both rotational and
reflectional symmetry. 4. The hairbrush has reflectional
M
O
symmetry.
Q
Q
5.
P
P
N
O
11. L
12.
13.
Q'
P'
S
6.
M
R'
P
P'
S
Q'
7. This figure has no lines of symmetry.
8.
P
R'
Q
Q
line symmetry and 72° rotational
symmetry
9.
R
R
Guided Problem Solving 9-3
line symmetry
1. The coordinates of point A, and three rotation
transformations. It is assumed that the rotations are
counterclockwise. 2. parallelogram, rhombus, square
86
Geometry
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
10.
D
Q
All rights reserved.
P
Geometry: All-In-One Answers Version A (continued)
10.
21.
line symmetry and 90° rotational
symmetry
11.
Guided Problem Solving 9-4
1. the coordinates of one vertex of a figure that is symmetric
about the y-axis 2. Line symmetry is the type of symmetry
line symmetry and 45° rotational
symmetry
All rights reserved.
12.
X
for which there is a reflection that maps a figure onto itself.
3. the coordinates of another vertex of the figure
4. images (and preimages) 5. reflection across the y-axis
6. (-3, 4) 7. Yes 8. (-6, 7)
O
X
180° rotational symmetry
13.
line symmetry
14.
Practice 9-5
1. L(-2, -2), M(-1, 0), N(2, -1), O(0, -1)
2. L(-30, -30), M(-15, 0), N(30, -15), O(0, -15) 3.
L(-12, -12), M(-6, 0), N(12, -6), O(0, -6)
4. 53 5. 12 6. 2 7. yes 8. no 9. no
R
10.
O
T
A
line symmetry and 45° rotational
symmetry
A
R
T
15.
A
12.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
R
11.
R
A
T
R
180° rotational symmetry
R
16.
O
A
line symmetry
17.
19.
18.
COOK
20.
H
O
A
X
All-In-One Answers Version A
T
A
T
13. P(-12, -12), Q(-6, 0), R(0, -6) 14. P(-12, 14),
Q(1 14, -14), R(1 34 , 2) 15. P(-21, 6), Q(3, 24),
R(-6, 6) 16. P(-2, 1), Q(-1, 0), R(0, 1)
Guided Problem Solving 9-5
1. A description of a square projected onto a screen by an
overhead projector, including the square’s area and the scale
factor in relation to the square on the transparency. 2. The
scale factor of a dilation is the number that describes the size
change from an original figure to its image. 3. the area of
the square on the transparency 4. smaller; The scale factor
16 > 1, so the dilation is an enlargement. 5. 1 ; 1
16 16
Geometry
87
Geometry: All-In-One Answers Version A
y
8.
4
tranparency has 1 3 1 5 1 times the area. 7. 3 ft2
16
16
256
256
8. The shape does not matter. Regardless of the shape, the
figure is being enlarged by a factor of 16 in two directions, so
that the screen image has 16 16 = 256 as large an area as
the figure on the transparency. 9. 2520 ft2
2
2 O
2
9.
m
m
B
y
T
S
4
2 O
B 2
E
2
y
4
B
C C
C
6 4 2 O
2
E
4
5.
y
m
J
S
4
y
4
S
2
4
T
2 O
2
T
y
T
8
4
6 x
B
y
13.
4
T
2
4
S
2
4 2 O
2
4
B
6
E
2
6 x
B
10
T
12.
E
E
4 x
4
m
S
B
6 E 2 O
2
y
B
S
2
J
4 x
2
11.
J
7.
T
2
m
O
2
6 x
4
10.
4.
2
8 x
6
T
2
B B B
6.
4
4
1. I. D II. C III. B IV. A 2. I. B II. A III. C IV. D
3.
2
2
4 x
4 2 O
2
4 E
2
4
6 x
S
14. reflection 15. rotation 16. glide reflection
17. translation
88
Geometry
All rights reserved.
Practice 9-6
E
S
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1 ;Being 1 as high and 1 as wide, the square on the
6. 256
16
16
(continued)
Geometry: All-In-One Answers Version A (continued)
Guided Problem Solving 9-6
7.
8.
1. assorted triangles and a set of coordinate axes
2. a transformation 3. the transformation that maps one
triangle onto another 4. They are congruent, which confirms
All rights reserved.
that they could be related by one of the isometries listed.
Corresponding vertices: E and P, D and Q, C and M.
5. counterclockwise; clockwise. The triangles have opposite
orientations, so the isometry must be a reflection or glide
reflection. 6. No; there is no reflection line that will work.
7. a glide reflection consisting of a translation (x, y) →
(x + 11, y) followed by a reflection across line y = 0
(the x-axis) 8. Yes, both are horizontal. 9. 180º rotation
with center Q 22 1 , 0 R
2
11.
Practice 9-7
1.
12. yes 13. yes 14. no 15. yes 16. no 17. no
translational symmetry
2.
line symmetry,
rotational symmetry,
translational symmetry,
glide reflectional symmetry
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
9.–11. Samples:
9.
10.
Guided Problem Solving 9-7
1. two different-sized equilateral triangles and a trapezoid
2. A tessellation is a repeating pattern of figures that
completely covers a plane, without gaps or overlaps.
3. Theorem 9-7 says that every quadrilateral tessellates. If the
three polygons can be joined to form a quadrilateral (using
each polygon at least once), that quadrilateral can be the basis
of a tessellation.
4.
3.
5. Possible answer:
translational symmetry
4.
line symmetry,
rotational symmetry,
translational symmetry,
glide reflectional symmetry
6. Check students’ work.
7. Answers may vary. Sample:
5.
line symmetry,
rotational symmetry,
translational symmetry,
glide reflectional symmetry
6.
9A: Graphic Organizer
rotational symmetry,
translational symmetry
All-In-One Answers Version A
1. Transformations 2. Answers may vary. Sample: reflections;
translations; compositions of reflections; and symmetry
3. Check students’ work.
Geometry
89
Geometry: All-In-One Answers Version A
(continued)
9B: Reading Comprehension
9E: Vocabulary Check
1. D 2. E, R, 5, and 20 3. The center of the county is
Transformation: A change in the position, size, or shape of a
geometric figure.
Preimage: The given figure before a transformation is
applied.
Image: The resulting figure after a transformation is applied.
Isometry: A transformation in which an original figure and
its image are congruent.
Composition: A transformation in which a second
transformation is performed on the image of the first
transformation.
the center of the square mile bounded by roads 12, 13, K
and L. 4. C 5. 10 miles 6. a
9C: Reading/Writing Math Symbols
1. reflection; opposite 2. rotation; same 3. glide reflection;
opposite 4. translation; same 5. rotation; same
6. translation; same
9D: Visual Vocabulary Practice/High-Use
Academic Words
1
2
P
O
All rights reserved.
9F: Vocabulary Review Puzzle
1. theorem 2. symbols 3. reduce 4. application
5. simplify 6. explain 7. characteristic 8. table
9. investigate
G
L
Y
H
E
T
R
Y
T
R
Y
D
R
O
N
I
D
3
S
Y
M
M
E
R
O
B
5
T
U
S
E
F
S
6
I
S
10
M
E
D
L
O
M
E
E
C
L
T
R
L
I
O
A
O
13
T
15
E
T
A
I
T
O
I
A
N
R
A
7
8
11
N
S
L
R
L
O
C
U
I
O
N
N
E
16
C
9
A
T
L
R
T
A
T
N
14
I
O
R
12
D
O
I
L
A
T
I
O
N
N
U
F
D
N
E
R
18
S
17
C
A
V
L
E
N
T
Chapter 10
Practice 10-1
1. 8 2. 8 3. 60 4. 240 5. 2.635 6. 1.92 7. 0.8125
8. 9 9. 1500 10. 8.75 11. 3.44 12. 7 13. 110 14. 30
15. 16 16. 119 17. 12 18. 48 19. 40 20. 56
Geometry
I
T
S
O
T
M
90
C
H
E
R
I
T
O
E
N
X
Guided Problem Solving 10-1
1. 21 bh 2. Check students’ work; B(7, 7) 3. a right angle
å
4. 7 5. OA 6. 7 7. 24.5 8. a square 9. 49 10. 12
11. 24.5 12. 12 b2
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4
Geometry: All-In-One Answers Version A (continued)
Practice 10-2
Practice 10-6
1. 48 cm2 2. 784 in.2 3. 11.4 ft2 4. 90 m2 5. 2400 in.2
6. 374 ft2 7. 160 cm2 8. 176.25 in.2 9. 54 ft2 10. 42.5
square units 11. 96 " 3 square units 12. 36 " 3 square
units 13. 45 cm2 14. 226.2 in.2 15. 49,500 m2
1. 32p 2. 16p 3. 7.8p 4. Samples: DFA, ABF
00
1 1
5. Samples: FE , FD 6. Samples: FEB , EDA 7. Samples:
0 0
FE , ED 8. ACB 9. FCD 10. 32 11. 43 12. 54
13. 39 14. 71 15. 90 16. 50 17. 220 18. 140 19. 220
20. 130 21. 6p in. 22. 16p cm 23. 94 p m
Guided Problem Solving 10-2
1. 12 h(b1 1 b2) 2. 3; 4; 2 3. 9 4. no
5. yes 6. 3
All rights reserved.
Practice 10-3
8 "3
8 "3
= 3 ;d = 3
1. x = 7; y = 3.5 " 3 2. a = 60; c
3. p = 4 " 3; q = 8; x = 30 4. m1 = 45; m2 = 45;
m3 = 90; m4 = 90 5. m5 = 60; m6 = 30;
m7 = 120 6. m8 = 60; m9 = 30; m10 = 60
7. 12 " 3 8. 36.75 " 3 9. 75 " 3 10. 120 in.2
11. 137 in.2 12. 97 in.2
Guided Problem Solving 10-3
1. 10 2. The angles are all congruent and the sides are all
congruent. 3. 360 4. 36 5. bisects 6. 2 7. 18 8. 180
9. 72 10. 180 11. 30; 15; 75
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 10-4
1. 4 : 5; 16 : 25 2. 5 : 3; 25 : 9 3. 3 : 4; 9 : 16 4. 1 : 2
2700
2
2
5. 9 : 5 6. 1 : 6 7. 8 : 3 8. 576
5 in. 9. 7 cm
1152
2
10. 25 in. 11. 2 : 3 12. 1 : 2 13. 2 : 5 14. 108 ft2
Guided Problem Solving 10-4
1. scale 2. Check students’ work. 3. Answers will vary.
Let a be the side opposite the 30° angle, b be the side opposite
the 50° angle, and c be the side corresponding to 200 yd. Then
200 yd
5 yd
a < 20 mm, b < 30 mm and c < 40 mm. 4.
5
40 mm
1 mm
5. < 90 mm 6. < 16 mm 7. < 320 mm2 8. 450 9. 8000
10. yes 11. 16,000
Practice 10-5
1. 174.8 cm2 2. 578 ft2 3. 1250.5 mm2 4. 192.6 m2
5. 1131.4 in.2 6. 419.2 cm2 7. 324.9 in.2 8. 162 cm2
9. 37.6 m2 10. 30.1 mi2 11. 55.4 km2 12. 357.6 in.2
13. 9.7 mm2 14. 384.0 in.2 15. 54.5 cm2 16. 9.1 ft2
17. 80.9 m2 18. 83.1 m2 19. 65.0 ft2 20. 119.3 ft2
21. 8 m2 22. 55.4 ft2
11
Guided Problem Solving 10-6
1. 11; 2 2. 11; 3; 33 3. circumference; 2 4. 31 5. 26 6. 41
7. 105 8. 102 9. 3027.15
Practice 10-7
p
49
1
1
1. 49p 2. 21.2 3. 49
6 p 4. 6 p - 21.2 5. 4 p or 4 6. 8
p
p
9
3
1
7. 16
p or 16
8. 16
- 81 9. 4p 10. 18
5 p 11. 8 p 12. 2 p
p
63
25
13. 15
2 p 14. 6p 15. 8 p 16. 3 p 17. 2 18. 3
19. 2 20. 10
Guided Problem Solving 10-7
? pr 2 2. 45 3. 25 4. 30 5. 27 6. a piece from
1. mAB
360
the outer ring 7. yes 8. a piece from the top tier
Practice 10-8
3
7
1
1. 8.7% 2. 10.9% 3. 15.3% 4. 10
5. 10
6. 25 7. 10
1
8. 1 9. 5 10. 22.2% 11. 4.0% 12. 37.5% 13. 20%
14. 50% 15. 40% 16. 33 13 %
Guided Problem Solving 10-8
1. outcomes 2. event 3. AF 4. FG 5. EG 6. AE;
Check students’ work. 7. 2 or about 67% 8. 2 or about
3
3
67%; yes 9. 3 or 60%
5
10A: Graphic Organizer
1. Area 2. Answers may vary. Sample: areas of
parallelograms and triangles; areas of trapezoids, rhombuses,
and kites; areas of regular polygons; perimeters and areas of
similar figures; trigonometry and area; circles and arcs; areas
of circles and sectors; geometric probability 3. Check
students’ work.
10B: Reading Comprehension
1. E 2. J 3. B 4. G 5. A 6. I 7. D 8. C 9. H 10. F
0
å å
11. a2 + b2 = c2 12. mAC 5 1058 13. MN RS
0 0
14. "225 5 15 15. #JKL 16. XY > PQ
Guided Problem Solving 10-5
17. ~ABCD 18. 42 = 16 19. b
1. subtract 2. 50 3. 12 ap 4. 30 5. 4 6. tan4308 7. 166
8. 116 9. yes 10. 151
10C: Reading/Writing Math Symbols
1. 2860 ft2 2. 3331 ft2 3. 86 bundles 4. 2823 squares 5. $9.00
6. $27.00 7. 45 minutes 8. 21.5 hours 9. $645 10. $1419
All-In-One Answers Version A
Geometry
91
(continued)
10D: Visual Vocabulary Practice
Guided Problem Solving 11-2
1. area of a triangle 2. radius 3. altitude of a parallelogram
4. area of a trapezoid 5. area of a kite or rhombus
6. similarity ratio 7. apothem 8. perimeter (of an octagon)
9. height of a triangle
1. a picture of a box, with dimensions 2. the amount of
cardboard the box is made of, in square inches 3. A front
and a back, each 4 in. 7 1 in.; a top and a bottom, each
2
4 in. 1 in.; and one narrow side, 1 in. 7 1 in. 4. Front and
2
back: each 30 in.2; top and bottom: each 4 in.2; one narrow
side: 7 1 in.2 5. 75 1 in.2 6. 75 1 in.2 is about half a square
2
2
2
foot (1 ft2 = 144 in.2), which is reasonable. The answer is an
approximation because the solution did not consider the
overlap of cardboard at the glued joints, nor does it factor in
the trapezoid-shaped cutout in the front and back sides.
7. 3 21 in.2
Semicircle: Half a circle.
Adjacent arcs: Arcs on the same circle with exactly one
point in common.
Apothem: The distance from the center to the side of a
regular polygon.
Circumference: The distance around a circle.
Concentric circles: Circles that lie on the same plane and
have the same center.
10F: Vocabulary Review Puzzle
1. adjacent arcs 2. semicircle 3. center 4. arc length
5. base 6. circle 7. central angle 8. congruent arcs
9. diameter 10. major arc 11. concentric circles
12. minor arc 13. Pythagorean triple 14. radius
15. apothem 16. sector 17. height 18. circumference
Highness, there is no royal road to geometry.
Chapter 11
Practice 11-1
1. 8 2. 5 3. 12 4. C 5. D 6. B 7. A 8. triangle
9. rectangle 10. rectangle
11. rectangle
Practice 11-3
1. 64 m2 2. 620 cm2 3. 188 ft2 4. 36p cm2 5. 800p m2
6. 300p in.2 7. 109.6 m2 8. 160.0 cm2 9. 387.7 ft2
10. 147.6 m2 11. 118.4 cm2 12. 668.2 ft2
Guided Problem Solving 11-3
1. a solid consisting of a cylinder and a cone 2. the surface
area of the solid 3. The top of the solid is a circular base of
the cylinder, A1 = pr2. The side of the solid is the lateral area
of the cylinder, A2 = 2prh. The bottom of the solid is the
lateral area of the cone, A3 = pr/. 4. r = 5 ft; h = 6 ft;
/ 5 "52 1 122 5 13 ft 5. Top: A1 = 78.5 ft2; side:
A2 = 188.5 ft2; bottom: 204.2 ft2 6. 471 ft2 7. About 0.43,
or a little less than half. This seems reasonable. 8. 628 ft2
Practice 11-4
12. rectangle
Guided Problem Solving 11-1
1. a two-dimensional network 2. Verify Euler’s Formula for
the network shown. 3. 4 4. 6; 9 5. 4 + 6 = 9 + 1; This is
true. 6. F becomes 3, V stays at 6, E becomes 8. Euler’s
Formula becomes 3 + 6 = 8 + 1, which is still true.
7. 4 + 4 = 6 + 2; This is true.
Practice 11-2
1. 62.8 cm2 2. 552.9 ft2 3. 226.2 cm2 4. 113.1 m2
5. 44.0 ft2 6. 84.8 cm2 7a. 320 m2 7b. 440 m2
8a. 576 in.2 8b. 684 in.2 9a. 216 mm2 9b. 264 mm2
10a. 48 cm2 10b. 60 cm2 11a. 1500 m2 11b. 1800 m2
12a. 2000 ft2 12b. 2120 ft2 13. 8p m2 14. 94.5p cm2
15. 290p ft2
92
Geometry
1. 1131.0 m3
5. 785.4 cm3
10. 1728 ft3
14. 943 in.3
2. 94,247.8 cm3 3. 7.1 in.3 4. 1781.3 in.3
6. 603.2 cm3 7. 120 in.3 8. 35 ft3 9. 42 m3
11. 784 cm3 12. 265 in.3 13. 76 ft3
15. 1152 m3
Guided Problem Solving 11-4
1. dimensions for a layer of topsoil, and two options for buying
the soil 2. which purchasing option is cheaper 3. 1 ft
3
4. 1400 ft3 5. 467 bags; $1167.50 6. 1400 ft3 < 52 yd3
7. $1164 (or slightly less if the topsoil volume is not rounded
up to the next whole number) 8. buying in bulk 9. The two
costs are quite close. It makes sense that buying in bulk would
be slightly cheaper for a good-sized backyard. 10. Buying
topsoil by the bag would be less expensive because it would
cost $335 and in bulk would cost at least $345.93.
Practice 11-5
1. 34,992 cm3 2. 400 in.3 3. 10,240 in.3 4. 4800 yd3
5. 150 m3 6. 48 cm3 7. 628.3 cm3 8. 314.2 in.3
9. 4955.3 m3 10. 1415.8 in.3 11. 1005.3 m3 12. 18.8 ft3
13. 20 14. 2 15. 6
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
10E: Vocabulary Check
All rights reserved.
Geometry: All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Guided Problem Solving 11-5
1. a description of a plumb bob, with dimensions 2. the plumb
bob’s volume 3. equilateral; 2 cm 4. a = "3 cm; B = 6"3
cm2 5. Prism: V1 = 36"3 cm3; pyramid: V2 = 6"3 cm3;
plum bob: V < 73 cm3 6. 1 ; this seems reasonable 7. 42 cm3
7
Practice 11-6
All rights reserved.
1. 2463.0 in.2 2. 6,157,521.6 m2 3. 12.6 cm2 4. 1256.6 m2
5. 50.3 ft2 6. 153.9 m2 7. 1436.8 mi3 8. 268,082.6 cm3
9. 7238.2 m3 10. 14.1 cm3 11. 2,572,354.6 cm3
12. 904,810.7 yd3 13. 546 ft2 14. 399 m2 15. 1318 cm2
16. 233 cm3 17. 7124 cm3 18. 5547 cm3 19. 42 cm3
Guided Problem Solving 11-6
1. the diameter of a hailstone, along with its density compared
to normal ice 2. the hailstone’s weight 3. V 5 4 pr3
3
4. 2.8 in. 5. 91.95 in.3 6. 1.7 lb 7. The density of normal
ice. This was given for comparison purposes only. 8. 0.01 oz
Practice 11-7
1. 7 : 9 2. 5 : 8 3. yes; 4 : 7 4. yes; 4 : 3 5. not similar
6. yes; 4 : 5 7. 125 cm3 8. 256 in.3 9. 432 ft3 10. 25 ft2
11. 90 m2 12. 600 cm2 13. 54,000 lb 14. 16 lb
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Guided Problem Solving 11-7
1. two different sizes for an image on a balloon, and the
volume of air in the balloon for one of those sizes 2. the
volume of air for the other size 3. The two sizes of the clown
face have a ratio of 4 : 8, or 1 : 2. 4. 1 : 8 5. 108 in.3 8 =
864 in.3 6. Yes, the two clown face heights are lengths,
measured in inches (not in.2 or in.3). 7. 351 cm3
11A: Graphic Organizer
1. Surface Area and Volume 2. Answers may vary. Sample:
space figures and nets; space figures and drawing; surface
areas; and volumes. 3. Check students’ work.
11B: Reading Comprehension
1. Area = 12 base times height; area of a triangle
2. slope = the difference of the y-coordinates divided by the
difference of the x-coordinates; slope of the line between
2 points 3. a squared plus b squared equals c squared, the
Pythagorean Theorem; the lengths of the sides of a right
triangle 4. area = pi times radius squared; area of a circle
5. volume = pi times radius squared times height; volume of a
cylinder 6. the sum of the x-coordinates of two points
divided by 2, and the sum of the y-coordinates of two points
divided by 2; midpoint of a line segment with endpoints
(x1, y1) and (x2, y2) 7. circumference = two times pi times
the radius; circumference of a circle 8. volume = one-third
pi times radius squared times height; volume of a cone
All-In-One Answers Version A
9. distance equals the square root of the quantity of the sum
of the square of the difference of the x-coordinates, squared,
plus the difference of the y-coordinates, squared; the distance
between two points (x1, y1) and (x2, y2) 10. area = one-half
the height times the sum of base one plus base two; area of a
trapezoid 11. a
11C: Reading/Writing Math Symbols
1. 527.8 in2 2. 136 ft2
11D: Visual Vocabulary Practice
1. prism 2. cross section 3. sphere 4. cylinder
5. cone 6. pyramid 7. polyhedron 8. altitude 9. base
11E: Vocabulary Check
Face: A surface of a polyhedron.
Edge: An intersection of two faces of a polyhedron.
Altitude: For a prism or cylinder, a perpendicular segment
that joins the planes of the bases.
Surface area: For a prism, cylinder, pyramid, or cone, the
sum of the lateral area and the areas of the bases.
Volume: A measure of the space a figure occupies.
11F: Vocabulary Review
1. polyhedron 2. edge 3. vertex 4. cross section 5. prism
6. altitude 7. right prism 8. slant height 9. cone
10. volume 11. sphere 12. hemispheres 13. similar figures
14. hypotenuse
Chapter 12
Practice 12-1
1. 32 2. 50 3. 72 4. 15 5. "91 6. 6 7. "634
8. "901 9. 4 "3 10. circumscribed 11. inscribed
12. circumscribed 13. 24 cm 14. 28 in. 15. 52 ft
Guided Problem Solving 12-1
1. The area of one is twice the area of the other.
2. inscribed; circumscribed
3.
4.
r
5. 2r 6. 4r2 7. "2r 8. 2r 2 9. 4r 2 = 2(2r 2): One area is
double the other area. 10. yes 11. The same: The area of
one circle is twice the area of the other circle.
Geometry
93
Geometry: All-In-One Answers Version A
0
0
3. r = #2 ; mAB 102.7 4. 3 5. 4.5
6. 3 7. 20.8 8. 13.7 9. 8.5
Q T;
010. 0
PR SU 11. A J; BC KL 12. Construct
radii OD and OB . FD = EB because a diameter perpendic41
0
(0, 5)
(0, 0)
ular to a chord bisects it, and chords CD and AB are congruent
(given). Then, by HL, OEB OFD, and by CPCTC,
OE = OF. 13. Because congruent arcs have congruent
chords, AB = BC = CA. Then, because an equilateral
triangle is equiangular, mABC = mBCA = mCAB.
Guided Problem Solving 12-2
y
18.
(3, 5)
x
1 2 3 4 5
y
19.
2
2
2
4
x
2
(2, 4)
4
Guided Problem Solving 12-3
1. C 2. 180 3. a diameter 4. Check students’ work.
5. Check students’ work. 6. Check students’ work.
7. Check students’ work. 8. 90 9. Check students’ work.
1. 87 2. 35 3. 45 4. 120 5. x = 45; y = 50; 6. 90;
y = 70 7. x = 58; y = 58; z = 63 8. x = 30; y = 66
9. x = 30; y = 30; z = 120 10. x = 16; y = 52
11. x = 138; y = 111; z = 111 12. x = 30; y = 60
13. 10 14. 14.8 15. 4.7 16. 4 17. 3.2 18. 6
3
5
4
3
2
1
Practice 12-3
Practice 12-4
x
5
1. perpendicular 2. bisects 3. 4; 3 4. right 5. radius
6. Pythagorean Theorem 7. 5 8. yes 9. 8; 4
1. A and D; B and C 2. ADB and CDB
3. ADB and CAD 4. 55 5. x = 45; y = 50;
z = 85 6. x = 90; y = 70 7. 180 8. 70 9. x = 120;
y = 60; z = 60 10. x = 120; y = 100; z = 140
11. x =0
63; y = 63; z = 54 12. x = 50; y = 80; z = 80
13a. mAE = 170 13b. mC = 85 13c. mBEC = 10
13d. mD = 85 14a. mA = 90 14b. mB = 80
14c. mC = 90 14d. mD = 100
(5, 0)
All rights reserved.
1. r = 13; mAB 134.8 2. r = 3 "5; mAB 53.1
12. (x + 3)2 + (y - 3)2 = 1 13. x2 + (y - 3)2 = 16
14. (x - 7)2 + (y + 2)2 = 4 15. x2 + (y + 20)2 = 100
16. (x + 4)2 + (y + 6)2 = 25
y
17.
4 6
20.
y
(1, 7)
6
4
(1, 1)
Guided Problem Solving 12-4
4 2
1. tangent 2. 360. 3. 12 4. 85 5. 2 6. 180 7. 95
8. 104; 86; 75 9. 360 10. yes; The sum of the measures of
the arcs should be 360. 11. Opposite angles are not
6
2
(5, 1)
2
4
x
2
4
supplementary.
Practice 12-5
1. C(0, 0), r = 6 2. C(2, 7), r = 7 3. C(-1, -6), r = 4
4. C(-3, 11), r = 2"3 5. x2 + y2 = 49
6. (x - 4)2 + (y - 3)2 = 64 7. (x - 5)2 + (y - 3)2 = 4
8. (x + 5)2 + (y - 4)2 = 14 9. (x + 2)2 + (y + 5)2 = 2
10. (x + 1)2 + (y - 6)2 = 5 11. x2 + y2 = 4
94
Geometry
21.
23.
24.
25.
26.
x2
(x
(x
(x
(x
+
+
+
-
y2 = 25
4)2 + (y
7)2 + (y
4)2 + (y
4)2 + (y
22. (x - 5)2 + (y - 9)2 = 9
+
+
+
-
3)2
2)2
3)2
1)2
=
=
=
=
61
80
4
16
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 12-2
(continued)
Geometry: All-In-One Answers Version A (continued)
13.
Guided Problem Solving 12-5
1. (2, 2); 5 2. 43 3. They are perpendicular. 4. - 34 5. 39
4
6. y 5 234 x 1 39
7.
13 8. y 5 23 x 1 39 ; yes
4
4
4
9. x-intercept: a = 25
; y-intercept: b = 25
3
4
0.5 cm
T
U
14.
5 mm
6 mm
Practice 12-6
1.
Guided Problem Solving 12-6
T
1. a line 2. Check students’ work. 3. They are
perpendicular. 4. - 1 5. 2 6. the midpoint of PQ 7. (3, 2)
2
All rights reserved.
1.5 cm
8. -4 9. y = 2x - 4 10. (2, 0); yes 11. y 5 21 x 1 2
2.
1 in.
P
12A: Graphic Organizer
1. Circles 2. Answers may vary. Sample: tangent lines; chords
Q
and arcs; inscribed angles; and angle measures and segment
lengths 3. Check students’ work.
3.
12B: Reading Comprehension
A
B
4.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Answers may vary. Sample answers:
1. AD, CE 2. BF, CD, AD 3. HD 4. AH
1 1 1
0 00
0 0
5. AF , AB , BC 6. CDF, DAB, DAC 7. AC , ED
1
8. /AOC 9. /DCE 10. DEA 11. /DAH
12. /AGB 13. /AHC 14. /DCE 15. AH 16. a
0.75 in.
0.5 in.
R 0.5 in.
S
12C:
Math Symbols
* )Reading/Writing
* )
1. BC ' DE 2. EF is tangent to (G. 3. WX > ZX
4. AB ' XY 5. HM = 4
0.75 in.
5.
6.
12D: Visual Vocabulary Practice
Y
X
B
A
7.
1. tangent to a circle 2. standard form of an equation of a
circle 3. secant 4. chord 5. circumscribed about
6. intercepted arc 7. inscribed in 8. inscribed angle
9. locus
H
F
E
P
J
C
Inscribed in: A circle inside a polygon with the sides of the
G
polygon tangent to the circle.
8. a third line parallel to the two given lines and midway
between them 9. a sphere whose center is the given
point and whose radius is the given distance 10. the points
within, but not on, a circle of radius 1 in. centered
at the given point
11.
12E: Vocabulary Check
12.
Chord: A segment whose endpoints are on a circle.
Point of tangency: The single point of intersection of a
tangent line and a circle.
Tangent to a circle: A line, segment, or ray in the plane of a
circle that intersects the circle in exactly one point.
Circumscribed about: circle outside a polygon with the
vertices of the polygon on the circle.
0.75 in.
2 cm
Q
All-In-One Answers Version A
R
S
0.75 in.
12F: Vocabulary Review
1. D 2. F 3. G 4. A 5. C 6. B 7. E 8. M 9. I
10. L 11. N 12. H 13. K 14. J
Geometry
95
Geometry: All-In-One Answers Version A
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 1-1
3 Finding a Counterexample Find a counterexample for the conjecture.
Patterns and Inductive Reasoning
Lesson Objective
1
Since 32 42 52, the sum of the squares of two consecutive numbers is
the square of the next consecutive number.
NAEP 2005 Strand: Geometry
Use inductive reasoning to make
conjectures
Topic: Mathematical Reasoning
Begin with 4. The next consecutive number is 4 1 5 .
Local Standards: ____________________________________
The sum of the squares of these two consecutive numbers is
42 52 16 25 41 .
Vocabulary.
The conjecture is that 41 is the square of the next consecutive number.
The square of the next consecutive number is (5 1)2 62 36 .
Inductive reasoning is reasoning based on patterns you observe.
A conjecture is a conclusion you reach using inductive reasoning.
A counterexample is an example for which the conjecture is incorrect.
All rights reserved.
All rights reserved.
So, since 42 52 and 62 are not the same, this conjecture is
Use the pattern to find the next two terms in the sequence.
384, 192, 96, 48, . . .
half
2
48
2
2000, $9.50 in 2001, and $11.00 in 2002. Make a conjecture about the price in 2003.
Write the data in a table. Find a pattern.
24
and 24 2
12
.
2 Using Inductive Reasoning Make a conjecture about the sum of the cubes
All rights reserved.
of the first 25 counting numbers.
Find the first few sums. Notice that each sum is a perfect square and that
the perfect squares form a pattern.
3
1
3
1 3
1 3
1 3
1 2
3
2
3
3
2 3
3
3
3
2 3 4
3
3
3
3
2 3 4 5
1 1 2
9 3 2
36 6 2
100 10 2
225 15 2
1
2
(1 2)
2
(1 2 3)
2
(1 2 3 4)
2
(1 2 3 4 5)
The sum of the first two cubes equals the square of the sum of the first
two
counting numbers. The sum of the first three cubes equals the
square of the sum of the first three counting numbers. This pattern
continues for the fourth and fifth rows. So a conjecture might be that
2000
2001
2002
$8.00
$9.50
$11.00
Each year the price increased by $
1.50
. A possible conjecture is that
the price in 2003 will increase by $
1.50
. If so, the price in 2003 would
be $11.00 $
the preceding term. The next two
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
terms are 48 96
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Each term is
192
2
.
4 Applying Conjectures to Business The price of overnight shipping was $8.00 in
Examples.
1 Finding and Using a Pattern Find a pattern for the sequence. 384
false
1.50
$
12.50
.
Quick Check.
1. Find the next two terms in each sequence.
a. 1, 2, 4, 7, 11, 16, 22,
29 ,
37
b. Monday, Tuesday, Wednesday,
,...
Thursday
,
Friday
,...
c.
,
,...
Answers may vary. Sample:
the sum of the cubes of the first 25 counting numbers equals the square of
the sum of the first 25 counting numbers, or (1 2 3 . . . 25) 2.
2
Geometry Lesson 1-1
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
1
12
Lesson 1-2
1 3
4
22
1 3 5
9
32
1 3 5 7
16
42
1 3 5 7 9 25
52
Lesson Objectives
1 Make isometric and orthographic
drawings
2 Draw nets for three-dimensional
figures
Drawings, Nets, and Other Models
NAEP 2005 Strand: Geometry
Topic: Dimension and Shape
Local Standards: ____________________________________
Vocabulary.
All rights reserved.
All rights reserved.
An isometric drawing of a three-dimensional object shows a corner view of the figure drawn
The sum of the first 35 odd numbers is 35 2, or 1225.
3. Finding a Counterexample Find a counterexample for the conjecture.
Some products of 5 and other numbers are shown in the table.
5 7 35
5 3 15
5 11 55
4. Suppose the price of two-day shipping was $6.00 in 2000, $7.00 in 2001, and
$8.00 in 2002. Make a conjecture about the price in 2003.
Each year the price increased by $1.00. A possible conjecture is that
the price in 2003 will increase by $1.00. If so, the price in 2003 would
be $8.00 $1.00 $9.00.
All-In-One Answers Version A
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1 Isometric Drawing Make an isometric drawing of
Since 5 2 10, not all products of 5 end in 5.
(But they all either end in 5 or 0.)
Geometry Lesson 1-1
three-dimensional figure.
A foundation drawing shows the base of a structure and the height of each part.
Examples.
Therefore, the product of 5 and any positive integer ends in 5.
4
on isometric dot paper.
An orthographic drawing is the top view, front view, and right-side view of a
A net is a two-dimensional pattern you can fold to form a three-dimensional figure.
5 13 65
5 9 45
5 25 125
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3
Name_____________________________________ Class____________________________ Date ________________
2. Make a conjecture about the sum of the first 35 odd numbers. Use your
calculator to verify your conjecture.
1
Geometry Lesson 1-1
the cube structure at right.
An isometric drawing shows three sides of a figure
corner
from a
view. Hidden edges are not
shown, so all edges are solid lines.
Daily Notetaking Guide
Geometry Lesson 1-2
Geometry
5
1
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
2 Orthographic Drawing Make an orthographic drawing of the
4 Identifying Solids from Nets Is the pattern a net for a cube?
isometric drawing at right.
If so, name two letters that will be opposite faces.
A
Orthographic drawings flatten the depth of a figure. An orthographic
three
drawing shows
views. Because
no edge of the isometric drawing is hidden in the top, front,
and right views, all lines are solid.
The pattern
a net because you
can
is
fold it to form a cube. Fold squares A and C up to form the back
and front of the cube. Fold D up to form a side. Fold E over to
form the top. Fold F down to form another side.
B
Fr
ht
t Rig
on
D
E
F
C
After the net is folded, the following pairs of letters are on opposite faces:
A and C are the back and front faces.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Top
Identify the square that represents the tallest part. Write the
number 2 in the back square to indicate that the back section
is 2 cubes high.
Write the number 1 in each of the two front squares to indicate
that each front section is 1 cube high.
1
1
Right
2
5 Drawing a Net Draw a net for the figure with a square base
and four isosceles triangle faces. Label the net with its dimensions.
Think of the sides of the square base as hinges, and “unfold” the
figure at these edges to form a net. The base of each of the four
square
isosceles triangle faces is a side of the
. Write in the
known dimensions.
ht
t Rig
on
Because the top view is formed from 3 squares, show 3 squares in the
foundation drawing.
faces.
top
8 cm
10
8 cm
cm
All rights reserved.
Fr
and
cm
isometric drawing.
To make a foundation drawing, use the top view of the
orthographic drawing.
bottom
D and F are the right and left side faces.
10
3 Foundation Drawing Make a foundation drawing for the
All rights reserved.
Right
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Top
All rights reserved.
B and E are the
Front
Quick Check.
1. Make an isometric drawing of the cube structure below.
Front
Geometry Lesson 1-2
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Geometry Lesson 1-2
Name_____________________________________ Class____________________________ Date ________________
2. Make an orthographic drawing from this isometric drawing.
5. The drawing shows one possible net for the Graham Crackers box.
14 cm
Fr
ht
on
t Rig
20 cm
Front
Top
7
Right
7 cm
AM
AH
GR KERS
AC
CR
AM
AH
GR KERS
AC
CR
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
6
20 cm
7 cm
14 cm
3. a. How many cubes would you use to make the structure in Example 3?
4
All rights reserved.
All rights reserved.
Draw a different net for this box. Show the dimensions in your diagram.
Answers may vary. Example:
14 cm
20 cm
b. Critical Thinking Which drawing did you use to answer part (a),
the foundation drawing or the isometric drawing? Explain.
7 cm
4. Sketch the three-dimensional figure that corresponds to the net.
8
2
Geometry Lesson 1-2
Geometry
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Answers may vary. Sample: the foundation drawing; you can just
add the three numbers.
Daily Notetaking Guide
Geometry Lesson 1-2
9
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 1-3
1
A plane is a flat surface that has no thickness.
Points, Lines, and Planes
Lesson Objectives
NAEP 2005 Strand: Geometry
Understand basic terms of geometry
Understand basic postulates of
geometry
2
Name_____________________________________ Class____________________________ Date ________________
Two points or lines are coplanar if they lie on the same plane.
Topic: Dimension and Shape
Local Standards: ____________________________________
A postulate or axiom is an accepted statement of fact.
Vocabulary and Key Concepts.
Examples.
t
Line t is the only line that passes through points A and B .
B
A
name three points that are collinear and three points
that are not collinear.
All rights reserved.
All rights reserved.
Through any two points there is exactly one line.
Points
*AE )
and
*BD )
intersect at C .
Z
Y
W
triangle
and are not collinear.
U
R
You can name a plane using any three or more points on that plane
that are not collinear.
Plane RST and plane STW intersect in ST .
Postulate 1-4
Through any three noncollinear points there is exactly one plane.
A point is a location.
Space is the set of all points.
A line is a series of points that extends in two opposite directions
without end.
Geometry Lesson 1-3
t
B
A
Collinear points are points that lie on the same line.
RSU
plane
RST
, plane
plane
STU
, and plane
, plane
T
S
Some possible names for the plane shown are the following:
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
*ST)
R
T W
,
RTU
.
RSTU
H
3 Finding the Intersection of Two Planes Use the diagram at right.
What is the intersection of plane HGC and plane AED?
G
E
F
As you look at the cube, the front face is on plane AEFB, the back face
is on plane HGC, and the left face is on plane AED. The back and left
faces of the cube intersect at
*HD)
vertically at
C
D
A
. Planes HGC and AED intersect
HD
B
.
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 1-3
11
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
4 Using Postulate 1-4 Shade the plane that contains X, Y, and Z.
Lesson 1-4
Z
Points X, Y, and Z are the vertices of one of the four triangular
faces of the pyramid. To shade the plane, shade the interior of
the triangle formed by X , Y , and Z .
Y
X
V
W
Quick Check.
Local Standards: ____________________________________
Segment AB
A segment is the part of a line consisting of two
AB
b. Name line m in three different ways.
)* )
Answers may vary. Sample: ZW , WY , YZ .
All rights reserved.
All rights reserved.
endpoints and all points between them.
no
)*
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Vocabulary.
1. Use the figure in Example 1.
a. Are points W, Y, and X collinear?
*
Segments, Rays, Parallel Lines and Planes
Lesson Objectives
1 Identify segments and rays
2 Recognize parallel lines
B
A
Endpoint
Endpoint
)
A ray is the part of a line consisting of one endpoint
Ray YX
YX
and all the points of the line on one side of the endpoint.
X
Y
Endpoint
c. Critical Thinking Why do you think
* ) arrowheads are used when drawing
a line or naming a line such as ZW ?
Opposite rays are two collinear rays with the same
Arrowheads are used to show that the line extends in
opposite directions without end.
Q
)
endpoint.
RQ
H
G
Z
Answers may vary. Sample: HEF, HEFG, FGH.
E
* )
3. Name two planes that intersect in BF .
F
H
ABF and CBF
E
G
F
C
D
A
4. a. Shade plane VWX.
Z
B
b. Name a point that is coplanar
with points V, W, and X.
Y
V
12
)
R
RS
S
are opposite rays.
Skew lines are noncoplanar; therefore, they are not parallel and do not intersect.
C
D
H
E
←→
AB is
←→
B
A
parallel
←→
AB and CG are
←
→
to EF.
skew
lines.
G
F
Parallel planes are planes that do not intersect.
G
A
J
D
Y
and
Parallel lines are coplanar lines that do not intersect.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2. List three different names for plane Z.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
m
and Z form a
If two planes intersect, then they intersect in exactly one line.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
lie on a line, so they are
2 Naming a Plane Name the plane shown in two different ways.
Postulate 1-3
10
W
, and
so no other set of three points is collinear. For example, X, Y,
B
E
S
Z
,
Any other set of three points in the figure do not lie on a line,
If two lines intersect, then they intersect in exactly one point.
D
Y
collinear.
Postulate 1-2
C
X
1 Identifying Collinear Points In the figure at right,
Postulate 1-1
A
B
C
A
Plane ABC
X
H
B
Plane ABCD is
parallel
to plane GHIJ.
I
C
W
Geometry Lesson 1-3
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 1-4
Geometry
13
3
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
2. Use the diagram in Example 2.
a. Name all labeled segments that are parallel to GF.
Examples.
1 Naming Segments and Rays Name the segments and rays in the figure.
A
HE, CB, DA
The labeled points in the figure are A, B, and C.
A segment is a part of a line consisting of two endpoints and all points
between them. A segment is named by its two endpoints. So the
BA (or AB)
segments are
and
.
BC (or CB)
B
A ray is a part of a line consisting of one endpoint and all the points of
the line on one side of that endpoint. A ray is named by its endpoint first, followed
) .
)
by any other point on the ray. So the rays are
and
BA
BC
C
B
A
Parallel segments lie in the same plane, and the lines that contain them
do not intersect. The three segments in the figure that are parallel to
AE are BF , CG and DH .
H
All rights reserved.
c. Name another pair of parallel segments and another pair of skew segments.
D
Name all segments that are parallel to AE. Name all segments that
are skew to AE.
AB, DC, AE, DH
All rights reserved.
2 Identifying Parallel and Skew Segments Use the figure at right.
b. Name all labeled segments that are skew to GF.
C
Answers may vary. Sample: CG, BF; EF, DH
3. Use the diagram to the right.
a. Name three pairs of parallel planes.
G
E
S
PSWT RQVU, PRUT SQVW, PSQR TWVU
W
F
Skew segments are segments that do not lie in the same plane. The four
segments in the figure that do not lie in the same plane as AE are
BC ,
GH .
do not intersect
Planes are parallel if they
opposite
. If the walls of your classroom
walls are parts of parallel planes. If the ceiling and
floor of the classroom are level, they are parts of parallel planes.
Quick Check.
)
)
1. Critical Thinking Use the figure in Example 1. CB and BC form a line. Are
they opposite rays? Explain.
No; they do not have the same endpoint.
14
Geometry Lesson 1-4
Name_____________________________________ Class____________________________ Date________________
TV
c. Name a line that is parallel to plane QRUV.
* )
Answers may vary. Sample: PS
15
Geometry Lesson 1-4
Name_____________________________________ Class____________________________ Date ________________
Measuring Segments
Lesson Objectives
1 Find the lengths of segments
* )
Daily Notetaking Guide
Daily Notetaking Guide
Lesson 1-5
* )
b. Name a line that is parallel to PQ .
All rights reserved.
3 Identifying Parallel Planes Identify a pair of parallel planes in your classroom.
are vertical,
V
U
Examples.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
1 Comparing Segment Lengths Find AB and BC.
A
Are AB and AC congruent?
Local Standards: ____________________________________
BC = ∆ –1
Vocabulary and Key Concepts.
B
C
4 3 2 1 0 1 2 3 4 5
AB = ∆-5 - (–1) « = ∆-4« =
4
- 4« = »–5» = 5
AB BC so AB and AC are
not congruent.
.
All rights reserved.
The points of a line can be put into one-to-one
correspondence with the real numbers so
that the distance between any two points is the absolute value of the difference of the
corresponding numbers.
All rights reserved.
Postulate 1-5: Ruler Postulate
Postulate 1-6: Segment Addition Postulate
2 Using the Segment Addition Postulate If AB 25,
find the value of x. Then find AN and NB.
AN
(
2x 6
)(
A
AB
) 25
1 25
3x B
is between A and C, then AB BC AC.
NB
x7
3x
If three points A, B, and C are collinear and B
C
( )6
NB x 7 ( 8 ) 7 R
AB 5 u
a
Q
b
u
A
B
a
b
a
coordinate of A
coordinate of B
Congruent () segments are segments with the same length.
2 cm
A
B
A
D
C
2 cm
C
B
AB CD
AB
CD
D
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the length of AB
24
x 8
AN 2x 6 2 8
A coordinate is a point’s distance and direction from zero on a number line.
2x 6
x7
A
N
Use the Segment Addition Postulate (Postulate 1-6) to write an equation.
B
Segment Addition Postulate
Substitute.
Simplify the left side.
Subtract 1 from each side.
Divide each side by 3 .
10
15
Substitute 8 for x.
AN 10 and NB 15 , which checks because the sum of the
segment lengths equals 25.
midpoint
A midpoint is a point that divides a segment into two congruent
segments.
A
|
AB
16
4
Geometry Lesson 1-5
Geometry
B
|
C
BC
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 1-5
17
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
and
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
FG
T
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
CD ,
Q
R
P
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
R
Find RM, MT, and RT.
Name_____________________________________ Class____________________________ Date ________________
2. EG 100. Find the value of x. Then find EF and FG.
8x 36
5x 9
3 Using the Midpoint M is the midpoint of RT.
M
4x – 20
T
E
Use the definition of midpoint to write an equation.
RM 8x 36
Substitute.
8x
Add
45 3 x
x
(
15
15
)9
) 36 84
Substitute
168
, which is half of
84
15
for x.
84
RT RM MT RM and MT are each
from each side.
Divide each side by 3 .
Segment Addition
168
Postulate
, the length of RT.
All rights reserved.
(
RM 5x 9 5
MT 8x 36 8
to each side.
5x
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
15
36
Subtract
x 15, EF 40; FG 60
All rights reserved.
45
G
midpoint
Definition of
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
5x 9
5x MT
2x + 30
F
3. Z is the midpoint of XY, and XY 27. Find XZ.
13.5
Quick Check.
1. Find AB. Find C, different from A, such that AB and AC are congruent.
A
B
7 6 5 4 3 2 1 0 1 2 3 4 5
AB |7 (2)| |5| 5
We also want BC 5. That means that C must lie 5 units on
Geometry Lesson 1-5
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
An angle () is formed by two rays with the same endpoint. The rays
Measuring Angles
Lesson Objectives
1 Find the measures of angles
2 Identify special angle pairs
TBQ
x°
A straight angle has a measurement of exactly 180.
Congruent angles are two angles with the same measure.
Examples.
O C
If AOC is a straight angle, then
mAOB mBOC 180
.
B
Classify each as acute, right, obtuse, or straight.
a.
80
17 0
70
12
0
60
13
0
50
y
x
10
A
80
C
0
17 0
Daily Notetaking Guide
B
10 0 1
10
20
180
90
1
16 0
10
0
80
7 0 10 0
6 0 11 0
20
50 0 1
3
4
y
x
17 0
All-In-One Answers Version A
13
0
50
20
20
60
16 0
10
90
1
0
15
0
30
14 0
7 0 10 0
6 0 11 0
20
50 0 1
3
10 0 1
10
80
12
70
0
60, acute
Daily Notetaking Guide
A
b.
C
O
Geometry Lesson 1-6
3
G
2 Measuring and Classifying Angles Find the measure of each /AOC.
A
20
.
4
A O C
lG
Finally, the name can be a point on one side, the vertex, and a point
on the other side of the angle: lAGC or lCGA .
40
B
The name can be the vertex of the angle:
.
0
A
mAOC.
14
BOC
30
m
15
0
If point B is in the interior of AOC, then
C
l3
The name can be the number between the sides of the angle:
0
15
0
30
14 0
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Postulate 1-8: Angle Addition Postulate
1 Naming Angles Name the angle at right in four ways.
20
B
16 0
10
17 0
180
O
AOB
A right angle has a measurement of exactly 90.
10
30
16 0
15
0
y
0
m
x 5 180
180
An obtuse angle has a measurement between 90 and 180.
13
0
50
x
straight angle
angle
,x,
17 0
60
1
D
40
10 0 1
10
80
12
70
0
14
90
17 0
20
10 0
20
10
80
obtuse
90
90
An acute angle has a measurement between 0 and 90.
30
20
50 0 1
3
11 0
x5
16 0
A
70
angle
right
90
15
0
60
14
0
40
C
.
mlCOD 5 u x 2 y u .
All rights reserved.
180
b. If OC is paired with x and OD is paired with y, then
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
)
)
and OB is paired with
)
angle
0,x,
x°
x°
x°
acute
All rights reserved.
Postulate 1-7: Protractor Postulate
)
)
) )
Let OA and OB be opposite rays in a plane. OA , OB , and all the rays with
* )
endpoint O that can be drawn on one side of AB can be paired with the
real numbers from 0 to 180 so that
a. OA is paired with 0
T 1 Q
Local Standards: ____________________________________
Vocabulary and Key Concepts.
)
B
are the sides of the angle and the endpoint is the vertex of the angle.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
0
15
0
30
14 0
4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson 1-6
19
Geometry Lesson 1-5
0
18
16 0
All rights reserved.
the other side of B from A. So C must lie at –2 5 3.
180
B
O
150, obtuse
Geometry Lesson 1-6
Geometry
21
5
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
3 Using the Angle Addition Postulate Suppose that m1 42
and mABC 88. Find m2.
Quick Check.
A
1. a. Name CED two other ways.
Use the Angle Addition Postulate (Postulate 1-8) to solve.
m 1 m 2 mABC
42
Angle Addition
m2 88
Substitute 42
for mABC.
m2 46
Subtract
42
l2, lDEC
Postulate
for m1 and
1
88
2
B
from each side.
b. Critical Thinking Would it be correct to name any of the
angles E? Explain.
C
No, 3 angles have E for a vertex, so you
need more information in the name to
distinguish them from one another.
⬔ 3 and ⬔ 4
1
All rights reserved.
numbered angles that are related as follows:
a. complementary
2
3
5
b. supplementary
All rights reserved.
4 Identifying Angle Pairs In the diagram identify pairs of
4
⬔ 1 and ⬔ 2; ⬔ 2 and ⬔ 3
2. Find the measure of the angle. Classify it as
acute, right, obtuse, or straight.
60; acute
c. vertical angles
⬔ 1 and ⬔ 3
G
3. If mDEG 145, find mGEF.
Yes; the markings show they are congruent.
B
C
D
b. B and ACD are supplementary
No; there are no markings.
c. m(BCA) m(DCA) 180
Yes; you can conclude that the angles are
supplementary from the diagram.
E
F
A
E
D
a. ⬔ AOB
b. ⬔ BOC or ⬔ DOE
O
B
C
5. Can you make each conclusion from the information in
the diagram? Explain.
a. 1 3
b. 4 and 5 are supplementary
c. m(1) m(5) 180
5
1
4
2
3
a. Yes; the markings show they are congruent.
b. Yes; you can conclude the angles are adjacent and
supplementary from the diagram.
c. Yes; you can conclude that the angles are
supplementary from the diagram and their sum is
180 by the Angle Addition Postulate.
d. AB BC
Yes; the markings show they are congruent.
22
D
4. Name an angle or angles in the diagram supplementary to
each of the following:
a. DOA
b. EOB
All rights reserved.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
conclusion from the diagram?
a. A C
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
35
5 Making Conclusions From a Diagram Can you make each
Geometry Lesson 1-6
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 1-6
23
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 1-7
2 Constructing Congruent Angles Construct Y so that Y G.
Basic Constructions
Step 1 Draw a ray with endpoint Y.
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Y
G
Local Standards: ____________________________________
75
Step 2 With the compass point on point G, draw an arc that intersects
both sides of G. Label the points of intersection E and F.
Vocabulary.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objectives
1 Use a compass and a straightedge to
construct congruent segments and
congruent angles
2 Use a compass and a straightedge to
bisect segments and angles
E
A straightedge is a ruler with no markings on it.
A compass is a geometric tool used to draw circles and parts of circles called arcs.
Perpendicular lines are two lines that intersect to form right angles.
A perpendicular bisector of a segment is a line, segment, or ray that is
A
D
C
B
All rights reserved.
All rights reserved.
Construction is using a straightedge and a compass to draw a geometric figure.
G
75
F
Step 3 With the same compass setting, put the compass point on
point Y. Draw an arc that intersects the ray. Label the point
of intersection Z.
perpendicular to the segment at its midpoint, thereby bisecting the segment
K
N
L
Examples.
1 Constructing Congruent Segments Construct TW congruent
K
to KM.
Step 1 Draw a ray with endpoint T.
M
T
Step 2 Open the compass the length of KM.
Step 3 With the same compass setting, put the compass point
on point T. Draw an arc that intersects the ray. Label the
point of intersection W.
T
W
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
J
coplanar angles.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
into two congruent segments.
An angle bisector is a ray that divides an angle into two congruent
Z
Y
Step 4 Open the compass to the length EF. Keeping the same
compass setting, put the compass point on Z. Draw an arc
that intersects with the arc you drew in Step 3. Label the
point of intersection X.
X
Y
)
Step 5 Draw YX to complete Y.
Z
X
KM TW
Y G
24
6
Geometry Lesson 1-7
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Y
Z
Geometry Lesson 1-7
25
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
3 Constructing the Perpendicular Bisector
A
Given: AB.
* )
* )
Construct: XY so that XY AB at the midpoint M of AB.
Quick Check.
B
1. Use a straightedge to draw XY. Then construct RS so that RS 2XY.
Step 1 Put the compass point on point A and draw a long arc.
greater
Be sure that the opening is
than 12AB.
X
A
B
* )
Step 3 Draw XY . The point of intersection of AB and XY is M,
midpoint
the
of AB.
M
Y
)
4 Finding Angle Measures WR bisects AWB so that mAWR x
R
All rights reserved.
4x 48
B
W
m
m
AWR
x BWR
x
16
16
(
Subtract
mAWB m
mAWB 26
4x
from each side.
3 .
Divide each side by
) 48 16
16
16
for x.
16
m
AWR
16
for
m⬔BWR.
Substitute
mBWR 4
4x 48
Substitute x for m⬔AWR and
4x 48
mAWR angle bisector
Definition of
48
3x
X
A
and mBWR 4x 48. Find mAWB.
x
Angle Addition
BWR
32
Substitute
16
All rights reserved.
All rights reserved.
.
6
North
4
Central
Cedar 2
represent Oak.
(x2, y2 )
represent
4 2
Oak
(1, 2)
(
(
1
d% 2 2 4 2
d% 4 20
16
))2 (
2 (
Elm
Use the Distance Formula.
2
))2
Jackson
(2, 4)
Symphony
City Plaza
(0, 0)
x
6
Substitute.
and (x 2, y 2 ) be
,
y1
y2
)
6
2
) 1
Substitute 8 for x1 and
6
for x2. Simplify.
3
) 3
Substitute 9 for y1 and
3
for y2. Simplify.
(1, 3)
.
2
The y-coordinate is
8 (
2
9 (
Midpoint Formula
.
2
2
6
2
x 1x y 1y
a 1 2 2 , 1 2 2 b be
coordinates of G.
(1, 4)
and the midpoint
x2
and
y2
d Use the Midpoint Formula. S
5
1 x2 d Multiply each side by 2 . S
10
(1, 5)
. Solve for
, the
Find the x-coordinate of G.
1
1 x2
2
2
3
x2
d Simplify. S
6
4 y2
2
4 y2
x2
The coordinates of G are (3, 6) .
Simplify within parentheses.
20
%www
4.472135955
Simplify.
Use a calculator.
To the nearest tenth, the subway ride from Oak to Symphony is
28
y
4 South
"(x2 2 x1) 2 1 (y2 2 y1) 2
d % 1 x2
2
Use the Midpoint Formula. Let (x 1, y 1 ) be
Symphony.
d
(8, 9)
D(1, 4). Find the coordinates of the other endpoint G.
(x1, y1 )
. Let
27
3 Finding an Endpoint The midpoint of DG is M(1, 5). One endpoint is
.
ride from Oak to Symphony? Round to the nearest tenth. Each unit
represents 1 mile.
(1, 2)
The coordinates of the midpoint M are
1 Applying the Distance Formula How far is the subway
Symphony has coordinates
x1
The x-coordinate is
Examples.
Oak has coordinates (1, 2) . Let
Geometry Lesson 1-7
The midpoint has coordinates
)
2
100°
L
Local Standards: ____________________________________
(y2 y1 )2
, mJKL 50°
(6, 3) .
(
y1 y2
mNKL Use the Midpoint Formula. Let (x 1, y 1 ) be
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
,
2
K
coordinates of its midpoint M.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
x1 x2
J
2 Finding the Midpoint AB has endpoints (8, 9) and (6, 3). Find the
Formula: The Midpoint Formula
The coordinates of the midpoint M of AB with endpoints A(x 1, y 1 ) and
B(x 2, y 2 ) are the following:
(
)
4. If mJKN 50°, then find mNKL and mJKL. KN bisects JKL.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Formula: The Distance Formula
The distance d between two points A(x 1, y 1 ) and B(x 2, y 2 ) is
M
T
Name_____________________________________ Class____________________________ Date ________________
Key Concepts.
S
Daily Notetaking Guide
The Coordinate Plane
(x2 x1 )2
Z
N
Daily Notetaking Guide
Lesson 1-8
Y
3. Draw ST. Construct its perpendicular bisector.
for m⬔AWR and for m⬔BWR.
Name_____________________________________ Class____________________________ Date________________
d Measure ⬔XYP and ⬔PYZ to see that they
are congruent.
P
Postulate
Geometry Lesson 1-7
Lesson Objectives
1 Find the distance between two points
in the coordinate plane
2 Find the coordinates of the midpoint
of a segment in the coordinate plane
F
bisector of XYZ.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
XY AB at the midpoint of AB, so XY is the
perpendicular bisector
of AB.
B
b. Explain how you can use your protractor to check that YP is the angle
B
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
B
)
X
* )
*
All rights reserved.
A
All rights reserved.
X
Y
* )
S
2. a. Construct F with mF 2mB.
Step 2 With the
compass setting, put the compass
same
point on point B and draw another long arc. Label the
intersect
points where the two arcs
as X and Y.
* )
R
Y
B
Geometry Lesson 1-8
All-In-One Answers Version A
4.5
miles.
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 1-8
Geometry
29
7
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Quick Check.
Lesson 1-9
1. a. Using the map in Example 1, find the distance between Elm and Symphony.
Perimeter, Circumference, and Area
Lesson Objectives
about 8.9 miles
1
2
NAEP 2005 Strand: Measurement
Find perimeters of rectangles and
squares, and circumferences of circles
Find areas of rectangles, squares, and
circles
Topic: Measuring Physical Attributes
Local Standards: ____________________________________
All rights reserved.
b. Maple is located 6 miles west and 2 miles north of City Plaza. Find the
distance between Cedar and Maple.
about 3.2 miles
All rights reserved.
Key Concepts.
Perimeter and Area
s
b
h
s
r
dO
h
b
Square with side length s.
Perimeter P Area A 4s
s2
Rectangle with base b and
height h.
C
Circle with radius r and
diameter d.
Perimeter P Circumference C Area A 2b 2h
pd
or C 2pr
bh
Area 2. Find the coordinates of the midpoint of XY with endpoints X(2, 5) and Y(6, 13).
pr 2
The area of a region is the sum of the areas of its non-overlapping parts.
Examples.
1 Finding Circumference G has a radius of
6.5
cm. Find the
circumference of G in terms of π. Then find the circumference to the
nearest tenth.
C2 p
Formula for circumference of a circle
(
6.5
C
13
π
C
13
)
6.5
Substitute
for r.
Exact answer
40.840704
13p
Use a calculator.
, or about
40.8
cm.
Daily Notetaking Guide
Daily Notetaking Guide
6.5 cm
G
r
The circumference of G is
Geometry Lesson 1-8
.
Postulate 1-10
C 2π
30
equal
If two figures are congruent, then their areas are
31
Geometry Lesson 1-9
Name_____________________________________ Class____________________________ Date________________
Date ________________
Name_____________________________________ Class____________________________ Date ________________
2 Finding Perimeter in the Coordinate Plane
4 Finding Area of an Irregular Shape Find the area of the figure
C
Draw a horizontal line to separate the figure into three
rectangle
non-overlapping figures: one
and two
8
AR 4
2
6
8
10
Find the length of each side. Add the lengths to find the
perimeter.
9 0
AB % 81
BC u
11
CD %
(
%(
u
144
12
9 2wwww
0 2 %www
12 w2
)
225 w %wwww
u u 2 u 2
)2 ( 0 12
)2
9
11
2 0
144
uu 2 u
Use the Distance Formula.
15
2
Ruler Postulate
BC
CD
DA
Perimeter 15
2
15
Perimeter 34
The perimeter of quadrilateral ABCD is
2
34
units.
3 Finding Area of a Circle Find the area of B in terms of π.
Aπ
A
8
( 5 )2
25
25
Add the areas.
Simplify.
125
125
2
ft .
Quick Check.
b. Find the circumference of a circle with a diameter of 18 m to the nearest tenth.
56.5m
2. Graph quadrilateral KLMN with vertices K(3, 3), L(1, 3),
M(1, 4), and N(3, 1). Find the perimeter of KLMN.
20 units
4
N(3, 1)
y
M(1, 4)
2
x
O
4 2
3. You are designing a rectangular banner for the front of
a museum. The banner will be 4 ft wide and 7 yd high.
How much material do you need in square yards?
2
4
2
K(3, 3) 4
L(1, 3)
2
91
3 yd
Formula for the area of a circle
1.5
)2
2.25
π
The area of B is
32
yd.
B
r2
(
25
d Simplify. S
36p m
Aπ
A
d Substitute. S
Use the Distance Formula.
225 w %wwww
AB
1.5
75
2
1. a. Find the circumference of a circle with a radius of 18 m in terms of π.
)2
Perimeter In B, r A
)( 5 )
The area of the figure is
Ruler Postulate
12
15
75
s
15
9 2 % 81
DA )2 (
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
(
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
AB %
All rights reserved.
4
(
All rights reserved.
D
2
x
12
5 ft
5 ft
d Use the formulas. S AS bh
5 ft
.
Find each area. Then add the areas.
6
A
O
squares
10 ft
Draw and label ABCD on a coordinate plane.
15 ft
to the right.
5 ft
B
5 ft
y
10
10 ft
12
Quadrilateral ABCD has vertices A(0, 0), B(9, 12),
C(11, 12), and D(2, 0). Find the perimeter.
Substitute
1.5
for r.
1.5 yd
4. Copy the figure in Example 4. Separate it in a different way. Find the area.
125 ft2
2.25p
Geometry Lesson 1-9
Geometry
yd 2.
Daily Notetaking Guide
All rights reserved.
(6, 9)
Postulate 1-9
Daily Notetaking Guide
Geometry Lesson 1-9
33
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3. The midpoint of XY has coordinates (4, 6). X has coordinates (2, 3).
Find the coordinates of Y.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
(4, 4)
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 2-1
2 Finding a Counterexample Find a counterexample to show that this
Conditional Statements
Lesson Objectives
1
Recognize conditional statements
Write converses of conditional
statements
2
conditional is false: If x 2 0, then x 0.
NAEP 2005 Strand: Geometry
If p ,
pSq
then its measure is 180.
then q .
If the measure of an angle is
If q ,
qSp
All rights reserved.
If an angle is a straight angle,
You read it
All rights reserved.
Because (1) 2 1, which is greater than 0, the hypothesis is
Symbolic Form
180, then it is a straight angle.
then p .
Because 1 0, the conclusion is
false
.
true
.
The counterexample shows that the conditional is
.
false
3 Writing the Converse of a Conditional Write the converse of the following
conditional.
If x 9, then x 3 12.
The converse of a conditional exchanges the hypothesis and the conclusion.
A conditional is an if-then statement.
Conditional
The truth value of a statement is “true” or “false” according to whether the statement is true
or false, respectively.
The converse of the conditional “if p, then q” is the conditional “if q, then p.”
Examples.
1 Identifying the Hypothesis and the Conclusion Identify the hypothesis
and conclusion: If two lines are parallel, then the lines are coplanar.
In a conditional statement, the clause after if is the hypothesis and the
clause after then is the conclusion.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The conclusion is the part of an if-then statement (conditional) that follows then.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The hypothesis is the part that follows if in an if-then statement.
All rights reserved.
and
. This counterexample must be an example
in which x 2 0 (hypothesis true) and x 0 or x 0 (conclusion false).
Because any negative number has a positive square, one possible
counterexample is 1.
Conditional Statements and Converses
Statement
Example
Converse
false
the conclusion is
Local Standards: ____________________________________
Vocabulary and Key Concepts.
Conditional
true
A counterexample is a case in which the hypothesis is
Topic: Mathematical Reasoning
Converse
Hypothesis
Conclusion
Hypothesis
Conclusion
x9
x 3 12
x 3 12
x9
So the converse is: If x 3 12, then x 9.
Quick Check.
1. Identify the hypothesis and the conclusion of this conditional statement:
If y 3 5, then y 8.
Hypothesis: y 3 5
Conclusion: y 8
2. Show that this conditional is false by finding a counterexample:
If the name of a state includes the word “New,” then the state borders an ocean.
Counterexample:
Hypothesis: Two lines are parallel.
New Mexico does not border an ocean. It is a counterexample, so the
conditional is false.
Conclusion: The lines are coplanar.
34
Geometry Lesson 2-1
Daily Notetaking Guide
Daily Notetaking Guide
35
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Example.
Lesson 2-2
4 Finding the Truth Value of a Converse Write the converse of the
Biconditionals and Definitions
Lesson Objectives
1 Write biconditionals
2 Recognize good definitions
conditional. Then determine the truth value of each.
If a 2 25, then a 5.
NAEP 2005 Strand: Geometry
Topics: Dimension and Shape; Mathematical Reasoning
Local Standards: ____________________________________
Conditional: If a2 25, then a 5.
Vocabulary and Key Concepts.
The converse exchanges the
and the
hypothesis
conclusion
.
Converse: If a 5, then a2 25.
Biconditional Statements
A biconditional combines p S q and q S p as p 4 q.
. A counterexample is a 5:
(5) 2 25, and 5 5.
Because
52
25, the converse is
true
.
All rights reserved.
false
All rights reserved.
The conditional is
Statement
Biconditional
Example
Symbolic Form
An angle is a straight angle if
You read it
p if and
p4q
only if q .
and only if its measure is 180°.
A biconditional statement is the combination of a conditional statement and its converse.
Quick Check.
A biconditional contains the words “if and only if.”
3. Write the converse of the following conditional:
If two lines are not parallel and do not intersect, then they are skew.
Examples.
If two lines are skew, then they are not parallel and do not intersect.
1 Writing a Biconditional Consider the true conditional statement. Write its
4. Write the converse of each conditional. Determine the truth value of the
conditional and its converse. (Hint: One of these conditionals is not true.)
a. If two lines do not intersect, then they are parallel.
Converse:
If two lines are parallel, then they do not intersect.
The conditional is
false
and the converse is
true
.
b. If x 2, then u x u 2.
Converse:
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
converse. If the converse is also true, combine the statements as a biconditional.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Geometry Lesson 2-1
Conditional: If x 5, then x 15 20.
To write the converse, exchange the hypothesis and conclusion.
Converse: If x 15 20, then x 5.
When you subtract 15 from each side to solve the equation, you get x 5. Because
both the conditional and its converse are
a
biconditional
using the phrase
true
, you can combine them in
if and only if
.
Biconditional: x 5 if and only if x 15 20.
2 Separating a Biconditional into Parts Write the two statements that
form this biconditional.
Biconditional: Lines are skew if and only if they are noncoplanar.
Write a biconditional as two conditionals that are converses of each other.
Conditional: If lines are skew, then they are noncoplanar.
If u x u 2, then x 2.
Converse: If lines are noncoplanar, then they are skew.
The conditional is
36
true
and the converse is
false
Geometry Lesson 2-1
All-In-One Answers Version A
.
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 2-2
Geometry
37
9
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
2. Write two statements that form this biconditional about integers greater than 1:
3 Writing a Definition as a Biconditional Show that this definition of
triangle is reversible. Then write it as a true biconditional.
A number is prime if and only if it has only two distinct factors, 1 and itself.
Definition: A triangle is a polygon with exactly three sides.
Statement:
true
The original conditional is
.
If a number is prime, then it has only two distinct factors, 1 and itself.
Conditional: If a polygon is a triangle, then it has exactly three sides.
true
The converse is also
.
Converse: If a polygon has exactly three sides, then it is a triangle.
Statement:
, they can be combined to
If a number has only two distinct factors, 1 and itself, then it is prime.
Biconditional: A polygon is a triangle if and only if it has exactly three sides.
4 Identifying a Good Definition Is the following statement a good
All rights reserved.
true
All rights reserved.
Because both statements are
form a biconditional.
definition? Explain.
An apple is a fruit that contains seeds.
3. Show that this definition of right angle is reversible. Then write it as a true
biconditional.
Definition: A right angle is an angle whose measure is 90.
Conditional:
The statement is true as a description of an apple.
If an angle is a right angle, then its measure is 90°.
Exchange “An apple” and “a fruit that contains seeds.” The converse reads:
A fruit that contains seeds is an apple.
.
false
is not
The original statement
statement
is not
, so the converse of the
a good definition because the
reversible.
Quick Check.
1. Consider the true conditional statement. Write its converse. If the converse
is also true, combine the statements as a biconditional.
Conditional: If three points are collinear, then they lie on the same line.
Converse:
If three points lie on the same line, then they are collinear.
true
The converse is
.
Converse:
If an angle has measure 90°, then it is a right angle.
The two statements are
true
All rights reserved.
counterexamples
statement is
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
and peaches. These are
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
There are many fruits that contain seeds but are not apples, such as lemons
.
Biconditional:
An angle is a right angle if and only if its measure is 90°.
4. Is the following statement a good definition? Explain.
A square is a figure with four right angles.
It is not a good definition because a rectangle has four right angles
and is not necessarily a square.
Biconditional:
Three points are collinear if and only if they lie on the same line.
38
Geometry Lesson 2-2
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 2-2
39
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 2-3
3 Using the Law of Syllogism Use the Law of Syllogism to draw a
Deductive Reasoning
conclusion from the following true statements:
If a quadrilateral is a square, then it contains four right angles.
If a quadrilateral contains four right angles, then it is a rectangle.
NAEP 2005 Strand: Geometry
Topic: Mathematical Reasoning
Local Standards: ____________________________________
The conclusion of the first conditional is the hypothesis of the second
Law of Syllogism
conditional. This means that you can apply the
Vocabulary and Key Concepts.
The Law of Syllogism: If
then
is true.
In symbolic form:
If p S q is a true statement and p is true, then q is true.
Law of Syllogism
All rights reserved.
conclusion
All rights reserved.
Law of Detachment
If a conditional is true and its hypothesis is true, then its
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objectives
1 Use the Law of Detachment
2 Use the Law of Syllogism
pSr
pSq
and
qSr
.
are true statements,
is a true statement.
So you can conclude:
If a quadrilateral is a square, then it is a rectangle.
4 Drawing Conclusions Use the Laws of Detachment and Syllogism to
draw a possible conclusion.
If the circus is in town, then there are tents at the fairground. If there are
tents at the fairground, then Paul is working as a night watchman. The
circus is in town.
If p S q and q S r are true statements, then p S r is a true statement.
Deductive reasoning is a process of reasoning logically from given facts to a conclusion.
Because the conclusion of the first statement is the
of the second statement, you can apply the
hypothesis
to write
Law of Syllogism
a new conditional:
1 Using the Law of Detachment A gardener knows that if it rains, the
garden will be watered. It is raining. What conclusion can he make?
The first sentence contains a conditional statement. The hypothesis is
it rains.
Because the hypothesis is true, the gardener can conclude that
the garden will be watered.
2 Using the Law of Detachment For the given statements, what can you
conclude?
Given: If A is acute, then mA 90°. A is acute.
A conditional and its hypothesis are both given as true.
By the
Law of Detachment
conclusion of the conditional, mA 90°, is
, you can conclude that the
true
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Examples.
is working as a night watchman.
If the circus is in town, then Paul
________________________________________
The third statement means that the hypothesis of the new conditional is true.
You can use the
Law of Detachment
to form the conclusion:
Paul is working as a night watchman.
Quick Check.
1. Suppose that a mechanic begins work on a car and finds that the car will not
start. Can the mechanic conclude that the car has a dead battery? Explain.
No, there could be other things wrong with the car, such as a faulty starter.
2. If a baseball player is a pitcher, then that player should not pitch a complete
game two days in a row. Vladimir Nuñez is a pitcher. On Monday, he pitches
a complete game. What can you conclude?
Answers may vary. Sample: Vladimir Nuñez should not pitch a complete
game on Tuesday.
40
10
Geometry Lesson 2-3
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 2-3
41
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 2-4
3. If possible, state a conclusion using the Law of Syllogism. If it is not possible
to use this law, explain why.
a. If a number ends in 0, then it is divisible by 10.
If a number is divisible by 10, then it is divisible by 5.
Reasoning in Algebra
Lesson Objective
1
NAEP 2005 Strand: Algebra and Geometry
Connect reasoning in algebra and
geometry
Topics: Algebraic Representations; Mathematical
Reasoning
If a number ends in 0, then it is divisible by 5.
Local Standards: ____________________________________
All rights reserved.
b. If a number ends in 6, then it is divisible by 2.
If a number ends in 4, then it is divisible by 2.
Not possible; the conclusion of one statement is not the hypothesis
of the other statement.
All rights reserved.
Key Concepts.
Properties of Equality
Addition Property
If a b, then a c b c .
Subtraction Property
If a b, then a c b c .
Multiplication Property If a b, then a c b c .
Division Property
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The Volga River is in Europe.
If a river is less than 2300 miles long, it is not one of the world’s ten longest
rivers.
If a river is in Europe, then it is less than 2300 miles long.
Conclusion:
All rights reserved.
The Volga River is less than 2300 miles long.
Conclusion:
The Volga River is not one of the world’s ten longest rivers.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4. Use the Law of Detachment and the Law of Syllogism to draw conclusions.
If a b and c 0, then
a
c
b
.
c
Reflexive Property
a a .
Symmetric Property
If a b, then b a .
Transitive Property
If a b and b c, then a c .
Substitution Property
If a b, then b can replace a in any expression.
Distributive Property
a(b c) ab
ac .
Properties of Congruence
Reflexive Property
AB AB
A ⬔A
Symmetric Property
If AB CD, then CD AB .
If A B, then B ⬔A .
Transitive Property
If AB CD and CD EF , then AB EF
.
If A B and B C, then A ⬔C .
42
Geometry Lesson 2-3
Daily Notetaking Guide
Daily Notetaking Guide
Examples.
Example.
1 Justifying Steps in Solving an Equation Justify each step used to solve
3 Using Properties of Equality and Congruence Name the property that
5x 12 32 x for x.
justifies each statement.
a. If x y and y 4 3x, then x 4 3x.
Given: 5x 12 32 x
5x 44 x
Addition Property of Equality
4x 44
Subtraction Property of Equality
x 11
The conclusion of the conditional statement is the same as the equation
y 4 3x (given) after x has been substituted for y (given). The
property used is the Substitution Property of Equality.
Division Property of Equality
AB BC AC
19
4 2x
B
C
Segment Addition Postulate
21
x
21
(4 2x)
A
All rights reserved.
15 x
points A, B, and C are collinear with point B between
points A and C. Solve for x if AC 21, AB 15 x, and
BC 4 2x. Justify each step.
All rights reserved.
b. If x 4 3x, then 4 2x.
2 Justifying Steps in Solving an Equation Suppose that
(15 x)
43
Name_____________________________________ Class____________________________ Date ________________
c. If P Q, Q R, and R S, then P S.
for
Transitive Property of Congruence
the first two parts of the hypothesis:
Simplify.
x 2
The conclusion of the conditional statement shows the result after x is
from each side of the equation in the hypothesis.
subtracted
The property used is the Subtraction Property of Equality.
Use the
Substitution Property of Equality
If P Q and Q R, then
Subtraction Property of Equality
Use the
⬔P 艑 ⬔R
.
Transitive Property of Congruence
for
P R and the third part of the hypothesis:
Quick Check.
1. Fill in each missing reason.
)
M
Given: LM bisects KLN.
(2x40)
4x
)
K
LM bisects KLN
mMLN mKLM Definition of
Angle
N
Bisector
4x 2x 40
Substitution Property of Equality
2x 40
Subtraction Property of Equality
x 20
L
Given
Division Property of Equality
15 x
2. Recall that the length of line AC is 21.
A
4 2x
B
C
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
If P R and R S, then
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Name_____________________________________ Class____________________________ Date________________
Geometry Lesson 2-4
⬔P 艑 ⬔S
.
The property used is the Transitive Property of Congruence.
Quick Check.
3. Name the property of equality or congruence illustrated.
a. XY XY
Reflexive Property of Congruence
b. If mA 45 and 45 mB, then mA mB
Transitive or Substitution Property of Equality
Find the lengths of AB and BC by substituting x 2 into the expressions in
the diagram. Check that AB BC 21.
AB 44
13
BC 8
AB BC 13
8
Geometry Lesson 2-4
All-In-One Answers Version A
21
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 2-4
Geometry
45
11
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 2-5
Proving Angles Congruent
Lesson Objectives
complementary angles
supplementary angles
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
2
Local Standards: ____________________________________
⬔1 and ⬔2 are complementary angles.
Vocabulary and Key Concepts.
Theorem 2-1: Vertical Angles Theorem
3
4
2
Theorem 2-2: Congruent Supplements Theorem
If two angles are supplements of the same angle (or of congruent angles),
congruent
All rights reserved.
.
1 2 and 3 4
then the two angles are
congruent
Theorem 2-4
right
All
A theorem is a conjecture that is proven.
A paragraph proof is a convincing argument that uses deductive reasoning in which
statements and reasons are connected in sentences.
Examples.
adjacent angles
1
4
3
2
1
⬔1 and ⬔2 are vertical angles,
2
4
⬔1 and ⬔2 are adjacent angles,
as are ⬔3 and
⬔4 .
⬔4 .
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
vertical angles
angle.
right
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 2-5
If two angles are congruent and supplementary, then each is a
as are ⬔3 and
Two angles are supplementary angles if
the sum of their measures is 180°.
.
angles are congruent.
3
Two angles are complementary angles if
the sum of their measures is 90°.
.
Theorem 2-3: Congruent Complements Theorem
If two angles are complements of the same angle (or of congruent angles),
then the two angles are
⬔3 and ⬔4 are supplementary angles.
1
All rights reserved.
congruent
Vertical angles are
1 Using the Vertical Angles Theorem Find the value of x.
The angles with labeled measures are vertical angles. Apply the
Vertical Angles Theorem to find x.
4x 101 2x 3
Vertical Angles Theorem
4x 2x 104
2x 104
Subtraction Property of Equality
x 52
Division Property of Equality
2 Proving Theorem 2-2 Write a paragraph proof of Theorem 2-2
using the diagram at the right.
Start with the given: 1 and 2 are supplementary, 3 and 2 are
supplementary angles
supplementary. By the definition of
Adjacent angles are two coplanar angles
m1 m2 180 and m3 m2 180. By substitution,
sides form two pairs of opposite rays.
that have a common side and a common
m1 m2 vertex but no common interior points.
m⬔2
Geometry Lesson 2-5
m⬔3
m⬔3
Name_____________________________________ Class____________________________ Date ________________
2
3
,
, subtract
m⬔2
from
, or ⬔ 1 ⬔ 3 .
Daily Notetaking Guide
Daily Notetaking Guide
1
. Using the
Subtraction Property of Equality
each side. You get m1 (2x 3)
(4x 101)
Addition Property of Equality
Vertical angles are two angles whose
46
4
3
1
All rights reserved.
Prove and apply theorems about
angles
47
Geometry Lesson 2-5
Name_____________________________________ Class____________________________ Date________________
Lesson 3-1
Quick Check.
Properties of Parallel Lines
Lesson Objectives
1 Identify angles formed by two lines
and a transversal
2 Prove and use properties of parallel
lines
1. Refer to the diagram for Example 1.
a. Find the measures of the labeled pair of vertical angles.
107°
b. Find the measures of the other pair of vertical angles.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Local Standards: ____________________________________
Vocabulary and Key Concepts.
c. Check to see that adjacent angles are supplementary.
107° 73° 180°
All rights reserved.
All rights reserved.
73°
Postulate 3-1: Corresponding Angles Postulate
If a transversal intersects two parallel lines, then corresponding
congruent
angles are
.
⬔1
12
Geometry Lesson 2-5
Geometry
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
No, the size of the angles does not affect the proof or the truth of
the theorem.
48
1
2
/
m
⬔1 ⬔2
Theorem 3-1: Alternate Interior Angles Theorem
If a transversal intersects two parallel lines, then alternate interior
congruent
angles are
.
2. Recall the proof of Theorem 2-2. Does the size of the angles in the diagram
affect the proof? Would the proof change if 1 and 3 were acute rather
than obtuse? Explain.
t
⬔3
a
b
t
4
3 2
1
5 6
Theorem 3-2: Same-Side Interior Angles Theorem
If a transversal intersects two parallel lines, then same-side interior
angles are supplementary .
m1 m2 180
Theorem 3-3: Alternate Exterior Angles Theorem
If a transversal intersects two parallel lines, then alternate exterior
angles are congruent .
⬔4
⬔5
Theorem 3-4: Same-Side Exterior Angles Theorem
If a transversal intersects two parallel lines, then same-side exterior
angles are supplementary .
m4 m6 180
Daily Notetaking Guide
Geometry Lesson 3-1
49
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
Name_____________________________________ Class____________________________ Date ________________
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
A transversal is
a line that intersects two coplanar lines at two
m
Alternate interior angles are nonadjacent interior angles that lie
In the diagram at right, ᐉm and pq. Find m1 and m2.
1 and the 42 angle are
corresponding angles
Because ᐉm, m1 by the
on opposite sides of the transversal.
⬔1
the steps
42
3 Using Algebra to Find Angle Measures
65
Alternate Interior Angles
Theorem
c
40
Alternate Interior Angles
Theorem
abc
65
b
b
3
1
The angle vertical to 1 is between the runway segments.
2 is between the runway segments and on the opposite side of the
alternate interior angles
are not
adjacent and lie between the lines on opposite sides of the transversal,
2 and the angle vertical to 1 are
alternate interior angles
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
2
180
/
40
65
Angle Addition
180
40
1 Applying Properties of Parallel Lines In the diagram of
Lafayette Regional Airport, the black segments are runways
and the gray areas are taxiways and terminal buildings.
.
a b c
In the diagram, ᐉm. Find the values of a, b, and c.
a
Compare 2 and the angle vertical to 1. Classify the angles
as alternate interior angles, same-side interior angles, or
corresponding angles.
/
m
m⬔2 180 .
42
The first
Examples.
transversal runway. Because
Angle Addition Postulate
for m1, the equation becomes
shows the reason for each step.
the steps to prove a theorem.
and the second column
All rights reserved.
⬔1 and ⬔7
corresponding angles.
lines.
5
42
from each side to find m2 138 .
42
Subtract
All rights reserved.
angles that lie on the same side of the
transversal and in corresponding positions relative to the coplanar
column shows
4
.
straight angle, m1 m2 180 by the
If you substitute
and ⬔4 are
same-side interior angles.
q
.
8 6
7
2 1
3
Because 1 and 2 are adjacent angles that form a
⬔1 and ⬔2 are
alternate interior angles.
of the transversal.
A two-column proof is a display that shows
42
Corresponding Angles Postulate
Same-side interior angles are interior angles that lie on the same side
Corresponding angles are
p
2 Finding Measures of Angles
t
56
13
42
78
/
distinct points.
Name_____________________________________ Class____________________________ Date ________________
m
Postulate
Substitution Property of Equality
75
Subtraction Property of Equality
Quick Check.
1. Use the diagram in Example 1. Classify 2 and 3 as alternate interior angles,
same-side interior angles, or corresponding angles.
same-side interior angles
2. Using the diagram in Example 2 find the measure of each angle. Justify each answer.
a. 3 42; Vertical angles are congruent
b. 4 138; Corresponding angles are congruent
c. 5 138; Same-side interior angles are supplementary
d. 6 42; Corresponding angles are congruent
e. 7 138; Alternate interior angles are congruent
f. 8 138; Vertical angles are congruent
3. Find the values of x and y. Then find the measures of the four angles
in the trapezoid.
50
Geometry Lesson 3-1
(y 50)°
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Lesson Objectives
1 Use a transversal in proving lines
parallel
Geometry Lesson 3-1
51
Name_____________________________________ Class____________________________ Date ________________
Proving Lines Parallel
A flow proof uses arrows to show the logical connections between the statements.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Reasons are written below the statements.
Local Standards: ____________________________________
Vocabulary and Key Concepts.
Examples.
Theorem 3-5: Converse of the Alternate Interior Angles Theorem
If two lines and a transversal form alternate interior angles that are
congruent, then the two lines are parallel.
2
ᐉ m .
All rights reserved.
m
then the two lines are parallel.
All rights reserved.
If two lines and a transversal form corresponding angles that are congruent,
1 Using Postulate 3-2
1
/
Postulate 3-2: Converse of the Corresponding Angles Postulate
5 6
1 4
2
m
3
If 1 2,
/
supplementary,
supplementary, then the two lines are parallel.
then ᐉ m .
Theorem 3-7: Converse of the Alternate Exterior Angles Theorem
If 3 5,
If two lines and a transversal form alternate exterior angles that are
then ᐉ m .
congruent, then the two lines are parallel.
Theorem 3-8: Converse of the Same-Side Exterior Angles Theorem
If 3 and 6 are
If two lines and a transversal form same-side exterior angles that are
supplementary,
supplementary, then the two lines are parallel.
then ᐉ m .
52
Geometry Lesson 3-2
All-In-One Answers Version A
Daily Notetaking Guide
and ⬔2
D1
are supplementary. Because
4K
2
congruent
supplements of the same angle are
⬔4 . Because
Converse of the Corresponding Angles
)
DK
)
⬔3
by the
Postulate.
2 Using Algebra
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
If 2 and 4 are
If two lines and a transversal form same-side interior angles that are
shows that ⬔4
3
C
E
It is given that 3 and 2 are supplementary. The diagram
and ⬔4 are congruent corresponding angles, EC
then ᐉ m .
Theorem 3-6: Converse of the Same-Side Interior Angles Theorem
Use the diagram at the right. Which lines, if any, must be parallel
if 3 and 2 are supplementary? Justify your answer with
a theorem or postulate.
(Congruent Supplements Theorem), ⬔3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson 3-2
y°
2x°
x 45, y 115; 90, 90, 115, 65
Find the value of x for which ᐉm.
The labeled angles are
alternate interior angles
.
If ᐉm, the alternate interior angles are
,
equal
and their measures are
equation
5x
5x
66 14
congruent
/
(5x 66)
. Write and solve the
m
3x.
66 14
3x
5x
80
3x
2x
80
Subtract
x
40
Divide each side by 2 .
Daily Notetaking Guide
(14 3x)
Add
66
to each side.
3x
from each side.
Geometry Lesson 3-2
Geometry
53
13
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date________________
Quick Check.
Lesson 3-3
1. Use the diagram from Example 1. Which lines, if any, must be parallel if
3 4? Explain.
Parallel and Perpendicular Lines
Lesson Objectives
1
u u
EC 储 DK ; Converse of Corresponding Angles Postulate
NAEP 2005 Strand: Geometry
Relate parallel and perpendicular
lines
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
a
x 18; 7(18) 8 118°, and 62° 118° 180°
Key Concepts.
(7x 8)
Theorem 3-9
62
All rights reserved.
b
All rights reserved.
2. Find the value of x for which a b. Explain how you can check your answer.
parallel to the same line
If two lines are
they are
parallel
a
b
, then
c
to each other.
a b .
perpendicular to the same line
In a plane, if two lines are
then they are
t
m
Theorem 3-10
parallel
,
n
to each other.
Geometry Lesson 3-2
Name_____________________________________ Class____________________________ Date ________________
side are parallel. And if an
outer
inner
inner and outer
Theorem 3-9
outer
outer
opposite side. But that outer edge is parallel to the
inner
All rights reserved.
Examples.
1 Real-World Connection
A picture frame is assembled as shown. Given the book’s
explanation for why the outer edges on opposite sides of
the frame are parallel, why must the inner edges on opposite
sides be parallel, too?
corners cut
to form 45
angles
55
Geometry Lesson 3-3
Parallel Lines and the
Triangle Angle-Sum Theorem
Lesson Objectives
1 Classify triangles and find the
measures of their angles
2 Use exterior angles of triangles
edge on the opposite side, then by
outer
ᐉ
m
Lesson 3-4
edge on one side is parallel to the
each inner edge is parallel to the
n
m
Name_____________________________________ Class____________________________ Date________________
edges of each
edge on the same side, and at the same time the
edge is parallel to the
⊥
Daily Notetaking Guide
Daily Notetaking Guide
The sides are constructed so that the
n
edge on the
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
edge on the
same side, so again by Theorem 3-9 , inner edges on opposite sides are
Vocabulary and Key Concepts.
1
k
Write a paragraph proof.
ᐉ
Given: In a plane, k l and k m. Also, m1 90°.
m
Prove: The transversal is perpendicular to line m.
All rights reserved.
2 Using Theorem 3-11
All rights reserved.
parallel.
Since m⬔1 90° , the transversal is perpendicular to line k .
Theorem 3-12: Triangle Angle-Sum Theorem
The sum of the measures of the angles of a triangle is
180
C
.
mA mB mC 180
B
A
Theorem 3-13: Triangle Exterior Angle Theorem
The measure of each exterior angle of a triangle equals the sum of the
2
1
measures of its two remote interior angles.
Since k 储 m , by Theorem 3-11 the transversal is also
3
perpendicular to line m.
Yes; 30° 60° 90°
60
60
60
60
30
30
2. From what is given in Example 2, can you also conclude that the transversal is
perpendicular to line l?
Yes, by Theorem 3-11 and the fact
that k 储 l
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. Can you assemble the framing at the right into a frame
with opposite sides parallel? Explain.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Quick Check.
acute
obtuse
right
An
A
An
triangle has
triangle has
triangle has
one obtuse angle.
three acute angles.
one right angle.
__________________
__________________
__________________
__________________ __________________ __________________
An equilateral
triangle has
three congruent sides.
________________________
________________________
isosceles
An
triangle has
at least two congruent sides.
__________________________
__________________________
An equiangular
triangle has
three congruent angles.
______________________
______________________
scalene
A
triangle has
no congruent sides.
_____________________
_____________________
An exterior angle of a polygon is an angle formed by a side
1
and an extension of an adjacent side.
Remote interior angles are the two nonadjacent interior
angles corresponding to each exterior angle of a triangle.
56
14
Geometry Lesson 3-3
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Exterior
angle
2
3
Remote
interior
angles
Geometry Lesson 3-4
57
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
54
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
m n .
Theorem 3-11
In a plane, if a line is perpendicular to one of two
parallel lines, then it is also perpendicular to the other.
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
Quick Check.
1 Applying the Triangle Angle-Sum Theorem
In triangle ABC, ACB is a right
angle, and CDAB. Find the values of a and c.
90
70
c
Angle Addition
20
Subtract
70
For 䉭CAB, (70 20) 70 b 180. Then 160 b 180 and b 20.
For 䉭CDB, 70 90 b 180. Then 160 b 180 and b 20.
b
a
A
a right angle
Definition of
B
D
Postulate
from each side.
2. a. Find m3.
a mADC c mADC a
90
a
20
110
a
Triangle Angle-Sum
180
90
Substitute
180
Simplify.
70
90
20
for m⬔ADC and
110
Subtract
Theorem
perpendicular lines
Definition of
180
All rights reserved.
To find a, use ADC.
for c.
All rights reserved.
c
90
70
c
Find c first, using the fact that ACB is a right angle.
mACB 1. Critical Thinking Describe how you could use either CAB or CDB to
find the value of b in Example 1.
C
45
90
45
b. Critical Thinking Is it true that if two acute angles of a triangle are
complementary, then the triangle must be a right triangle? Explain.
3
True, because the measures of the two complementary angles add to
90, leaving 90 for the third angle.
from each side.
2 Applying the Triangle Exterior Angle Theorem Explain what happens
All rights reserved.
x
(180180
)
(180180
x )
x The exterior angle and the angle formed by the back of the chair and the
adjacent angles
armrest are
straight angle
decreases
measure
the armrest
58
, which together form a
increases
. As one measure
increases
, the other
. The angle formed by the back of the chair and
as you make a lounge chair recline more.
Geometry Lesson 3-4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
x
Arm
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Back
to the angle formed by the back of the chair and the armrest as you make a
lounge chair recline more.
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
The Polygon Angle-Sum Theorems
Lesson Objectives
1 Classify polygons
2 Find the sums of the measures of the
interior and exterior angles of
polygons
An equilateral polygon has all sides congruent.
An equiangular polygon has all angles congruent.
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
A regular polygon is both equilateral and equiangular.
Local Standards: ____________________________________
Examples.
1 Classifying Polygons Classify the polygon at the right by its sides. Identify it
Vocabulary and Key Concepts.
as convex or concave.
Starting with any side, count the number of sides clockwise around the figure.
Because the polygon has 12 sides, it is a dodecagon. Think of the
.
Theorem 3-15: Polygon Exterior Angle-Sum Theorem
The sum of the measures of the exterior angles of a polygon,
360
one at each vertex, is
3
All rights reserved.
(n 2)180
The sum of the measures of the angles of an n-gon is
All rights reserved.
Theorem 3-14: Polygon Angle-Sum Theorem
polygon as a star. Draw a diagonal connecting two adjacent points of the star.
That diagonal lies
concave
4
360
1
.
2 Finding a Polygon Angle Sum Find the sum of the measures of the angles
5
of a decagon.
A polygon is a closed plane figure with at least three sides that are segments. The sides
Sum (n 2)(180)
intersect only at their endpoints, and no two adjacent sides are collinear.
B
B
A
A
C
E
C
E
D
not a
D
D
Not a polygon
A polygon
C
closed
;
figure
Not a polygon
A concave polygon has at least one diagonal
with points outside of the polygon.
;
two sides
intersect
between endpoints.
A convex polygon does not have diagonal points outside
of the polygon.
E
A
D
R
Y
T
Diagonal
P
M
K
W
Q
G
S
A convex
polygon
A concave
polygon
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
B
sides, so n 10
A decagon has
A
the polygon, so the dodecagon is
outside
.
2
.
For the pentagon, m1 m2 m3 m4 m5 © Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson 3-5
59
Geometry Lesson 3-4
(
10
10
.
Polygon Angle-Sum
)
2 (180)
Substitute
8 ? 180
Subtract.
Simplify.
1440
10
3 Using the Polygon Angle-Sum Theorem Find mX in quadrilateral
Geometry Lesson 3-5
All-In-One Answers Version A
Daily Notetaking Guide
Y
Z
XYZW.
The figure has 4 sides, so n 4 .
100°
X
mX mY mZ mW mX mY (4 2)(180)
Polygon Angle-Sum
100
360
Substitute.
mX mY 190
360
Simplify.
mX mY 170
Subtract
170
Substitute
170
Simplify.
90
mX m⬔X
2m⬔X
mX 60
Theorem
for n.
Daily Notetaking Guide
85
190
W
Theorem
from each side.
m⬔X
for m⬔Y.
Divide each side by 2 .
Geometry Lesson 3-5
Geometry
61
15
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
4 Applying Theorem 3-15 A regular hexagon is inscribed in a rectangle.
Lesson 3-6
Explain how you know that all the angles labeled 1 have equal measures.
2
2
1
Lines in the Coordinate Plane
1
Lesson Objectives
congruent
The hexagon is regular, so all its angles are
supplement
exterior angle is the
1
2
1
, all the angles marked 1
congruent
Graph lines given their equations
Write equations of lines
Topics: Patterns, Relations, and Functions;
Algebraic Representations
supplements
are adjacent angles that form a straight angle. Because
of congruent angles are
. An
of a polygon’s angle because they
NAEP 2005 Strand: Algebra
Local Standards: ____________________________________
1
2
2
have equal measures.
Vocabulary.
Slope-intercept
form
All rights reserved.
Quick Check.
1. Classify each polygon by its sides. Identify each as convex or concave.
a.
b.
All rights reserved.
The slope-intercept form of a linear equation is
y 2x 3
y mx b.
1
Ax By C.
Standard
The point-slope form for a nonvertical line is
form
y y 1 m(x x 1 ).
hexagon; convex
O
4
2x y 1
4
3
octagon; concave
x
y 2 2(x 1)
Point-slope
form
Examples.
2. a. Find the sum of the measures of the angles of a 13-gon.
1 Graphing Lines Using Intercepts Use the x-intercept and y-intercept to
You can solve the equation (n 2)180 720.
3. Pentagon ABCDE has 5 congruent angles. Find the measure of each angle.
108
4. a. In the figure from Example 4, find m1 by using the Polygon Exterior
Angle-Sum Theorem.
graph 5x 6y 30.
To find the x-intercept, substitute
0 for y and solve for x.
To find the y-intercept, substitute
0 for x and solve for y.
5x 6y 30
5x 6y 30
( ) 30
( ) 6y 30
5x 6 0
5 0
5x 0 30
5x 0 6y 30
6y 30
x 6
30
y 5
The y-intercept is 5 .
The x-intercept is 6 .
A point on the line is
( 6 , 0).
(
A point on the line is 0, 5
Plot (6, 0) and (0, 5). Draw the line containing the two points.
(6, 0)
6
Geometry Lesson 3-5
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
3
y
4
2x
63
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3
Geometry Lesson 3-6
point-slope form of the line that contains the points G(4, 9) and H(1, 1).
6x 3y 12
3
(0, 5)
4 Writing an Equation of a Line Given Two Points Write an equation in
Step 1 Transform the equation to slope-intercept form.
12
8x
Example.
6x 3y 12 to slope-intercept form. Then graph the resulting equation.
6x
6
Name_____________________________________ Class____________________________ Date ________________
2 Transforming to Slope-Intercept Form Transform the equation
3y 4
4
30; no, it is not formed by extending one side of the polygon.
12
2
2
b. Find m2. Is 2 an exterior angle? Explain.
6x
).
y
2
2 O
60
3y All rights reserved.
b. Critical Thinking The sum of the measures of the angles of a given
polygon is 720. How can you use Sum (n 2)180 to find the number
of sides in the polygon?
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1980
62
y
The standard form of a linear equation is
Step 1 Find the slope.
Add
6x
to each side.
m
Divide each side by 3 .
Simplify.
4
y
(1, 6)
(0, 4)
All rights reserved.
6
All rights reserved.
Step 2 Use the y-intercept and the slope to plot two points and draw
the line containing them.
y1
x2
x1
1 m
The y-intercept is 4 and the slope is 2 .
y2
1
(
Use the formula for slope.
9
)
Substitute
4
(4, 9)
for (x1, y1) and
(1, 1)
for (x2, y2).
10
Simplify.
5
2
2
Step 2 Select one of the points. Write the equation in point-slope form.
2 O
2
4x
y
2
y
(
y1
9
(
m x
)
2
y 9 2
3 Using Point-Slope Form Write an equation in point-slope form
x1
(x (x 4
4
)
)
)
Point-slope form.
Substitute
2
for m and
(4, 9)
for (x1, y1).
Simplify.
(
y1
(
m x
)
6
8
y 6 8
x1
(x (x 3
3
)
)
)
Use point-slope form.
Substitute
8
for m and
(3, 6)
for (x1, y1).
Simplify.
Quick Check.
1. Graph each equation.
a. y 12x 2
b. 2x 4y 8
y
O
1
x
16
1
2
4
c. 5x y 3
x
2
O
3
64
O y
y
2
x
2
Geometry Lesson 3-6
Geometry
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y
y
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
of the line with slope 8 that contains P(3, 6).
Quick Check.
2. Write an equation of the line with slope 1 that contains the point P(2, 4).
y 4 1(x 2)
3. Write an equation of the line that contains the points P(5, 0) and Q(7, 3).
3
y 0 3
2(x 5) or y 3 2 (x 7)
Daily Notetaking Guide
Geometry Lesson 3-6
65
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 3-7
2 Writing Equations of Parallel Lines Write an equation in point-slope
Slopes of Parallel and Perpendicular Lines
Lesson Objectives
1
Relate slope and parallel lines
Relate slope and perpendicular lines
2
form for the line parallel to 6x 3y 9 that contains (5, 8).
NAEP 2005 Strand: Measurement
Step 1 To find the slope of the line, rewrite the equation in slope-intercept form.
Topic: Measuring Physical Attributes
6x 3y 9
Local Standards: ____________________________________
3 y 6x 9
y
Key Concepts.
Any two vertical lines are parallel.
Slopes of Perpendicular Lines
is 1.
slopes
Examples.
x 5y 4
x 4 5y
4
x
5
Subtract 4 from each side.
y
5
5 .
Divide each side by
1
y 4
x
5
5
The line x 5y 4 has slope
1
5
.
The line y 5x 4 has slope
5
.
are not
parallel because their slopes are
not equal
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
slope-intercept
form. Write the equation x 5y 4 in slope-intercept form.
The lines
x1 )
(
) 2 ( x ( 5 ))
2 (x 5 )
m x
8
y 8 Substitute 2 for m and
(5, 8)
for (x1, y1).
Simplify.
b. y 12x 5 and 2x 4y 20
No; the lines have the same slope and
y-intercept, so they are the same line.
2. Write an equation for the line parallel to y x 4 that contains (2, 5).
y 5 (x 2)
.
Geometry Lesson 3-7
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Geometry Lesson 3-7
67
Name_____________________________________ Class____________________________ Date________________
Lesson 3-8
Example.
Lesson Objectives
1 Construct parallel lines
2 Construct perpendicular lines
3 Writing Equations for Perpendicular Lines Write an equation for a line
perpendicular to 5x 2y 1 that contains (10, 0).
Step 1 To find the slope of the given line, rewrite the equation in
slope-intercept form.
Constructing Parallel and Perpendicular Lines
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
5x 2y 1
2y 5 x 1
5x
Example.
from each side.
1 Constructing ᐉm Examine the diagram at right. Explain
2
5
2
The line 5x 2y 1 has slope
.
Step 2 Find the slope of a line perpendicular to 5x 2y 1. Let m be the slope
of the perpendicular line.
5
2
m 1
m 1
The product of the slopes of perpendicular lines is
?
( )
2
5
m
y 0 y
2
5
2
5
2
Multiply each side by .
5
Simplify.
5
(
x
10
(x 10
y1
m(x x1
), to write an
2
)
Substitute
)
Simplify.
for m and
5
(10, 0)
how to construct 1 congruent to H. Construct the angle.
N 1
Use the method learned for constructing congruent angles.
m
Step 1 With the compass point on point H, draw an arc that
intersects the sides of ⬔H .
/
H
J
Step 2 With the same compass setting, put the compass point
on point N . Draw an arc.
1 .
2
Step 3 Use point-slope form, y equation for the new line.
All rights reserved.
Divide each side by 2 .
for (x1, y1).
Quick Check.
3. Write an equation for the line perpendicular to 5y x 10 that contains (15, 4).
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2
Subtract
1
x
All rights reserved.
5
y © Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y1
(
y-intercepts are different.
x 5y 4 parallel? Explain.
66
y
y
Yes; Each line has slope 1
2 and the
1
3 .
.
a. y 12x 5 and 2x 4y 9
1 Determining Whether Lines are Parallel Are the lines y 5x 4 and
2
1. Are the lines parallel? Explain.
Any horizontal line and vertical line are perpendicular.
The equation y 5x 4 is in
from each side.
Quick Check.
of two lines have a product of 1, the lines are perpendicular.
slopes
If the
All rights reserved.
of two distinct nonvertical lines are equal, the lines are parallel.
If two nonvertical lines are perpendicular, the product of their
6x
Subtract
Divide each side by
Step 2 Use point-slope form to write an equation for the new line.
are equal.
slopes
All rights reserved.
If two nonvertical lines are parallel, their
slopes
3
The line 6x 3y 9 has slope
Slopes of Parallel Lines
If the
2x
Step 3 Put the compass point below point N where the arc
* )
intersects HN . Open the compass to the length where
the arc intersects line ᐉ . Keeping the same compass
setting, put the compass point above point N where the
* )
arc intersects HN . Draw an arc to locate a point.
Step 4 Use a straightedge to draw line m through the point
you located and point N.
Quick Check.
1. Use Example 1. Explain why lines ᐉ and m must be parallel.
If corresponding angles are congruent, the lines are parallel by the
Converse of Corresponding Angles Postulate.
y 4 5(x 15)
68
Geometry Lesson 3-7
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 3-8
Geometry
69
17
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
Example.
Example.
2 Constructing a Special Quadrilateral Construct a quadrilateral with both
the
3 Perpendicular From a Point to a Line * Examine
)
R
construction. At what special point does RG meet line ᐉ?
pairs of sides parallel.
Step 1 Draw point A and two rays with endpoints at A. Label point B on
one ray and point C on the other ray.
/
Point R is the same distance from point E as it is from point F
)
one
because the arc was made with
Step 2 Construct a ray parallel to AC through point B.
E
F
compass opening.
G
Point G is the same distance from point E as it is from point F
the same
because both arcs were made with
B
*
)
compass opening.
C
)
Step 3 Construct a ray parallel to AB through point C.
)
All rights reserved.
A
All rights reserved.
This means that RG intersects line ᐉ at the
*
)
midpoint
of
perpendicular bisector
that RG is the
Quick Check.
* )
of
* )
, and
EF
EF
.
* ) * )
3. a. Use a straightedge to draw EF . Construct FG so that FG EF at
point F.
Step 4 Label point D where the ray parallel to AC intersects the ray
)
parallel to AB . Quadrilateral ABDC has both pairs of opposite
sides parallel.
D
B
Quick Check.
2. Draw two segments. Label their lengths c and d. Construct a quadrilateral
with one pair of parallel sides of lengths c and 2d.
d
c
d
d
F
E
* )
* )
* )
All rights reserved.
C
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
G
A
* ) * )
b. Draw a line CX and a point Z not on CX . Construct ZB so that ZB CX .
Z
C
c
X
B
Geometry Lesson 3-8
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 3-8
71
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 4-1
3 Finding Congruent Triangles Can you conclude that ABC CDE?
Congruent Figures
E
List corresponding vertices in the same order.
Lesson Objective
1 Recognize congruent figures and their
corresponding parts
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
If ABC CDE, then BAC The diagram above shows BAC The statement ABC CDE
Local Standards: ____________________________________
Notice that BC Vocabulary and Key Concepts.
Also, CBA two angles of another triangle, then the third angles
B
C
E
F
C F
Congruent polygons are polygons that have corresponding
D
sides congruent and corresponding angles congruent.
1 Naming Congruent Parts ABC QTJ. List the
Q
T
B
congruent corresponding parts.
List the corresponding sides and angles in the same order.
Angles: A Q
B T
Sides: AB > QT
BC
TJ
C J
AC QJ
C
A
J
2 Using Congruency XYZ KLM, mY 67, and mM 48. Find mX.
Use the Triangle Angle-Sum Theorem and the definition of congruent
polygons to find mX.
mX mY mZ 180
Triangle Angle-Sum Theorem
mZ m⬔M
Corresponding angles of congruent triangles are congruent.
mZ 48
Substitute 48 for m⬔M.
48
115
180
Substitute.
180
Simplify.
mX 65
Geometry Lesson 4-1
Geometry
4
D
A
EC
.
.
ACB
.
ABC
EDC
.
Quick Check.
WY O MK
lWSY O lMVK
WS O MV
lSWY O lVMK
YS O KV
lWYS O lMKV
2. It is given that WYS MKV. If mY 35, what is mK? Explain.
GHI
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
DEF
Examples.
18
DEC
are congruent. The correct way to state this is
Angles:
G
F
E
72
DE , and AC and BAC 3
true.
C
Since all of the corresponding sides and angles are congruent, the triangles
Sides:
I
mX , not DCE.
is not
3
4
1. WYS MKV. List the congruent corresponding parts. Use three letters
for each angle.
H
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
are congruent.
All rights reserved.
All rights reserved.
D
A
If two angles of one triangle are congruent to
67
.
DEC
Using Theorem 4-1, you can conclude that ECD Theorem 4–1
mX DC , BA CDE
DCE
B
mK 35
Corresponding angles of congruent triangles are congruent and
have the same measure.
3. Can you conclude that JKL MNL? Justify your answer.
N
J
L
K
M
No. The corresponding sides are not necessarily equal.
Subtract 115 from each side.
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 4-1
73
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
70
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
G
Triangle Congruence
by SSS and SAS
Lesson 4-2
Example.
4 Proving Triangles Congruent
Lesson Objective
Given: CG DG, CN DN, C D, GN # CD
Prove: CNG DNG
1
NAEP 2005 Strand: Geometry
Prove two triangles congruent using
the SSS and SAS Postulates
Topic: Transformation of Shapes and Preservation of
Properties
Local Standards: ____________________________________
C
Key Concepts.
1. Given
2.
Given
3. GN > GN
3.
Reflexive Property of Congruence
4. C D
4.
Given
5. CNG DNG
6.
5.
⬔CGN 艑 ⬔DGN
Right angles are congruent
All rights reserved.
Reasons
1. CG > DG
2. CN > DN
All rights reserved.
Statements
D
N
7. Definition of
Prove: AEB DCB
All rights reserved.
C
B
A
D
E
Statements
Reasons
1.
⬔A 艑 ⬔D; ⬔E 艑 ⬔C
1.
Given
2.
⬔ABE 艑 ⬔DBC
2.
Vertical angles are congruent.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4. Given: A D, E C, AE DC, EB CB, BA BD
F
PQR
BCA
R
F
D
B
E
3.
AE O CD; AB O BD; EB O BC
3.
Given
4.
䉭AEB 艑 䉭DCB
4.
Definition of Congruent Triangles
Geometry Lesson 4-1
C
FDE
A
Examples.
A
1 Proving Triangles Congruent
Given: M is the midpoint of XY , AX > AY
Prove: AMX AMY
Write a paragraph proof.
You are given that M is the midpoint of XY , and AX AY .
Midpoint M implies that MX MY . AM > AM by the
X
SSS Postulate
.
Daily Notetaking Guide
Daily Notetaking Guide
Y
M
, so
Reflexive Property of Congruence
AMX AMY by the
Geometry Lesson 4-2
75
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
2 Using SAS AD > BC . What other information do you
Lesson 4-3
B
A
need to prove ADC BCD by SAS?
Reflexive Property of Congruence
.
⬔ADC ⬵ ⬔BCD
SAS
Local Standards: ____________________________________
.
Key Concepts.
R
3 Are the Triangles Congruent? From the information given,
Postulate 4-3: Angle-Side-Angle (ASA) Postulate
can you prove RSG RSH? Explain.
Given: RSG RSH, SG > SH
It is given that RSG RSH and SG > SH.
RS RS by the
.
Reflexive Property of Congruence
H
G
Two pairs of corresponding sides and their included angles are
congruent, so RSG RSH by the
SAS
All rights reserved.
ADC BCD by
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation
of Properties
, you can prove
All rights reserved.
you know
C
D
You now have two pairs of corresponding congruent sides. Therefore, if
Triangle Congruence
by ASA and AAS
Lesson Objective
1 Prove two triangles congruent using
the ASA Postulate and the AAS
Theorem
It is given that AD > BC. Also, DC > CD by the
If two angles and the included side of one triangle are congruent
P
to two angles and the included side of another triangle, then
the two triangles are congruent
S
HGB
B
K
.
G
NKP
N
H
Postulate.
F
Theorem 4-2: Angle-Angle-Side (AAS) Theorem
J
Quick Check.
C
If two angles and a nonincluded side of one triangle are congruent
1. Given: HF HJ, FG JK,
H is the midpoint of GK.
to two angles and the corresponding nonincluded side of another
Prove: FGH JKH
Statements
H
K
Reasons
1. HF > HJ, FG > JK
1. Given
2. H is the midpoint of GK.
2. Given
3. GH > HK
3.
Definition of Midpoint
4. 䉭FGH ⬵ 䉭JKH
4.
SSS Postulate
2. What other information do you need to prove ABC CDA by SAS?
⬔DCA 艑 ⬔BAC
A
8
3. From the information given, can you prove AEB DBC? Explain.
Given: EB > CB , AE > DB
No; you don’t know that ⬔AEB ⬵ ⬔DBC or that AB O DC .
B
6
D
7
7
C
A
E
B
C
D
the triangles are congruent
triangle, then
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
G
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
GHF
are congruent.
Quick Check.
74
Q
P
Postulate 4-2: Side-Angle-Side (SAS) Postulate
If two sides and the included angle of one triangle
are congruent to two sides and the included angle
of another triangle, then the two triangles
triangles
congruent
H
congruent.
6. If two angles of one triangle are congruent to two
angles of another triangle, then the third angles are
(Theorem 4-1 ).
congruent
7. CNG DNG
Postulate 4-1: Side-Side-Side (SSS) Postulate
If the three sides of one triangle are
congruent to the three sides of another
G
triangle, then the two triangles are
CDM
M
.
G
Example.
C
and I is not congruent to C. Name the triangles
that are congruent by the ASA Postulate.
The diagram shows N A D and
FN > CA > GD.
X
D
O
1 Using ASA Suppose that F is congruent to C
F
G
T
N
A
I
If F C, then F C G .
Therefore, FNI CAT GDO by
T
D
XGT
ASA
.
Quick Check.
1. Using only the information in the diagram, can you conclude that INF is
congruent to either of the other two triangles? Explain.
No; Only one angle and one side are shown to be congruent. At least
one more congruent side or angle is necessary to prove congruence
with SAS, ASA, or AAS.
76
Geometry Lesson 4-2
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 4-3
Geometry
77
19
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
Quick Check.
2. Write a proof.
A
Given: /A > /B, AP > BP
Prove: APX BPY
P
It is given that A B and AP > BP. APX BPY by
B
A
Given: B D, AB6CD
Prove: ABC CDA
Because AB6CD, BAC DCA
Angles Theorem. Then ABC congruent. 䉭 NML 艑 䉭 NPO by ASA.
B
.
3 Planning a Proof Write a plan for a proof that uses AAS.
3. Write a flow proof.
P
CDA
AC , so ABC CDA by
C
D
SRP ⬔QRP
R
Given: S Q, RP bisects SRQ.
Prove: SRP QRP
.
4 Writing a Proof Write a two-column proof that uses AAS.
Definition of angle bisector
S Q
B
A
RP Reasons
1. B D, AB6CD
1. Given
2. lBAC O lDCA
2. If lines are parallel, then
alternate interior
2. angles are congruent.
3. AC > AC
3. Reflexive Property of Congruence
4. ABC CDA
4. AAS Theorem
Geometry Lesson 4-3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
D
Statements
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Given
Given: B D, AB6CD
Prove: ABC CDA
78
Q
S
if a pair of
AAS
RP bisects SRQ
Given
by the Alternate Interior
corresponding sides are congruent. By the Reflexive Property,
AC O
⬔MNL 艑 ⬔PNO because vertical angles are
All rights reserved.
ASA
M
It is given that NM O NP and ⬔M 艑 ⬔P.
X
two pairs of corresponding angles and their included sides are
congruent, APX BPY by
P
N
RP
Reflexive Property of Congruence
kSRP O kQRP
All rights reserved.
Theorem. Because
Vertical Angles
All rights reserved.
the
L
Given: NM > NP, M P
Prove: NML NPO
Y
AAS
4. Recall Example 4. Explain how you could prove ABC CDA
using ASA.
It is given that lB O lD. You know that lBAC O lDCA by the
Alternate Interior Angles Theorem. By Theorem 4-1, lBCA O lDAC .
By the Reflexive Property of Congruence, AC O AC . Therefore,
kABC O kCDA by ASA.
Daily Notetaking Guide
Daily Notetaking Guide
79
Geometry Lesson 4-3
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 4-4
2 Using Right Triangles According to legend, one of Napoleon’s followers used
Using Congruent Triangles: CPCTC
Lesson Objective
1 Use triangle congruence and CPCTC
to prove that parts of two triangles are
congruent
congruent triangles to estimate the width of a river. On the riverbank, the officer
stood up straight and lowered the visor of his cap until the farthest thing he could
see was the edge of the opposite bank. He then turned and noted the spot on his
side of the river that was in line with his eye and the tip of his visor.
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
Local Standards: ____________________________________
D
P arts
are C ongruent
All rights reserved.
of C ongruent
C orresponding
.
Examples.
All rights reserved.
CPCTC stands for:
T riangles
What congruence statements besides 3 4 can you
prove from the diagram, in which SL > SR and
1 2 are given?
and LSC RSC by
SAS
. 3 4 because
corresponding parts of congruent triangles are congruent.
When two triangles are congruent, you can form congruence
statements about three pairs of corresponding angles and
three pairs of corresponding sides. List the congruence
statements.
12
Stretcher
S
Shaft
Sides:
SL > SR
Given
SC SC
Reflexive Property of Congruence
CL CR
Other congruence statement
Angles:
1 2
Given
3 4
Corresponding Parts of Congruent Triangles
lCLS O lCRS
Other congruence statement
right angles
.
DEG and DEF are the angles that the officer makes with the ground.
R
6
L
5
Given: DEG and DEF are right angles; EDG EDF.
The given states that DEG and DEF are
What conditions must hold for that to be true?
Rib
So the officer must stand
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
by the Reflexive Property of Congruence,
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
SC
C
34
E
The officer then paced off the distance to this spot and declared that
distance to be the width of the river!
1 Congruence Statements The diagram shows the frame of an umbrella.
SC G
F
Vocabulary.
ground must be
perpendicular
level or flat
to the ground, and the
.
Quick Check.
1. Given: Q R, QPS RSP
Prove: SQ > PR
P
It is given that ⬔Q 艑 ⬔R, ⬔QPS 艑 ⬔RSP. PS O SP by the Reflexive
Property of 艑. kQPS O kRSP by ASA. SQ O PR by CPCTC.
R
Q
S
2. Recall Example 2. About how wide was the river if the officer paced off
20 paces and each pace was about 212 feet long?
The congruence statements that remain to be proved are
50 feet
lCLS O lCRS and CL O CR.
80
20
Geometry Lesson 4-4
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 4-4
81
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2 Writing a Proof Using ASA Write a paragraph proof.
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 4-5
Name_____________________________________ Class____________________________ Date ________________
The legs of an isosceles triangle
Isosceles and Equilateral Triangles
The vertex angle of an isosceles
triangle is formed by the two
are the congruent sides.
Lesson Objective
1
NAEP 2005 Strand: Geometry
Use and apply properties of isosceles
triangles
Vertex
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
Vocabulary and Key Concepts.
The base of an isosceles triangle
Legs
is the third, or non-congruent,
Base
side.
Theorem 4-3: Isosceles Triangle Theorem
B
A
Theorem 4-4: Converse of Isosceles Triangle Theorem
C
All rights reserved.
A B
All rights reserved.
the angles opposite those sides are congruent.
Example.
1 Using the Isosceles Triangle Theorems Explain why ABC is
isosceles.
X
ABC and XAB are
*AB )
ABC ←
→
. Because XA 6 BC ,
.
B
the Transitive Property of Congruence
Theorem 4-5
You can use
C
A
bisects AB.
CD
D
B
Corollary to Theorem 4-3
Y
If a triangle is equilateral, then
the triangle is equiangular.
X Y Z
Z
X
Corollary to Theorem 4-4
Y
If a triangle is equiangular, then
the triangle is equilateral.
XY
YZ
Z
X
ZX
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
and
AB
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
CD '
A corollary is a statement that follows immediately from a theorem.
82
Geometry Lesson 4-5
Name_____________________________________ Class____________________________ Date________________
2y 90
Therefore, x 90
Triangle-Sum Theorem
180
Substitute.
180
Simplify.
90
Subtract
y
45
Divide each side by 2 .
and y 45
90
a regular hexagon. Suppose that a segment is drawn between the endpoints
of the angle marked x. Find the angle measures of the triangle that is formed.
The measure of x is 120.
Because the garden is a regular hexagon, the sides have
x
of the angle measures of a triangle is
xyy
180
2y 180
2y 60
y
30
So the angle measures of the triangle are
flagstone
walk
120
,
30
, and
.
30
Quick Check.
2. Suppose mL 43. Find the values of x and y.
M
x 90, y 47
3. Use the diagram from Example 3. The path around the
garden is made up of rectangles and equilateral triangles.
What is the measure of z, the angle at the outside corner
of the path?
y
N
63
x
O
L
x 90
Definition of perpendicular
mN mL
Isosceles Triangle Theorem
y
x
O
L
Geometry Lesson 4-5
N
83
Congruence in Right Triangles
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
Geometry Lesson 4-5
All-In-One Answers Version A
Theorem 4-6: Hypotenuse-Leg (HL) Theorem
If the hypotenuse and a leg of one right triangle are congruent to the
hypotenuse and leg of another right triangle, then
the triangles are congruent.
Hypotenuse
Legs
Examples.
by the HL Theorem” for the diagram. Is the student correct? Explain.
The diagram shows the following congruent parts.
CA >
MA
CPA lMPA
PA PA
C
hypotenuse
Since AC is the
of right triangle CPA, and MA is the
leg
and PA is a
hypotenuse
P
M
leg
and PA is a
of right triangle MPA, the triangles are congruent by
HL Theorem
The student
Daily Notetaking Guide
A
1 Proving Triangles Congruent One student wrote “CPA MPA
the
150
84
The bisector of the vertex
angle of an isosceles triangle is
the perpendicular bisector of
the base.
right angle.
Label each unknown angle y:
120
MO ' LN
railroad
tie steps
z
.
180
M
Examples.
2 Using Algebra Suppose that mL y. Find the values of x and y.
The legs of a right triangle are the two shortest sides, or the sides that are not opposite the
y
.
The measure of the angle marked x is 120, and the sum
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
congruent
S
The hypotenuse of a right triangle is its longest side, or the side opposite the right angle.
raised
bed
garden
.
By the Isosceles Triangle Theorem, the unknown angles
are
V
Vocabulary and Key Concepts.
.
isosceles
W
R
from each side.
3 Using Isosceles Triangles The garden shown at the right is in the shape of
equal length, so the triangle is
No; neither lRUV nor lRVU can be
shown to be congruent to lR.
Local Standards: ____________________________________
2y All rights reserved.
All rights reserved.
90
1. Can you deduce that RUV is isosceles? Explain.
Lesson Objective
1 Prove triangles congruent using the
HL Theorem
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
yy
T
U
Quick Check.
Lesson 4-6
Transitive Property of Equality
mN mNMO mMON 180
to
AC .
By the definition of an isosceles triangle, ABC is isosceles.
Name_____________________________________ Class____________________________ Date ________________
Given
mN y
C
.
ACB
Daily Notetaking Guide
Daily Notetaking Guide
mL y
, ABC the Converse of the Isosceles Triangle Theorem
conclude that AB >
The bisector of the vertex angle of an isosceles triangle is the
of the base.
←
→
The diagram shows that XAB ACB. By
B
A
BC
XAB
A
angles formed
alternate interior
by XA , BC , and the transversal
the sides opposite those sides are congruent.
perpendicular bisector
and the legs.
Base Angles
←
→ ←
→
If two angles of a triangle are congruent, then
The base angles of an isosceles
triangle are formed by the base
C
If two sides of a triangle are congruent, then
AC
congruent sides (legs).
Angle
is
Daily Notetaking Guide
.
correct.
Geometry Lesson 4-6
Geometry
85
21
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
2 Flow Proof—Using the HL Theorem XYZ is isosceles. From vertex X,
Quick Check.
a perpendicular is drawn to YZ, intersecting YZ at point M. Explain why
XMY XMZ.
1.
R
L
3
S
O
3
X
3
M
5
5
5
Q
P
Z
XMY and
XMZ are
right angles
right triangles
Definition of
perpendicular lines
Definition of right triangle
2. Write a paragraph proof of Example 2.
XMY XMZ
XY XZ
Definition of
HL Theorem
3 Two-Column Proof—Using the HL Theorem
C
D
Given: ABC and DCB are right angles, AC > DB
Prove: ABC DCB
Statements
E
Reasons
1. ABC and DCB are
B
A
1. Given
right angles
2. ABC and DCB
2.
Definition of Right Triangle
3.
Given
are right triangles
3. AC >
DB
BC O CB
4.
5. ABC 86
4. Reflexive Property of Congruence
DCB
5.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Reflexive Property of Congruence
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
XM XM
isosceles
䉭LMN 艑 䉭OQP
It is given that XM is perpendicular to YZ . This means that lYMX and
⬔ZMX are right angles, since that is the definition of perpendicular lines.
It follows, then, that 䉭YMX and 䉭ZMX are right triangles, since they
each contain a right angle. It is given that 䉭XYZ is isosceles, so XY O XZ
by the definition of isosceles. XM O XM by the Reflexive Property of
Congruence. Therefore, since the hypotenuses are congruent and one leg
is congruent, 䉭XMY 艑 䉭XMZ by the HL Theorem.
3. You know that two legs of one right triangle are congruent to two legs of
another right triangle. Explain how to prove the triangles are congruent.
Since all right angles are congruent, the triangles are congruent by SAS.
If the hypotenuse and a leg of one right triangle are
congruent to the hypotenuse and leg of another right
triangle, then the triangles are congruent (HL Theorem).
Geometry Lesson 4-6
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Lesson Objectives
1 Identify congruent overlapping
triangles
2 Prove two triangles congruent by first
proving two other triangles congruent
Examples.
3 Using Two Pairs of Triangles Write a paragraph proof.
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Plan: XPW YPZ by AAS if WXZ SAS
congruent by
Examples.
and
HG
F
FG
E
D
, respectively.
2 Using Common Parts Write a plan for a proof that
Y
Z
does not use overlapping triangles.
All rights reserved.
These parts are
W
are congruent. By the
Angle Addition Postulate, ZWM YXM. So ZWM YXM by
ASA
.
.
Quick Check.
1. The diagram shows triangles from the scaffolding that workers used
when they repaired and cleaned the Statue of Liberty.
a. Name the common side in ACD and BCD.
CD
A
B
F
C
D
E
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
vertical angles
.
CPCTC
SAS
congruence holds by
.
Converse of the Isosceles Triangle Theorem
CPCTC
AAS
YPZ
are congruent. Therefore,
vertical angles
XPW YPZ by
Plan: CBE CDA by
Look at MWX. MW > MX by the
ZMW YMX because
because
by CPCTC, and XPW ZYW
C
Write a two-column proof to show that CBE CDA.
X
You can prove these triangles congruent using ASA as follows:
So ZW > YX by
. Therefore, XWZ YZW by
Reflexive Property of Congruence
SAS. WXZ Z
W
Given: CA > CE, BA > DE
Label point M where ZX intersects WY.
䉭ZWM 艑 䉭YXM
P
.
4 Separating Overlapping Triangles
M
Given: ZXW YWX, ZWX YXW
Prove: ZW > YX
ZW > YX by CPCTC if
. These angles
, XWZ YZW. WZ > ZW by the
right angles
All rights reserved.
EG
lZYW
Proof: You are given XW > YZ. Because XWZ and YZW are
G
H
are shared by DFG and EHG.
Parts of sides DG and
Y
if XWZ YZW. These triangles are
CPCTC
are congruent by
DFG and EHG share.
Identify the overlapping triangles.
X
Given: XW > YZ, XWZ and YZW are right angles.
Prove: XPW YPZ
Local Standards: ____________________________________
1 Identifying Common Parts Name the parts of the sides that
87
Name_____________________________________ Class____________________________ Date ________________
Using Corresponding Parts
of Congruent Triangles
Lesson 4-7
Geometry Lesson 4-6
B
if CBE CDA. This
if CB > CD.
X
A
D
E
Proof:
Statement
Reason
1. BCE DCA
1.
Reflexive Property of Congruence
2. CA CE , BA DE
2. Given
3. CA CE , BA DE
3. Congruent sides have equal measure.
4. CA BA CE DE
4.
Subtraction Property of Equality
5. CA BA CB
5.
Segment Addition Postulate
CE DE CD
6. CB CD
6. Substitution
7. CB > CD
7.
Definition of Congruence
8. CBE CDA
8.
SAS
9. CBE CDA
9.
CPCTC
b. Name another pair of triangles that share a common side. Name the
common side.
Answers may vary. Sample: 䉭ABD and 䉭CBD; BD
88
22
Geometry Lesson 4-7
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 4-7
89
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Given
T
Which two triangles are congruent by the HL Theorem? Write a correct
congruence statement.
All rights reserved.
XMY and
XMZ are
XM ⊥ YZ
All rights reserved.
M
All rights reserved.
N
Y
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date________________
Quick Check.
Lesson 5-1
A
2. Write a paragraph proof.
Given: ACD BDC
Prove: CE > DE
B
Midsegments of Triangles
Lesson Objective
E
1
C
D
NAEP 2005 Strand: Geometry
Use properties of midsegments to
solve problems
Topics: Relationships Among Geometric Figures
Local Standards: ____________________________________
It is given that 䉭ACD 艑 䉭BDC, so lADC O lBCD by CPCTC. This
means that 䉭CED is isosceles by the Converse of Isosceles Triangle
Theorem. Then CE O DE by the definition of isosceles triangle.
Vocabulary and Key Concepts.
Q
3. Write a two-column proof.
Given: PS > RS, PSQ RSQ
Prove: QPT QRT
P
R
T
All rights reserved.
All rights reserved.
Theorem 5-1: Triangle Midsegment Theorem
S
Given
2.
QS O QS
2.
Reflexive Property of Congruence
3.
䉭PSQ 艑 䉭RSQ
3.
SAS
4.
PQ O RQ
4.
CPCTC
5.
lPQT O lRQT
5.
CPCTC
6.
QT O QT
6.
Reflexive Property of Congruence
7.
䉭QPT 艑 䉭QRT
7.
SAS
Examples.
A
D
F
E
Reason
Given
2.
AD O AD
2.
Reflexive Property of Congruence
3.
䉭ACD 艑 䉭AED
3.
AAS
4.
CD O ED
4.
CPCTC
5.
lBDC O lFDE
5.
Vertical Angles are Congruent
6.
䉭BDC 艑 䉭FDE
6.
ASA
7.
BD O FD
7.
CPCTC
Geometry Lesson 4-7
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
B
1.
90
2 Identifying Parallel Segments Find mAMN and mANM. MN and BC
Triangle Midsegment
Theorem, so
AMN B by the
Corresponding Angles
Postulate.
75
m⬔AMN
22
P
Definition of perimeter
24
NP 22
NP 46
24
Substitute
60
60
Simplify.
NP 14
Subtract
22
24
for MN and
Y
for MP.
46
N
Z
from each side.
Use the Triangle Midsegment Theorem to find YZ.
1
MP YZ Triangle Midsegment Theorem
2
1
22
44
YZ
YZ
2
Substitute
22
Multiply each side by
for MP.
2
.
Geometry Lesson 5-1
91
A
E
M
N
B
D
by the
Theorem. mANM Isosceles Triangle
60
1. AB 10 and CD 18. Find EB, BC, and AC.
because congruent angles have the same measure.
In AMN, AM AN , so mANM M
NP MN MP Quick Check.
A
corresponding
angles. MN6BC by the
Because the perimeter of MNP is 60, you can find NP.
Name_____________________________________ Class____________________________ Date ________________
75
C
75
C
EB 9; BC 10; AC 20
B
to swim the length of the lake, as shown in the photo. Explain
how Dean could use the Triangle Midsegment Theorem to
measure the length of the lake.
To find the length of the lake, Dean starts at the middle
of the left edge of the lake and paces straight along that end of the
lake. He counts the number of strides (35). Where the lake ends,
he sets a stake. He paces the same number of strides (35) in the
same direction and sets another stake. The first stake marks the
midpoint
of one side of a triangle. Dean paces from the
second stake straight to the middle of the other end of the lake.
He counts the number of his strides (236).
65
Y
118
strides.
236
35
35
From this midpoint of the second side of the triangle,
?
he returns to the midpoint of the first side, counting the
number of strides (128). Dean has paced a triangle. He
has also formed a
midsegment
whose third side is the
By the
length
of a triangle
of the lake.
Triangle Midsegment
Theorem, the segment connecting the two midpoints
is
half
the distance across the lake. So, Dean
multiplies the length of the midsegment by 2 to find
U
Z
V
65; UV6XY so ⬔VUZ and ⬔YXZ are corresponding and congruent.
118
128
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the number of strides back
half
toward the second stake. He paces
2. Critical Thinking Find mVUZ. Justify your answer.
X
Dean finds the midsegment of the second side by pacing exactly
All rights reserved.
All rights reserved.
by substitution.
3 Applying the Triangle Midsegment Theorem Dean plans
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
are cut by transversal AB, so AMN and B are
The perimeter of MNP is 60. Find NP and YZ.
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
mAMN its length.
A midsegment of a triangle is a segment connecting the midpoints of two sides.
1 Finding Lengths In XYZ, M, N, and P are midpoints.
lCAD O lEAD, lC O lE
1.
half
to the third side, and is
to prove a theorem.
1.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
Reason
PS O RS, lPSQ O lRSQ
Statement
parallel
is
A coordinate proof is a form of proof in which coordinate geometry and algebra are used
Statement
1.
4. Write a two-column proof.
Given: CAD EAD, C E
Prove: BD > FD
If a segment joins the midpoints of two sides of a triangle, then the segment
3. CD is a new bridge being built over a lake as shown. Find the length of
the bridge.
963 ft
2640 ft
C
Bridge
963 ft
D
1320 ft
?
the length of the lake.
92
Geometry Lesson 5-1
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 5-1
Geometry
93
23
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 5-2
Examples.
Bisectors in Triangles
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
1 Applying the Perpendicular Bisector Theorem Use the map of
NAEP 2005 Strand: Geometry
Use properties of perpendicular
bisectors and angle bisectors
Washington, D.C. Describe the set of points that are equidistant from the
Lincoln Memorial and the Capitol.
Topics: Relationships Among Geometric Figures
Local Standards: ____________________________________
Vocabulary and Key Concepts.
All rights reserved.
If a point is on the perpendicular bisector of a segment, then it is
equidistant from the endpoints of the segment.
Theorem 5-3: Converse of the Perpendicular Bisector Theorem
If a point is equidistant from the endpoints of a segment, then it is
All rights reserved.
Theorem 5-2: Perpendicular Bisector Theorem
on the perpendicular bisector of the segment.
Converse of the Perpendicular Bisector
The
Theorem
states: If a point is equidistant from the endpoints of a segment, then it is on
the perpendicular bisector of the segment.
If a point is on the bisector of an angle, then it is
segment from the point to the line.
* )
* )
D is 3 in. from AB and AC
0
1
2
3
4
5
6
94
.
C
B
Geometry Lesson 5-2
Name_____________________________________ Class____________________________ Date ________________
*
2 Using the Angle Bisector Theorem
C
Find x, FB, and FD in the diagram at right.
FD
FB
Theorem
7x 37 2x 5
7x 2x 5x
B
Angle Bisector
42
Add
42
37
Subtract
x 8.4
Substitute.
FD 7( 8.4 ) 37 21.8
Substitute.
2x
1. Use the information given in the diagram. CD is the perpendicular
C
bisector of AB. Find CA and DB. Explain your reasoning.
* )
CA 5; DB 6; CD is the perpendicular bisector of AB; therefore
CA CB and DA DB.
5
.
Geometry Lesson 5-2
6
)
95
Concurrent Lines,
Medians, and Altitudes
Lesson Objective
1 Identify properties of
perpendicular bisectors and
angle bisectors
2 Identify properties of medians
and altitudes of a triangle
B
D
NAEP 2005 Strand: Geometry
Topics: Relationships Among Geometric Figures
Local Standards: ____________________________________
Vocabulary and Key Concepts.
2. a. According to the diagram, how far is K from EH ? From ED ?
2x
K
C
10
E
(x 20)
All rights reserved.
D
10; 10
All rights reserved.
)
E
Name_____________________________________ Class____________________________ Date________________
5
A
7x 37
from each side.
Lesson 5-3
)
F
to each side.
Divide each side by
FB 2( 8.4 ) 5 21.8
D
2x 5
A
Substitute.
Daily Notetaking Guide
Daily Notetaking Guide
Quick Check.
of the segment whose endpoints are the Lincoln Memorial and the Capitol.
H
Theorem 5-6
The perpendicular bisectors of the sides of a triangle are
concurrent at a point equidistant from the vertices.
Theorem 5-7
The bisectors of the angles of a triangle are
)
b. What can you conclude about EK ?
concurrent at a point equidistant from the sides.
)
EK is the angle bisector of lDEH.
Theorem 5-8
The medians of a triangle are concurrent at a point that is two-thirds the distance from
c. Find the value of x.
20
d. Find mDEH.
80
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
each vertex to the midpoint of the opposite side.
DC 2
DJ
3
EC 2
EG
3
FC D
2
H
FH
3
G
Theorem 5-9
The lines that contain the altitudes of a triangle are concurrent.
C
E
J
F
Concurrent lines are three or more lines that meet in one point.
The point of concurrency is the point at which concurrent lines intersect.
A circle is circumscribed about a polygon when the vertices
Circumscribed
of the polygon are on the circle.
S
The circumcenter of a triangle is the point of concurrency of
circle
QC SC RC
the perpendicular bisectors of a triangle.
Median
C
A median of a triangle is a segment that
has as its endpoints a vertex of the
Q
R
triangle and the midpoint of the opposite
side.
96
24
Geometry Lesson 5-2
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
circumcenter
Geometry Lesson 5-3
97
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The distance from a point to a line is the length of the perpendicular
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
angle, then it is on the angle bisector.
D
from the Lincoln Memorial and
perpendicular bisector
the Capitol must be on the
If a point in the interior of an angle is equidistant from the sides of the
A
equidistant
A point that is
equidistant from the sides of the angle.
Theorem 5-5: Converse of the Angle Bisector Theorem
All rights reserved.
Theorem 5-4: Angle Bisector Theorem
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
101
of a triangle are
bisectors of the angles
incenter
concurrent at a point
Park
rd
V
va
vertex to the line containing the opposite side.
Z
le
ou
Theorem 5-7 states that the
U
B
The roads form a triangle around the park.
I
An altitude of a triangle is the perpendicular segment from a
a
os
ip
ar
M
X
bisectors of the triangle.
fountain equidistant from three straight roads that enclose
a park. Explain how they can find the location.
XI YI ZI
Y
T
The incenter of a triangle is the point of currency of the angle
2 Applying Theorem 5-7 City planners want to locate a
circle
Inscribed
tangent to the circle.
Hig
hw
ay
A circle is inscribed in a polygon if the sides of the polygon are
Name_____________________________________ Class____________________________ Date ________________
Andover Road
from the sides.
equidistant
inside
a side
.
Obtuse Triangle
Altitude is
outside
.
.
All rights reserved.
Right Triangle
Altitude is
Acute Triangle
Altitude is
All rights reserved.
The city planners should find the point of concurrency of the
of the triangle formed
angle bisectors
by the three roads and locate the fountain there.
3 Finding Lengths of Medians M is the centroid of WOR, and WM 16.
W
Find WX.
The centroid of a triangle is the point of intersection of the medians of the triangle.
centroid
The
The medians of a triangle are concurrent at a point that is
(1, 7)
, XY lies on the vertical line x 1 .
The perpendicular bisector of XY is the horizontal line that
7
passes through (1, 1 2 ) or
(1, 4)
(1, 1)
(5, 1)
6
(3, 4)
4
, so the
2
equation of the perpendicular bisector of XY is y 4 .
Because X has coordinates
y
8Y
and Y has
X
Z
O
and Z has
2
4
6
x
, XZ lies on the horizontal line
y 1 . The perpendicular bisector of XZ is the vertical line that
(1 2 5, 1)
or
(3, 1)
, so the equation
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
(1, 1)
Because X has coordinates
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
circumscribes XYZ.
All rights reserved.
Y
R
X
of the opposite side. (Theorem 5-8)
1 Finding the Circumcenter Find the center of the circle that
passes through
O
the distance from each vertex to the midpoint
two-thirds
Examples.
coordinates
Z
M
of the triangle.
coordinates
is the point of concurrency of the medians
of a triangle.
The orthocenter of a triangle is the point of intersection of the lines containing the altitudes
Because M is the
WM 16
24
2WX
3
of WOR, WM centroid
.
2
WX
Theorem
WX
Substitute
5-8
3
2
16
for WM.
3
WX
Multiply each side by
3
2
.
of the perpendicular bisector of XZ is x 3 .
Draw the lines y 4 and x 3 . They intersect at the point
(3, 4)
. This point is the center of the circle that
circumscribes XYZ.
98
Geometry Lesson 5-3
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
y
Lesson Objectives
5
1
Write the negation of a statement
and the inverse and contrapositive
of a conditional statement
2
Use indirect reasoning
NAEP 2005 Strand: Geometry
Topics: Mathematical Reasoning
Local Standards: ____________________________________
(4, 3)
Vocabulary and Key Concepts.
5
5
All rights reserved.
All rights reserved.
5x
b. Critical Thinking In Example 1, explain why it is not necessary to find
the third perpendicular bisector.
Theorem 5-6: All the perpendicular bisectors of the sides of a
triangle are concurrent.
2. a. The towns of Adamsville, Brooksville, and Cartersville want to
build a library that is equidistant from the three towns. Show on
the diagram where they should build the library. Explain.
Brooksville
Cartersville
b. What theorem did you use to find the location?
The perpendicular bisectors of the sides of a triangle are concurrent
at a point equidistant from the vertices (Theorem 5-6).
K
8
All-In-One Answers Version A
If an angle is a straight angle,
then its measure is 180.
Symbolic Form
Negation (of p)
An angle is not a straight angle.
p
Not
Inverse
If an angle is not a straight angle,
then its measure is not 180.
p S q
If not
p, then not q.
Contrapositive
If an angle’s measure is not 180,
then it is not a straight angle.
q S p
If not
q, then not p.
pSq
You Read It
If
p,
then
q.
p.
Step 1 State as an assumption the
opposite (negation) of what you want to prove.
Step 2 Show that this assumption leads to a contradiction.
Step 3 Conclude that the assumption must be
false and that what you want to prove must be true.
The negation of a statement has the opposite truth value from that of the original statement.
The inverse of a conditional statement negates both the hypothesis and the conclusion.
The contrapositive of a conditional switches the hypothesis and the conclusion and negates
both.
In indirect reasoning all possibilities are considered and then all but one are proved false.
L
Geometry Lesson 5-3
Example
Conditional
Equivalent statements have the same truth value.
median; MY is a segment drawn from vertex M to the midpoint
of the opposite side.
100
Statement
A conditional and its contrapositive always have the same truth value.
Y
4. Is MY a median, an altitude, or neither? Explain.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Adamsville
Draw segments connecting the towns. Build the library at the
intersection point of the perpendicular bisectors of the segments.
3. Using the diagram in Example 3, find MX. Check that WM MX WX.
Negation, Inverse, and Contrapositive Statements
Writing an Indirect Proof
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Inverses, Contrapositives,
and Indirect Reasoning
Lesson 5-4
Quick Check.
1. a. Find the center of the circle that you can
circumscribe about the triangle with
vertices (0, 0), (8, 0), and (0, 6).
99
Geometry Lesson 5-3
X
M
Daily Notetaking Guide
The remaining possibility must be true.
An indirect proof is a proof involving indirect reasoning.
Daily Notetaking Guide
Geometry Lesson 5-4
Geometry
101
25
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
4 Identifying Contradictions Identify the two statements that contradict
Examples.
each other.
I. P, Q, and R are coplanar.
II. P, Q, and R are collinear.
III. mPQR 60
1 Writing the Negation of a Statement Write the negation of “ABCD is
not a convex polygon.”
opposite
The negation of a statement has the
truth value. The
not
negation of is not in the original statement removes the word
.
contradict
Two statements
each other when they cannot both be
true at the same time. Examine each pair of statements to see whether they
contradict each other.
The negation of “ABCD is not a convex polygon” is
All rights reserved.
of the conditional statement “If ABC is equilateral, then it is isosceles.”
To write the inverse of a conditional, negate both the hypothesis and the
conclusion.
Hypothesis
Conclusion
T
If ABC is
then it is isosceles.
Negate both.
not equilateral
,
not isosceles
II and III
P, Q, and R are coplanar, and
mPQR 60.
P, Q, and R are collinear, and
mPQR 60.
Three points that lie on the
same line are both coplanar
and collinear, so these two
statements do not
contradict each other.
Three points that lie on an
angle are coplanar, so these
two statements do not
contradict each other.
If three distinct points are
collinear, they form a straight
angle, so mPQR cannot
equal 60. Statements II and
do
III
contradict
each other.
.
5 Indirect Proof Write an indirect proof.
Prove: ABC cannot contain two obtuse angles.
T
T
Step 1 Assume as true the
Switch and negate both.
not isosceles
, then it is
not equilateral
.
3 The First Step of an Indirect Proof Write the first step of an indirect proof.
Prove: A triangle cannot contain two right angles.
In the first step of an indirect proof, you assume as true the
negation
of what you want to prove.
Because you want to prove that a triangle cannot contain two right angles,
can
you assume that a triangle
contain two right angles.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Conclusion
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Hypothesis
If ABC is equilateral, then it is isosceles.
Contrapositive: If ABC is
I and III
P, Q, and R are coplanar and
collinear.
T
then it is
To write the contrapositive of a conditional, switch the hypothesis and
conclusion, then negate both.
Conditional:
I and II
The first step is “Assume that a triangle contains two right angles.”
opposite
of what you want to prove.
That is, assume that ABC contains two
Let lA and lB be
Step 2 If A and B are
obtuse
obtuse
, mA mB mC obtuse
Triangle Angle-Sum
This contradicts the
which states that mA mB mC Step 3 The assumption in Step 1 must be
contain two
angles.
.
false
180
.
All rights reserved.
T
Inverse:
T
If ABC is equilateral,
Conditional:
All rights reserved.
“ABCD is a convex polygon.”
2 Writing the Inverse and Contrapositive Write the inverse and contrapositive
Theorem,
180
.
. ABC cannot
angles.
obtuse
Quick Check.
1. Write the negation of each statement.
a. mXYZ 70.
The measure of ⬔XYZ is not more than 70.
b. Today is not Tuesday.
Today is Tuesday.
Geometry Lesson 5-4
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Lesson 5-5
2. Write (a) the inverse and (b) the contrapositive of Maya Angelou’s
statement “If you don’t stand for something, you’ll fall for anything.”
a.
b.
103
Geometry Lesson 5-4
Inequalities in Triangles
Lesson Objectives
1 Use inequalities involving angles of
triangles
2 Use inequalities involving sides of
triangles
If you stand for something, you won’t fall for anything.
NAEP 2005 Strand: Geometry
Topics: Relationships Among Geometric Figures
Local Standards: ____________________________________
If you won’t fall for anything, then you stand for something.
Key Concepts.
All rights reserved.
The shoes cost more than $20.
Comparison Property of Inequality
All rights reserved.
3. You want to prove each statement true. Write the first step of an indirect proof.
a. The shoes cost no more than $20.
If a b c and c 0, then a b .
Corollary to the Triangle Exterior Angle Theorem
The measure of an
b. mA mB
m⬔1
III. FG > KL
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
II. FG ' KL
I and II;
Two segments cannot be parallel as well as perpendicular. I and II
contradict each other.
Two segments can be parallel and congruent. I and III do not contradict
each other.
Two segments can be perpendicular and congruent. II and III do not
contradict each other.
5. Critical Thinking You plan to write an indirect proof showing that X is an
obtuse angle. In the first step you assume that X is an acute angle. What
have you overlooked?
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4. Identify the two statements that contradict each other. Explain.
I. FG6KL
exterior
m2 and
m⬔1
3
angle of a triangle is greater
interior
than the measure of each of its remote
mlA # mlB
angles.
2
m3
Theorem 5-10
If two sides of a triangle are not congruent, then
1
Y
the larger angle lies opposite the longer side.
If XZ XY, then mY mZ.
X
Theorem 5-11
If two angles of a triangle are not congruent, then
Z
B
C
the longer side lies opposite the larger angle.
If mA mB, then BC AC.
A
L
Theorem 5-12: Triangle Inequality Theorem
The sum of the lengths of any two sides of a triangle is
greater than the length of the third side.
M
LM MN LN
MN LN LM
N
LN LM MN
⬔X could be a right angle.
104
26
Geometry Lesson 5-4
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 5-5
105
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
102
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
4 Finding Possible Side Lengths In FGH, FG 9 m and GH 17 m.
Examples.
Describe the possible lengths of FH.
1 Applying the Corollary Explain why m4 m5.
4 is an
B
5
angle of XYC. The Corollary
exterior
to the Exterior Angle Theorem states that the measure
greater
of an exterior angle of a triangle is
4
X
A
than the measure of each of its remote interior angles.
The Triangle Inequality Theorem: The sum of the lengths of any two sides of
greater
a triangle is
than the length of the third side.
Y
2
1
Solve three inequalities.
3
All rights reserved.
angles
formed by transversal BC . Because AB6XY, you can conclude
that 2 ⬔5 by the Corresponding Angles Postulate. Thus,
m2 m⬔5 . Substituting m⬔5 for m2 in the
inequality m4 m2 produces the inequality m⬔4 m⬔5 .
List the angles from largest to smallest.
Theorem 5-10: If two sides of a triangle are not congruent, then
the larger angle lies opposite the longer side.
All rights reserved.
12
Y
20
R
. The opposite angle, ⬔R , is the
smallest. From largest to smallest, the angles are ⬔G , ⬔Y , ⬔R .
3 Using the Triangle Inequality Theorem Can a triangle have sides with the
given lengths? Explain.
a. 2 cm, 2 cm, 4 cm
Triangle Inequality
According to the
Theorem, the
sum of the lengths of any two sides of a triangle is
greater
than the length of the third side.
22 ⬐ 4
The sum of 2 and 2
No , a triangle
cannot
is not
greater than 4.
can
1. Use the diagram in Example 1. If CY XY, explain why m4 m1.
A
27 ft
C
21 ft
18 ft
B
3. Can a triangle have sides with the given lengths? Explain.
a. 2 m, 7 m, and 9 m
no; 2 7 is not greater than 9
b. 4 yd, 6 yd, and 9 yd
yes; 4 6 9, 6 9 4, and 4 9 6
Daily Notetaking Guide
Daily Notetaking Guide
Lesson 6-1
Examples.
Classifying Quadrilaterals
Lesson Objective
1 Define and classify special types of
quadrilaterals
1 Classifying by Coordinate Methods Determine the most precise name
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
for the quadrilateral with vertices Q(4, 4), B(2, 9), H(8, 9), and A(10, 4).
Graph quadrilateral QBHA.
Local Standards: ____________________________________
Find the slope of each side.
94
Slope of QB 2 (4)
99
Slope of BH 8 2
y
B
Key Concepts.
H
8
Special Quadrilaterals
(
6
Parallelogram
Rhombus
Trapezoid
Isosceles
Trapezoid
All rights reserved.
Rectangle
Q
All rights reserved.
Quadrilateral
Quadrilateral
of
irs es
2 pairs of
pa
1 pair of
id
No llel s
parallel
sides parallel sides
ra
pa
Square
A
4 2 O
2
4
6
Slope of HA 10
5
2
0
25
2
0
.
not equal
.
trapezoid
One pair of opposite sides is parallel, so QBHA is a
.
Next, use the distance formula to see whether any pairs of sides are congruent.
kite
A
is a
quadrilateral with two pairs
adjacent sides
of ______________________
opposite
congruent and no ________
sides
________________________
congruent.
rectangle
A
is a parallelogram with
right angles
four______________________
__________________________.
trapezoid
A
is a quadrilateral with
exactly one pair of
________________________
parallel sides
________________________
________________________ .
4 4
4 equal
BH is parallel to QA because their slopes are
square
A
is a parallelogram with
congruent sides ___
four__________________
______________ and four
right
angles
___
___________________
.
isosceles trapezoid
An _____________________
________________________
is a trapezoid whose
nonparallel sides are
congruent.
Geometry Lesson 6-1
All-In-One Answers Version A
Daily Notetaking Guide
(2 (4))2 (9 4 )2 4 25 "29
(10 8)2 ( 4 9 )2 4 25 "29
10
BH (8 (2))2 ( 9 9 )2 0 100
2
2
QA (4 10 ) ( 4 4 ) 0 196
QB HA © Pearson Education, Inc., publishing as Pearson Prentice Hall.
rhombus
A
is a parallelogram with
congruent sides
four_____________________
______________________.
10 8
Slope of QA x
8
)
4 9
2
HA is not parallel to QB because their slopes are
parallelogram
A
is a quadrilateral with
both pairs of opposite
parallel
sides ___________________
________________________ .
107
Geometry Lesson 5-5
Name_____________________________________ Class____________________________ Date ________________
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Quick Check.
15 12 8
Name_____________________________________ Class____________________________ Date________________
108
FH
m.
have sides with these lengths.
Geometry Lesson 5-5
Kite
26
9 x 15
8 12 15
Yes , a triangle
106
m and shorter than
FH
4. A triangle has sides of lengths 3 in. and 12 in. Describe the possible lengths
of the third side.
have sides with these lengths.
b. 8 in., 15 in., 12 in.
8 15 12
8
17
26
⬔A, ⬔C, ⬔B
. The opposite angle, ⬔G ,
is the largest. The shortest side is
9
9
FH 8
2. List the angles of ABC in order from smallest to largest.
14
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Find the angle opposite each side.
G
12
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
No two sides of RGY are congruent, so the largest angle lies
opposite the longest side.
20
FH 17 m⬔1 m⬔3 by the Isosceles Triangle Theorem
m⬔4 m⬔3 by the Corollary to the Triangle Exterior Angle Theorem
m⬔4 m⬔1 by Substitution
2 Applying Theorem 5-10 In RGY, RG 14, GY 12, and RY 20.
The longest side is
8
FH must be longer than
All rights reserved.
corresponding
17
FH C
The remote interior angles are ⬔2 and ⬔3 , so
m4 m2. 2 and 5 are
FH 9 Because QB HA
an isosceles trapezoid
, QBHA is
14
.
U
2 Using the Properties of Special Quadrilaterals In parallelogram RSTU,
mR 2x 10 and mS 3x 50. Find x.
T
RSTU is a parallelogram.
Given
ST6
Definition of parallelogram
RU
(3x 50)
(2 x 10)
R
S
mR mS If lines are parallel, then interior
angles on the same side of a transversal are
180
(2x 10) (3x 50) 180
5x
2x 10
for m⬔R and
40
180
Simplify.
5x
Subtract 40 from each side.
x
Daily Notetaking Guide
Substitute
140
28
supplementary
3x 50
.
for m⬔S.
Divide each side by 5.
Geometry Lesson 6-1
Geometry
109
27
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Quick Check.
Lesson 6-2
1. a. Graph quadrilateral ABCD with vertices A(3, 3), B(2, 4),
C(3, 1), and D(2, 2).
b. Classify ABCD in as many ways as possible.
y
B
A
ABCD is a quadrilateral because it has four sides. It is a
parallelogram because both pairs of opposite sides are
parallel. It is a rhombus because all four sides are congruent.
It is a rectangle because it has four right angles and its
opposite sides are congruent. It is a square because it has
four right angles and its sides are all congruent.
Properties of Parallelograms
Lesson Objectives
1
2
x
NAEP 2005 Strand: Geometry
Use relationships among sides and
among angles of parallelograms
Use relationships involving diagonals
of parallelograms and transversals
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
C
D
c. Which name gives the most information about ABCD? Explain.
All rights reserved.
All rights reserved.
Vocabulary and Key Concepts.
Square; the name square means a figure is both a rectangle
and a rhombus, since its sides are all congruent and it has
four right angles.
Theorem 6-1
C
A
Opposite angles of a parallelogram are
congruent
D
.
R
S
Theorem 6-3
bisect
The diagonals of a parallelogram
4b 2
S
N
3a 8
T
each other.
T
BD B
A
D
C
transversal.
E
F
DF
All rights reserved.
3b 2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
L
5a 4
U
Theorem 6-4
If three (or more) parallel lines cut off congruent segments on one
transversal, then they cut off congruent segments on every
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2. Find the values of the variables in the rhombus. Then find the
lengths of the sides.
a 2, b 4, LN ST NT SL 14
B
sides of a parallelogram are congruent.
Opposite
Theorem 6-2
Consecutive angles of a polygon share a common side.
K
J
L
M
In JKLM, J and M are consecutive angles,
as are J and K . J and L are not consecutive.
Examples
Geometry Lesson 6-1
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
1 Using Algebra Find the value of x in ABCD. Then find mA.
x 15 135 x
135
2x 120
60
Add x to each side.
Subtract
15
60
Divide each side by 2 .
15 75
Substitute
60
D
for x.
.
supplementary
180
75
mA C
Consecutive angles of a parallelogram are
mA mB 180
mA Place a corner of the top of the card on the first line of the lined paper.
Place a corner that forms a consecutive angle with the first corner on the
sixth line.
Mark the points where the lines intersect one side of the card.
Mark the points where the lines intersect the opposite side of the card.
Connect the marks on opposite sides using a straightedge.
(135 x)
from each side.
x
Subtract
for mB.
75
Substitute
105
75
from each side.
All rights reserved.
mB segments on every transversal.
Explain how to divide a blank card into five equal rows using Theorem 6-4
and a sheet of lined paper.
.
All rights reserved.
15 off congruent segments on one transversal, then they cut off congruent
(x 15)
Opposite angles of a
parallelogram are
congruent
2x
Name_____________________________________ Class____________________________ Date ________________
B
A
2 Using Algebra Find the values of x and y in KLMN.
2x 5
5y
14y
H
The diagonals of a parallelogram
bisect
each other.
Substitute
for x in
7y 16
the second equation to solve for y.
) 5 5y
32
5 5y
27
Distribute.
5y
27 Simplify.
9y
3 y
Subtract
14y
Divide each side by
( ) 16
x7 3
x 5
from each side.
9 .
Substitute 3 for y in the first equation
to solve for x.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
x 7y 16
G
M
2x 5 5y
7y 16
(3y 37)
(6y 4)
E
N
14y
F
A
7y 16
(
1. Find the value of y in EFGH. Then find mE, mF, mG, and
mH.
L
x
2
Quick Check.
y 5 11; mlE 5 70, mlF 5 110, mlG 5 70, and mlH 5 110
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
K
111
Geometry Lesson 6-2
2. Find the values of a and b.
X
a
a 16, b 14
b
10
W
←
→
←
→
←
→
Y
8
2a b
2
Z
←
→
3. In the figure, DH 6 CG 6 BF 6 AE, AB BC CD 2, and
EF 2.5. Find EH.
D
7.5
C
Simplify.
B
So x 5 and y 3 .
A
H
2
G
2
F
2.5
2
E
3 Using Theorem 6-4 Theorem 6-4 states If three (or more) parallel lines cut
112
28
Geometry Lesson 6-2
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 6-2
113
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
110
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Proving That a Quadrilateral
Is a Parallelogram
Lesson 6-3
2 Is the Quadrilateral a Parallelogram? Can you prove the quadrilateral is a
parallelogram from what is given? Explain.
Lesson Objective
1
NAEP 2005 Strand: Geometry
Determine whether a quadrilateral is
a parallelogram
a. Given: AB > CD
Prove: ABCD is a parallelogram.
Topic: Geometry
Local Standards: ____________________________________
All you know about the quadrilateral is that only
one pair of opposite sides is
.
congruent
Therefore, you cannot conclude that the quadrilateral
is a parallelogram.
Key Concepts.
the quadrilateral is a parallelogram.
of a quadrilateral are congruent,
Theorem 6-7
If the diagonals of a quadrilateral
then the quadrilateral is a parallelogram.
each other,
bisect
Theorem 6-8
If one pair of opposite sides of a quadrilateral is both
parallel
and
B
8x 12
y9
If the diagonals of quadrilateral ABCD bisect each other,
Theorem 6-7
then ABCD is a parallelogram by
.
Write and solve two equations to find values of x and y for
which the diagonals bisect each other.
2y 80
G
10x 24
C
D
Diagonals of parallelograms
bisect each other.
10x 24 8x 12
2x
If x 114
24 12
Collect the variables on one side.
2x 36
Solve.
x
18
18
and y 2y 80
y9
F
The sum of the measures of the angles of a polygon is
(n 2)180 where n represents the number of sides,
so the sum of the measures of the angles of a
360
quadrilateral is 4 2 180 .
105
G
75
)
E
75
H
congruent, then the quadrilateral is a parallelogram.
Because both pairs of opposite angles are congruent, the quadrilateral
is a parallelogram by
Theorem 6-6
.
y 80 9
y
89
89 , then ABCD is a parallelogram.
Geometry Lesson 6-3
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 6-3
115
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Quick Check.
Lesson 6-4
1. Find the values of a and c for which PQRS must be a
parallelogram.
Q
R
(a 40)
a
3c 3
a 70, c 2
c1
P
Special Parallelograms
Lesson Objectives
1 Use properties of diagonals of
rhombuses and rectangles
2 Determine whether a parallelogram is
a rhombus or a rectangle
NAEP 2005 Strand: Geometry
Topic: Geometry
Local Standards: ____________________________________
S
2. Can you prove the quadrilateral is a parallelogram? Explain.
a. Given: PQ > SR, PQ 6SR
Prove: PQRS is a parallelogram.
Q
Yes; a pair of opposite sides are parallel
and congruent (Theorem 6-8).
R
Rhombuses
Theorem 6-9
B
Each diagonal of a rhombus bisects two angles of the rhombus.
AC bisects BAD, so 1 2
AC
bisects BCD
, so 3 4
x
AC
S
b. Given: DH > GH, EH > FH
Prove: DEFG is a parallelogram.
D
E
H
G
F
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
P
The diagonals of a rhombus are
'
A
perpendicular
D
A
congruent
C
.
Rectangles
Theorem 6-11
AC
C
4
B
BD
The diagonals of a rectangle are
3
1
2
Theorem 6-10
x
No; the diagonals do not necessarily bisect
each other—the figure could be a trapezoid,
with DG and EF being parallel but of
unequal length.
All rights reserved.
Key Concepts.
All rights reserved.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
Examples.
for which ABCD must be a parallelogram.
D
Theorem 6-6 states If both pairs of opposite angles of a quadrilateral are
, then the quadrilateral is a parallelogram.
1 Finding Values for Parallelograms Find values for x and y
A
(
congruent
C
b. Given: mE mG 75°, mF 105°
Prove: EFGH is a parallelogram.
If x represents the measure of the unmarked angle,
360
105
x 75 105 75 , so x =
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
opposite angles
then the quadrilateral is a parallelogram.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 6-6
If both pairs of
All rights reserved.
All rights reserved.
Theorem 6-5
If both pairs of opposite sides of a quadrilateral are congruent, then
B
D
A
D
B
C
.
BD
Parallelograms
Theorem 6-12
If one diagonal of a parallelogram bisects two angles of the parallelogram, then
the parallelogram is a rhombus.
Theorem 6-13
If the diagonals of a parallelogram are perpendicular, then
the parallelogram is a rhombus.
Theorem 6-14
If the diagonals of a parallelogram are congruent, then
the parallelogram is a rectangle.
116
Geometry Lesson 6-3
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 6-4
Geometry
117
29
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
Example.
1 Finding Angle Measures Find the measures of the
3 Identifying Special Parallelograms The diagonals of ABCD are such that
numbered angles in the rhombus.
Theorem 6-9 states that each diagonal of a rhombus
bisects two angles of the rhombus, so m1 78.
B
78
the diagonals of the rhombus are perpendicular
so m2 90
2
3
Theorem 6-10 states that
,
A
AC 16 cm and BD 8 cm. Can you conclude that ABCD is a rhombus or
a rectangle? Explain.
C
1
ABCD
4
D
be a rectangle, because AC BD and the diagonals of a
cannot
rectangle are
(Theorem 6-11). ABCD may be a rhombus,
congruent
AC ' BD
but it may also not be one, depending on whether
(Theorem 6-10).
.
m3 90 78 . Because mABD 78,
complementary
All rights reserved.
3 and ABD must be
.
12
Finally, because BC DC, the
Isosceles Triangle Theorem
allows you to conclude that 1 4. So m4 .
78
All rights reserved.
Because the four angles formed by the diagonals all must have measure 90,
Quick Check.
2. Find the length of the diagonals of rectangle GFED if FD 5y 9
and GE = y 5.
F
E
G
D
81
2
2 Finding Diagonal Length One diagonal of a rectangle has length 8x 2.
The other diagonal has length 5x 11. Find the length of each diagonal.
congruent
By Theorem 6-11, the diagonals of a rectangle are
8x 2
Subtract
3 x
( )2
5x 11 5( 3 ) 11 8x 2 8 3
5x
from each side.
Subtract 2 from each side.
Divide each side by 3 .
26
Substitute.
26
The length of each diagonal is
Substitute.
26
.
Quick Check.
1. Find the measures of the numbered angles in the rhombus.
50
1
m⬔1 90, m⬔2 50, m⬔3 50, m⬔4 40
4
118
Geometry Lesson 6-4
2
3
3. A parallelogram has angles of 30°, 150°, 30°, and 150°. Can you conclude
that it is a rhombus or a rectangle? Explain.
No; there is not enough information to conclude that the
parallelogram is a rhombus. It cannot be a rectangle because it has
no right angles.
Daily Notetaking Guide
Daily Notetaking Guide
All rights reserved.
2
Geometry Lesson 6-4
119
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 6-5
Examples.
1 Finding Angle Measures in Trapezoids WXYZ is
Trapezoids and Kites
Lesson Objective
1 Verify and use properties of
trapezoids and kites
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
are
156
Trapezoids
Theorem 6-15
diagonals
of an isosceles trapezoid are congruent.
All rights reserved.
All rights reserved.
Theorem 6-16
The
mW 180
mW .
Z
a leg of a trapezoid
180
Vocabulary and Key Concepts.
congruent
Y
W
Two angles that share
mX mW The base angles of an isosceles trapezoid are
X
156
an isosceles trapezoid, and mX 156. Find mY,
mZ, and mW.
.
supplementary
Substitute.
24
Subtract
156
from each side.
congruent
Because the base angles of an isosceles trapezoid are
mY mX 156 and mZ m⬔W
,
24 .
A
2 Using Isosceles Trapezoids Half of a spider’s web is shown at the right,
C
formed by layers of congruent isosceles trapezoids. Find the measures
of the angles in ABDC.
The base angles of a trapezoid are two angles that share a base of the trapezoid.
B
D
Trapezoid ABDC is part of an isosceles triangle whose vertex is at
angles
Base
angles
Leg
Kites
Theorem 6-17
The diagonals of a kite are
perpendicular
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Leg
Base
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the center of the web. Because there are 6 adjacent congruent
vertex angles at the center of the web, together forming a
straight
angle, each vertex angle measures 180 , or 30 .
6
180
By the Triangle Angle-Sum Theorem, mA mB 30 so mA mB 150
,
.
Because ABDC is part of an isosceles triangle, mA mB,
so 2(mA) 150
and mA mB 75
.
Another way to find the measure of each acute angle is to divide the
difference between 180 and the measure of the vertex angle by 2:
180 2 30 75 .
2
Because the bases of a trapezoid are parallel, the two angles that share a leg
are
supplementary
, so mC mD 180 75 =
105
.
Quick Check.
1. In the isosceles trapezoid, mS 70. Find mP, mQ, and mR.
P
Q
110, 110, 70
S
120
30
Geometry Lesson 6-5
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
70
Geometry Lesson 6-5
R
121
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3x
3x
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
11 9 .
Diagonals of a rectangle are congruent.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
5x 11
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date________________
Example.
Lesson 6-6
3 Finding Angle Measures in Kites Find m1, m2, and m3 in the kite.
Lesson Objective
S
1
1
3
Example.
1 Proving Congruency Show that TWVU is a parallelogram by
U
RU RS
Definition of a kite
m1 72
Isosceles Triangle Theorem
m3 mRDU 72 180
mRDU 90
perpendicular
90
72 180
162
180
m3 18
Triangle Angle-Sum Theorem
Diagonals of a kite are perpendicular.
If both pairs of opposite sides of a quadrilateral are
congruent
from each side.
Quick Check.
(
(
WV (
TU (
3. Find m1, m2, and m3 in the kite.
m⬔1 90, m⬔2 46, and m⬔3 44
46
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
The measure of both inner angles is 85.
The measure of both outer angles is 95.
(
E
TU . Use the distance formula.
) (
) (
)
a , b , W a c , b d , V c e , d , and U
) (
) )2 ( d 0 )2 c 2 d 2
2
2
ac
ce ) ( bd d ) 2
2
a e ) ( b 0 ) (a e) 2 b 2
VU 2. The middle ring of the piece of ceiling shown is made from congruent isosceles
trapezoids. Imagine a circular glass ceiling made from the ceiling pieces with
18 angles meeting at the center. What are the measures of the two sets of base
angles of the trapezoids in the middle ring?
x
U(e, 0)
O
You can prove that TWVU is a parallelogram by showing that
TW V(c e, d)
.
Theorem 6-5
Use the coordinates T
Simplify.
162
W(a c, b d)
C
T(a, b)
, then the quadrilateral is a
parallelogram by
TW VU and WV Substitute.
Subtract
A
y
proving pairs of opposite sides congruent.
.
All rights reserved.
Diagonals of a kite are
All rights reserved.
m2 90
m3 Topic: Position and Direction
T
2
72
m3 NAEP 2005 Strand: Geometry
Name coordinates of special figures
by using their properties
Local Standards: ____________________________________
D
R
Placing Figures in
the Coordinate Plane
ac
a
ce
e
2
2
b
bd
Because TW VU and WV TU, TWVU is a
c2
(
)
e , 0 .
d2
parallelogram
(a e) 2 b 2
.
Quick Check.
1. Use the diagram above. Use a different method: Show that TWVU is a
parallelogram by finding the midpoints of the diagonals.
Midpoint of TV 5 aa 1 c 1 e , b 1 d b midpoint of UW . Thus, the
2
2
diagonals bisect each other, and TWVU is a parallelogram.
1
3
122
Geometry Lesson 6-5
Geometry Lesson 6-6
123
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Example.
Lesson 6-7
2 Naming Coordinates Use the properties of parallelogram
y
OCBA to find the missing coordinates. Do not use any new
variables.
(0, 0)
The vertex O is the origin with coordinates O
.
B
Proofs Using Coordinate Geometry
Lesson Objective
1 Prove theorems using figures in the
coordinate plane
(p, q)
A
NAEP 2005 Strand: Geometry
Topics: Position and Direction; Mathematical Reasoning
Local Standards: ____________________________________
Because point A is p units to the left of point O, point B
Vocabulary and Key Concepts.
is also p units to the left of point C , because OCBA is a
p x
.
Because AB 6 CO and CO is horizontal, AB is
C (x, 0)
O
. So point B has the same second
coordinate, q , as point A.
The missing coordinates are O
(0, 0)
and B
(p x, q)
Theorem 6-18: Trapezoid Midsegment Theorem
All rights reserved.
also horizontal
x
.
All rights reserved.
parallelogram. So the first coordinate of point B is
parallel
(1) The midsegment of a trapezoid is
Quick Check.
lengths of the bases
.
opposite sides of the trapezoid.
2. Use the properties of parallelogram OPQR to find the missing
coordinates. Do not use any new variables.
y R(b, c)
Q(?, ?)
R
Q(s b, c)
M
T
P(s, 0)
Geometry Lesson 6-6
All-In-One Answers Version A
A
N
P
x
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
O
124
to the bases.
(2) The length of the midsegment of a trapezoid is half the sum of the
The midsegment of a trapezoid is the segment that joins the midpoints of the nonparallel
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Daily Notetaking Guide
Daily Notetaking Guide
MN 6
TP , MN 6
1
RA , and MN 2 (
TP
RA )
Examples.
1 Planning a Coordinate Geometry Proof Examine trapezoid
y
R(2b, 2c) A(2d, 2c)
TRAP. Explain why you can assign the same y-coordinate to points
R and A.
M
In a trapezoid, only one pair of sides is parallel. In TRAP,
TP 6
must be
RA . Because TP lies on the horizontal x-axis, RA also
horizontal
T
N
x
P(2a, 0)
.
The y-coordinates of all points on a horizontal line are the same, so
points R and A have the same y-coordinates.
Daily Notetaking Guide
Geometry Lesson 6-7
Geometry
125
31
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
2 Using Coordinate Geometry Use coordinate geometry to prove that the
Quick Check.
quadrilateral formed by connecting the midpoints of rhombus ABCD is
a rectangle.
Draw quadrilateral XYZW by connecting the midpoints of ABCD.
Given: MN is the midsegment of trapezoid TRAP.
1
Prove: MN 6 TP, MN 6 RA, and MN = 2 (TP + RA)
y
M
a. Find the coordinates of midpoints M and N. How do the multiples
T
of 2 help?
D(0, 2b)
Z(a, b)
y
R(2b, 2c) A(2d, 2c)
1. Complete the proof of Theorem 6-18.
W(a, b)
C(2a, 0)
N
x
P(2a, 0)
M(b, c) and N(a d, c); starting with multiples of 2 eliminates
fractions when using the midpoint formula.
A(2a, 0)
All rights reserved.
B(0, 2b)
From Lesson 6-6, you know that XYZW is a parallelogram.
congruent
If the diagonals of a parallelogram are
All rights reserved.
x
X(a, b)
Y(a, b)
b. Find and compare the slopes of MN, TP, and RA.
0, 0, 0; they are all parallel.
, then the
parallelogram is a rectangle from Theorem 6–14.
To show that XYZW is a rectangle, find the lengths of its diagonals, and then
equal
compare them to show that they are
(a a)2 (b b)2
( 2a )2 ( 2b )2
4a2 4b2
YW XZ YW
Because the diagonals
are
parallelogram XYZW is a
rectangle
126
congruent,
.
Geometry Lesson 6-7
d. In parts (b) and (c), how does placing a base along the x-axis help?
The base along the x-axis allows calculating length by subtracting
x-values.
2. Use the diagram from Example 2. Explain why the proof using A(2a, 0),
B(0, 2b), C(2a, 0), and D(0, 2b) is easier than the proof using A(a, 0),
B(0, b), C(a, 0), and D(0, b).
Using multiples of 2 in the coordinates for A, B, C, and D eliminates
the use of fractions when finding midpoints, since finding midpoints
requires division by 2.
Daily Notetaking Guide
Daily Notetaking Guide
All rights reserved.
MN d a b, TP 2a, RA 2d 2b. So RA TP 2d 2a 2b
which is twice MN. So the midsegment is half the sum of the bases.
(a a) (b (b))
( 2a )2 ( 2b )2
4a2 4b2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
c. Find and compare the lengths MN, TP, and RA.
.
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
XZ 2
Geometry Lesson 6-7
127
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 7-1
2 Properties of Proportions
b
Complete: If a4 5 12
b , then 12 5
NAEP 2005 Strand: Geometry
Topic: Position and Direction
ab 5
Local Standards: ____________________________________
ab
(3) ac b
d
b
(4) a 1
b c1d
d
A proportion is a statement that two ratios are equal.
All rights reserved.
d
c
(2) ba All rights reserved.
c
a
b d is equivalent to
bc
c
a
b d and a : b c : d are examples of proportions.
6
1
4
24 16 4 is an example of an extended proportion.
a.
25 n
5
35
5n 5 2
(
)
35
Simplify.
n5
14
Divide each side by 5 .
3x 5
A scale drawing is a drawing in which all lengths are proportional to corresponding actual
lengths.
A scale is the ratio of any length in a scale drawing to the corresponding actual length.
The lengths may be in different units.
Examples.
1 Finding Ratios A scale model of a car is 4 in. long. The actual car is 15 ft
long. What is the ratio of the length of the model to the length of the car?
Write both measurements in the same units.
180
15 ft 15 12 in. in.
length of model
4 in.
4 in.
4
length of car 5 15 ft 5 180 in. 5 180 5
(
2x
x 1 1
Geometry
)
1 2
Subtract
Property
Cross-Product
Property
Distributive
Property
2x
from each side.
Quick Check.
1. A photo that is 8 in. wide and 531 in. high is enlarged to a poster that is 2 ft
wide and 113 ft high. What is the ratio of the height of the photo to the height
of the poster?
1:3
n
2. Write two expressions that are equivalent to m
4 5 11.
4 5 11, m 1 4 5 n 1 11
Answers may vary. Sample: m
n
4
11
1
45
The ratio of the length of the scale model to the length of the car is
Geometry Lesson 7-1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
a d b c
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
a
c
b d
Cross-Product
70
x5 2
extremes
32
a
5n 5
to the product of the means.
means
.
3 Solving for a Variable Solve each proportion.
The Cross-Product Property states that the product of the extremes of a proportion is equal
a:b c:d
12a
Divide each side by
Simplify.
1
x
b. x 1
3 52
3x 5 2
An extended proportion is a statement that three or more ratios are equal.
128
12a
4
b
12 5
Properties of Proportions
.
Cross-Product Property
48
12a
Vocabulary and Key Concepts.
(1) ad 48
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Ratios and Proportions
Lesson Objective
1 Write ratios and solve proportions
1
:
45 .
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 7-1
129
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Example.
Lesson 7-2
4 Using Proportions Two cities are 3 12 in. apart on a map with the scale
1 in. 50 mi. Find the actual distance.
Let d represent the actual distance.
map distance (in.)
actual distance (mi.) 5
Similar figures have the same shape but not necessarily the same size. Two polygons are
)
31
2
Cross-Product
175
Property
Simplify.
175
The cities are actually
miles apart.
All rights reserved.
d5
(
Quick Check.
The mathematical symbol for similarity is ⬃ .
The similarity ratio is the ratio of the lengths of corresponding sides of similar figures.
similar to the original rectangle.
6
b. n 18
165n
z 0.75
The golden ratio is the ratio of the length to the width of any golden rectangle, about
1.618 : 1.
4. Recall Example 4. You want to make a new map with a scale of 1 in. 35 mi.
Two cities that are actually 175 miles apart are to be represented on your map.
What would be the distance in inches between the cities on the new map?
5 inches
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
n3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
similar if corresponding angles are congruent and corresponding sides are proportional.
A golden rectangle is a rectangle that can be divided into a square and a rectangle that is
3. Solve each proportion.
a. 5z 5 20
3
Example.
1 Understanding Similarity 䉭ABC 䉭XYZ.
Y
⬔B ⬔ Y and m⬔Y 78, so m⬔B 78
because congruent angles have the same measure.
A
C
X
BC 5
b. YZ
XZ
AC
78
42
Z
BC AC .
corresponds to XZ, so YZ
XZ
Quick Check.
1. Refer to the diagram for Example 1. Complete:
BC and YZ
42
AB
XY
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 7-2
131
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
Quick Check.
2 Determining Similarity Determine whether the
K
parallelograms are similar. Explain.
Check that corresponding sides are proportional.
B
1
2
1
2
DA MJ
N
A
Corresponding sides of the two parallelograms
proportional.
are
2
4
4
Y
21
D
1
14
J
M
2
Check that corresponding angles are congruent.
⬔B corresponds to ⬔ K , but m⬔B ⫽ m⬔K, so corresponding
are not
angles
congruent.
Although corresponding sides
parallelograms
are not
are
proportional, the
Because 䉭ABC 䉭YXZ, you can write and solve
a proportion.
x
Corresponding sides are
XZ
proportional
12
x .
12
C
X
40
Z
50
Solve for x.
15
Simplify.
4 Using the Golden Ratio The dimensions of a rectangular tabletop are in
the Golden Ratio. The shorter side is 40 in. Find the longer side.
Let ᐍ represent the longer side of the tabletop.
1.618
ᐉ
40 5 1
64.72
ᐉ5
Write a proportion using the
The table is about
132
Cross-Product
65
24
M
50
Substitute.
40
x5
B
30
x
12
40
P
Y
A
BC
X
9
3 Using Similar Figures If 䉭ABC 䉭YXZ, find the value of x.
50
16
Yes; corresponding angles are congruent and corresponding sides are
proportional.
.
the corresponding angles are not congruent
AC 5
YZ
Z
18
12
3. Refer to the diagram for Example 3. Find AB.
similar because
Golden Ratio
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2
4
C
120
2
CD
LM 100
All rights reserved.
BC
KL 2. Sketch 䉭XYZ and 䉭MNP with ⬔X ⬔M, ⬔Y ⬔N, and ⬔Z ⬔P.
Also, XY 12, YZ 14, ZX 16, MN 18, NP 21, and PM 24. Can
you conclude that the two triangles are similar?
L
All rights reserved.
2
4
AB 5
JK
1
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Geometry Lesson 7-1
B
Complete each statement.
a. m⬔B m⬔A 130
Topics: Transformation of Shapes and Preservation of
Properties; Measuring Physical Attributes
Vocabulary.
Substitute.
50
d 5 50
Identify similar polygons
Apply similar polygons
Local Standards: ____________________________________
1
d
1
2
NAEP 2005 Strand: Geometry and Measurement
1 in.
50 mi
All rights reserved.
31
2
Similar Polygons
Lesson Objectives
4. A golden rectangle has shorter sides of length 20 cm. Find the length of the
longer sides.
about 32.4 cm
.
Property
in. long.
Geometry Lesson 7-2
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 7-2
Geometry
133
33
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 7-3
2 Using Similarity Theorems Explain why the triangles
Proving Triangles Similar
W
must be similar. Write a similarity statement.
Lesson Objectives
Use AA, SAS, and SSS similarity
statements
Apply AA, SAS, and SSS similarity
statements
Topic: Transformation of Shapes and Preservation of
Properties
1
VY
12
VW 5 24 Local Standards: ____________________________________
VZ 5
and VX
2
18
36
Z
18
1
V
,
2
12
Y
so corresponding sides are proportional.
36
X
Vocabulary and Key Concepts.
R
PLM
P
T
S
L
M
All rights reserved.
䉭TRS 䉭
Therefore, 䉭YVZ 䉭WVX
Side-Angle-Side Similarity (SAS)
by the
All rights reserved.
Postulate 7-1: Angle-Angle Similarity (AA⬃) Postulate
If two angles of one triangle are congruent to
two angles of another triangle, then the
triangles are similar.
Theorem.
3 Using Similar Triangles Joan places a mirror 24 ft from the base of a
T
tree. When she stands 3 ft from the mirror, she can see the top of the tree
reflected in it. If her eyes are 5 ft above the ground, how tall is the tree?
TR represents the height of the tree, Point M represents the mirror, and
point J represents Joan’s eyes.
Both Joan and the tree are
Theorem 7-1: Side-Angle-Side Similarity (SAS⬃) Theorem
If an angle of one triangle is congruent to an angle of a second triangle,
and the sides including the two angles are proportional, then
so m⬔JOM m⬔TRM, and therefore
.
lJOM O lTRM
The light reflects off the mirror at the same angle at which it hits the
mirror, so ⬔JMO ⬔ TMR .
the triangles are similar.
Indirect measurement is a way of measuring things that are difficult to measure directly.
Examples.
1 Using the AA⬃ Postulate MX ' AB. Explain why the triangles
M
are similar. Write a similarity statement.
K
.
⬔A > ⬔ B because their measures are equal.
䉭AMX 䉭BKX
Angle-Angle Similarity (AA)
by the
134
Postulate.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the triangles are similar.
BXK
J
5
O 3 M
24
R
Geometry Lesson 7-3
135
Use similar triangles to find the height of the tree.
Theorem 7-2: Side-Side-Side Similarity (SSS⬃) Theorem
If the corresponding sides of two triangles are proportional, then
Because MX ' AB, ⬔AXM and ⬔BXK are
right angles
, so ⬔AXM > ⬔
to the ground,
perpendicular
nJOM , nTRM
TR
AA
Postulate
RM
TR
JO
Corresponding sides of similar triangles are proportional.
OM
All rights reserved.
2
24
⬔YVZ > ⬔WVX because they are vertical angles.
24
Substitute.
3
5
24
TR 5
Solve for TR.
3
TR 5
40
58
A
Geometry Lesson 7-3
40
The tree is
58
feet tall.
B
X
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Quick Check.
Lesson 7-4
1. In Example 1, you have enough information to write a similarity statement.
Do you have enough information to find the similarity ratio? Explain.
Similarity in Right Triangles
Lesson Objective
1 Find and use relationships in similar
right triangles
No; none of the side lengths are given.
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
Local Standards: ____________________________________
2. Explain why the triangles at the right must be similar.
Write a similarity statement.
G
8
F
9
A
6
B
6
All rights reserved.
Theorem 7-3
12
E
All rights reserved.
nABC , nEFG
Vocabulary and Key Concepts.
8
AC 5 CB 5 AB 5 3 , so the triangles are similar by the SSS~ Theorem;
EF
4
EG
GF
C
The altitude to the hypotenuse of a right triangle divides the triangle into
similar
two triangles that are
to the original triangle and
to each other.
similar
Corollary 1 to Theorem 7-3
The length of the altitude to the hypotenuse of a right triangle is the
geometric mean
of the lengths of the segments
of the hypotenuse.
3. In sunlight, a cactus casts a 9-ft shadow. At the same time,
a person 6 ft tall casts a 4-ft shadow. Use similar triangles
to find the height of the cactus.
x ft
6 ft
4 ft
9 ft
13.5 ft
The altitude to the hypotenuse of a right triangle separates the hypotenuse
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Corollary 2 to Theorem 7-3
so that the length of each leg of the triangle is the geometric mean of the
adjacent hypotenuse segment
length of the
and the length of the
hypotenuse
.
The geometric mean of two positive numbers a and b is the positive number x such
a 5 x.
that x
b
Examples.
1 Finding the Geometric Mean Find the geometric mean of 3 and 12.
x
3
x5
x2 5
Write a proportion.
12
36
x5
x5 6
!36
Cross-Product Property.
Find the positive square root.
The geometric mean of 3 and 12 is 6 .
136
34
Geometry Lesson 7-3
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 7-4
137
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
NAEP 2005 Strand: Geometry
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Cup
2 Finding Distance At a golf course, Maria drove her ball 192 yd straight
y
toward the cup. Her brother Gabriel drove his ball straight 240 yd, but not
toward the cup. The diagram shows the results. Find x and y, their remaining
distances from the cup.
Use Corollary 2 of Theorem 7-3 to solve for x.
240
Cross-Product
Property
y
Distributive
Property
If a line is parallel to one side of a triangle and intersects the other two sides,
then it divides those sides proportionally.
Theorem 7-5: Triangle-Angle-Bisector Theorem
If a ray bisects an angle of a triangle, then it divides the opposite side into two segments that
y
Substitute 108
are proportional to the other two sides of the triangle.
for x.
108
300
y
y
Corollary to Theorem 7-4
Simplify.
y
Cross-Product
!32,400
Find the positive square root.
Property
y 180
Maria’s ball is
yd from the cup, and Gabriel’s ball is
108
yd
180
from the cup.
Quick Check.
1. Find the geometric mean of 15 and 20.
2. Recall Example 2. Find the distance
between Maria’s ball and Gabriel’s ball.
10"3
144 yd
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
32,400
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
108
y2
If three parallel lines intersect two transversals, then the segments intercepted on the
transversals are proportional.
Name_____________________________________ Class____________________________ Date ________________
R
6
x
9
15
y5
x5
15
A
10
6
Substitute.
6
y
72
y
7.2
Cross-Product
Property
Solve for y.
Geometry Lesson 7-5
Example.
139
G
40
24
I
12.5
x
N
NA
Corollary to the Side-Splitter Theorem
75
15y C
10
Theorem
3 Using the Triangle-Angle-Bisector Theorem Find the value of x.
Cross-Product
x
IG GH
Solve for x.
12.5
Corollary to the Side-Splitter Theorem
y
12.5
Substitute
y
for x.
Cross-Products
112.5
7.5
and y 5
15
24
IK
Triangle-Angle-Bisector
Theorem.
HK
x
Substitute.
30
( 30 ) x
40
Property
Solve for y.
40
24
H
30
K
Property
All rights reserved.
15
Side-Splitter
12.5
y
T
9
U
5
x
x
9
.
12
M
CN
Name_____________________________________ Class____________________________ Date ________________
All rights reserved.
6
x5
UT
y
B
CM MB
S
5
x
Solve for x.
18
Quick Check.
7.5
3. Find the value of y.
3.6
5.76
1. Use the Side-Splitter Theorem to find the value of x.
5
x 1.5
2.5
x
2. Solve for x and y.
x ⫽ 225
13 ; y ⫽ 28.6
16.5
15
y
26
x
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Quick Check.
1.5
y
5
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Because RSTU is a trapezoid, RS 6
c
d
Examples.
Daily Notetaking Guide
Daily Notetaking Guide
2 Using the Corollary to the Side-Splitter Theorem The segments
joining the sides of trapezoid RSTU are parallel to its bases. Find x and y.
a
b5
d
1 Using the Side-Splitter Theorem Find y.
12
y
Geometry Lesson 7-4
c
a
b
10
5
Local Standards: ____________________________________
★
Solve for x.
Write a proportion.
y
138
Topic: Transformation of Shapes and Preservation of
Properties
Theorem 7-4: Side-Splitter Theorem
20,736
x
108 192
All rights reserved.
Use the Side-Splitter Theorem
Use the Triangle-Angle-Bisector
Theorem
Key Concepts.
Use Corollary 2 of Theorem 7-3 to solve for y.
y
NAEP 2005 Strand: Geometry
192
x 5 108
x 192
1
All rights reserved.
192 x 240
Proportions in Triangles
Lesson Objectives
192
57,600
36,864
Lesson 7-5
x
M
2
Write a proportion.
192 (x 1 192) 5 2402
192x G
All rights reserved.
x 192
240
Name_____________________________________ Class____________________________ Date________________
8
30
140
Geometry Lesson 7-5
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 7-5
Geometry
141
35
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 8-1
1
Example.
The Pythagorean Theorem and Its Converse
Lesson Objectives
1 Pythagorean Triples Find the length of the hypotenuse of #ABC.
NAEP 2005 Strand: Geometry
Use the Pythagorean Theorem
Use the Converse of the Pythagorean
Theorem
2
Name_____________________________________ Class____________________________ Date ________________
a2 1 b2 5 c2
Local Standards: ____________________________________
Î 169
a
.
b
a2 b2 c 2
Theorem 8-2: Converse of the Pythagorean Theorem
All rights reserved.
All rights reserved.
hypotenuse
is equal to the square of the length of the
c
If the square of the length of one side of a triangle is equal to the sum of
c
Take the square root.
A
5
C
Simplify.
The length of the hypotenuse is 13 . The lengths of the sides, 5, 12, and 13 ,
form a Pythagorean triple because they are
whole
numbers that
satisfy a2 1 b2 5 c2.
Quick Check.
the squares of the lengths of the other two sides, then the triangle is a
right
12
Simplify.
13 c
legs
Theorem.
Substitute 5 for a, and 12 for b.
169 c2
Theorem 8-1: Pythagorean Theorem
Pythagorean
Use the
52 122 c2
Vocabulary and Key Concepts.
In a right triangle, the sum of the squares of the lengths of the
B
Do the lengths of the sides of #ABC form a Pythagorean triple?
Topic: Relationships Among Geometric Figures
1. A right triangle has a hypotenuse of length 25 and a leg of length 10. Find the
length of the other leg. Do the lengths of the sides form a Pythagorean triple?
triangle.
5!21; no
obtuse
is
c
a
.
C
If c 2 a 2 b 2, the triangle is
obtuse
A
b
.
Theorem 8-4
B
If the square of the length of the longest side of a triangle is less than
the sum of the squares of the lengths of the other two sides, the triangle
acute
is
a
c
C
b
.
If c 2 a 2 b 2, the triangle is
acute
A
.
All rights reserved.
B
the sum of the squares of the lengths of the other two sides, the triangle
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
If the square of the length of the longest side of a triangle is greater than
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 8-3
A Pythagorean triple is a set of nonzero whole numbers a, b, and c that satisfy
the equation a 2 ⫹ b 2 ⫽ c 2.
Geometry Lesson 8-1
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Examples.
3 Is It a Right Triangle? Is this triangle a right triangle?
answer in simplest radical form
12
Use the
82 b2 122
Substitute 8 for a, 12 for c.
64 b2 144
Simplify.
b2 80
b
42
62
0
72
Substitute 4 for a, 6 for b, and 7 for c.
16
36
0
49
Simplify.
52
2
49
Subract 64 from both sides.
Because a2 b2 ⫽ c2, the triangle
Take the square root.
(
5
)
Simplify.
b 4 !5
c2 0 a2 1 b2
Geometry
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
36
Compare c 2 with a 2 b 2.
202 0
10 2
15 2
Substitute the greatest length for c .
400
0
100
225
Simplify.
400
.
325
Because c2 ⬎ a2 b2, the triangle is a(n)
Geometry Lesson 8-1
a right triangle.
and 20 represent the lengths of the sides of a triangle. Classify the triangle
as acute, obtuse, or right.
2. The hypotenuse of a right triangle has length 12. One leg has length 6. Find the
length of the other leg. Leave your answer in simplest radical form.
6 !3
is not
7m
6m
4 Classifying Triangles as Acute, Obtuse, or Right The numbers 10, 15,
Quick Check.
144
4m
a2 1 b2 0 c2
All rights reserved.
b Î 16
8
Theorem.
All rights reserved.
b Î 80
Pythagorean
143
Name_____________________________________ Class____________________________ Date ________________
Example.
2 Using Simplest Radical Form Find the value of b. Leave your
a2 1 b2 5 c2
Geometry Lesson 8-1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
142
obtuse
triangle.
Quick Check.
3. A triangle has sides of lengths 16, 48, and 50. Is the triangle a right triangle?
no
4. A triangle has sides of lengths 7, 8, and 9. Classify the triangle by its angles.
acute
Daily Notetaking Guide
Geometry Lesson 8-1
145
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 8-2
2 Applying the 45-45-90 Triangle Theorem The distance from
Special Right Triangles
Lesson Objectives
NAEP 2005 Strand: Geometry
Use the properties of 45-45-90
triangles
Use the properties of 30-60-90
triangles
1
2
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
and the
hypotenuse "2 ?
45 s
s2
length of the hypotenuse is "2 times the length of a leg.
45
s
leg
All rights reserved.
congruent
In a 45-45-90 triangle, both legs are
All rights reserved.
Theorem 8-5: 45-45-90 Triangle Theorem
hypotenuse 2
?
60
s
shorter leg
1 Finding the Length of the Hypotenuse Find the value of the
variable. Use the 45-45-90 Triangle Theorem to find the
hypotenuse.
hypotenuse "2
"12
45
h
5 6
? leg
Simplify.
45
4(3)
h 5( 2 ) 3
h5
5 6
"3
The length of the hypotenuse is
10"3
Divide each side by "2 .
67.882251
Use a calculator.
Each side of the playground is about
68
ft.
3 Using the Length of One Side Find the value of each variable.
Use the 30-60-90 Triangle Theorem to find the lengths of the legs.
hypotenuse 2 ? shorter leg
4"3
"3
y
Simplify.
60
y "3 ? shorter leg
y "3 ? 2 !3
30°-60°-90°
Substitute
y 2 ? "3 ? "3
y
30
4 3
Divide each side by 2 .
2
x 2
Examples.
h "2 ? 5!6
leg 30 s3
2s
.
shorter leg
longer leg "3 ?
Triangle Theorem
2"3
x
for shorter leg.
Simplify.
6
2"3
The length of the shorter leg is
6
, and the length of the longer leg is
.
Quick Check.
1. Find the length of the hypotenuse of a 45-45-90 triangle with legs of length 5 !3.
5!6
2. A square garden has sides of 100 ft long. You want to build a brick path along
a diagonal of the square. How long will the path be? Round your answer to the
nearest foot.
141 ft
.
Geometry Lesson 8-2
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 8-2
147
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Example.
4 Applying the 30-60-90 Triangle Theorem A rhombus-shaped
Lesson 8-3
25
garden has a perimeter of 100 ft and a 60 angle. Find the area of the
garden to the nearest square foot.
Because a rhombus has four sides of equal length, each side is
25
ft
h
25
ft.
Local Standards: ____________________________________
Vocabulary.
The tangent of acute ⬔A in a right triangle is the ratio of the length of the leg opposite
The height h is the longer leg of the right triangle. To find the height h, you
can use the properties of 30-60-90 triangles. Then apply the
area formula.
Divide each side by 2 .
h 12.5 "3
longer leg "3 ? shorter leg
A bh
Use the formula for the area of a
A
A
(
25
)(12.5!3)
541.26588
To the nearest square foot, the area is
Substitute
25
for b and
parallelogram
12.5"3
A
Leg
⬔A
length of leg opposite lA
length of leg adjacent to lA
tangent of ⬔A opposite
adjacent
ft 2.
8
x 4, y 4 !3
30
60
x
y
4. A rhombus has 10-in. sides, two of which meet to form the indicated angle.
Find the area of each rhombus. (Hint: Use a special right triangle to find height.)
a. a 30 angle
b. a 60 angle
Geometry Lesson 8-2
All-In-One Answers Version A
Daily Notetaking Guide
.
Examples.
1 Writing Tangent Ratios Write the tangent ratios for ⬔A and ⬔B.
B
20
C
tan A tan B 50 !3 in.2
⬔A
opposite
C
adjacent to
You can abbreviate the equation as tan A 3. Find the value of each variable.
148
Leg
for h.
Use a calculator.
541
B
.
Quick Check.
50 in.2
All rights reserved.
12.5
hypotenuse 2 ? shorter leg
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2
lA to the length of the leg adjacent to lA.
All rights reserved.
25
shorter leg NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
60°
Because a rhombus is also a parallelogram, you can use the area formula
A bh . Draw the altitude h, and then solve for h.
25 2 ? shorter leg
The Tangent Ratio
Lesson Objective
1 Use tangent ratios to determine side
lengths in triangles
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
shorter leg
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
. The length of the longer leg
is "3 times the length of the
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
96
96 ft
"2 ? leg
hypotenuse "2
x
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
shorter leg
the length of the
146
s
twice
In a 30-60-90 triangle, the length of the hypotenuse is
10
96 "2 ? leg
4 !3 2 ? x
Theorem 8-6: 30-60-90 Triangle Theorem
h
s
leg Key Concepts.
h 5
one corner to the opposite corner of a square playground is 96 ft.
To the nearest foot, how long is each side of the playground?
The distance from one corner to the opposite corner, 96 ft, is the
length of the hypotenuse of a 45-45-90 triangle.
29
21
A
opposite
adjacent
opposite
adjacent
Daily Notetaking Guide
BC
AC
AC
BC
20
21
21
20
Geometry Lesson 8-3
Geometry
149
37
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
2 Using a Tangent Ratio To measure the height of a tree, Alma walked
Quick Check.
125 ft from the tree and measured a 32 angle from the ground to the top
of the tree. Estimate the height of the tree.
1. a. Write the tangent ratios for ⬔K and ⬔J.
J
3, 7
7 3
3
L
right
The tree forms a
K
7
32
125 ft
angle with the ground, so you can use the
Use the tangent ratio.
125
(
height 125
)
tan 32°
78.108669
32
78
The tree is about
Solve for height.
All rights reserved.
height
tan 32 125
All rights reserved.
tangent ratio to estimate the height of the tree.
b. How is tan K related to tan J?
They are reciprocals.
Use a calculator.
2. Find the value of w to the nearest tenth.
a.
b.
10
54
feet tall.
3 Using the Inverse of Tangent Find m⬔R to the nearest degree.
c.
w
1.0
w
w
2.5
28
57
33
S
T
47
tan R 47
41
1
m⬔R tan
47
⫺1
( )
47
41
1.14634146
48.900494
41
1.14634146
1.9
13.8
3.8
3. Find m⬔Y to the nearest degree.
100
P
T
68
41
Y
All rights reserved.
41
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
R
So m⬔R 49 .
Geometry Lesson 8-3
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Lesson 8-4
Name_____________________________________ Class____________________________ Date ________________
2 Using the Cosine Ratio A 20-ft wire supporting a flagpole forms a
Sine and Cosine Ratios
Lesson Objective
1 Use sine and cosine to determine side
lengths in triangles
151
Geometry Lesson 8-3
35 angle with the flagpole. To the nearest foot, how high is the flagpole?
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
The flagpole, wire, and ground form a
wire as the
Local Standards: ____________________________________
with the
right triangle
20 ft
35
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
150
. Because you know an angle and the
hypotenuse
measure of its hypotenuse, you can use the
ratio
cosine
to find the height of the flagpole.
Vocabulary.
cosine of ⬔A B
leg opposite lA
hypotenuse
This can be abbreviated sin A hypotenuse
opposite
hypotenuse
Leg
⬔A
C
.
A
Leg
adjacent
hypotenuse
.
A
T
sin T cos T 38
adjacent
hypotenuse
20
5
16
4
5
20
16
20
5
adjacent
cos G hypotenuse 12
3
20
Geometry Lesson 8-4
Geometry
16
R
20
12
G
4
opposite
hypotenuse
sin G 152
hypotenuse
3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
.
sin G, and cos G. Write your answers in simplest terms.
12
16.383041
16
Solve for height.
Use a calculator.
feet tall.
1.5
Examples.
1 Writing Sine and Cosine Ratios Use the triangle to find sin T, cos T,
35
cos 35°
C
⬔A
(90 x)⬚
?
long and hypotenuse 4.0 units long. Find the measures of its acute angles to
the nearest degree.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
An example of a trigonometric identity is sin x cos
20
The flagpole is about
adjacent to
An identity is an equation that is true for all allowed values of the variable.
opposite
20
Use the cosine ratio.
20
height 3 Using the Inverse of Cosine and Sine A right triangle has a leg 1.5 units
opposite
leg adjacent to lA
hypotenuse
This can be abbreviated cos A All rights reserved.
The cosine of ⬔A is the ratio of the length of the leg adjacent to ⬔A to the hypotenuse.
All rights reserved.
hypotenuse.
sine of ⬔A height
cos 35 The sine of ⬔A is the ratio of the length of the leg opposite ⬔A to the length of the
B
4.0
Use the inverse of the cosine function to find m⬔A.
1.5
cos A m⬔A
⫺1
4.0
cos1
(
)
67.975687
0.375
m⬔A 0.375
0.375
68
Use the cosine ratio.
Use the inverse of the cosine.
Use a calculator.
Round to the nearest degree.
To find m⬔B, use the fact that the acute angles of a right triangle are
complementary
.
m⬔A m⬔B 90
Definition of complementary angles
m⬔B 90
Substitute.
m⬔B 22
Simplify.
68
The acute angles, rounded to the nearest degree, measure
5
Daily Notetaking Guide
Daily Notetaking Guide
68
and
22 .
Geometry Lesson 8-4
153
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Quick Check.
Lesson 8-5
1. a. Write the sine and cosine ratios for ⬔X and ⬔Y.
X
48
48
64
sin X ⫽ 64
80, cos X ⫽ 80, sin Y ⫽ 80, cos Y ⫽ 80
Angles of Elevation
and Depression
NAEP 2005 Strand: Measurement
Lesson Objective
1
48
80
Z
64
Topic: Measuring Physical Attributes
Use angles of elevation and
depression to solve problems
Local Standards: ____________________________________
Y
Vocabulary.
An angle of elevation is the angle formed by a horizontal line and the line of sight to an
sin X ⫽ cos Y when lX and lY are complementary.
All rights reserved.
All rights reserved.
object above the horizontal line.
b. In general, how are sin X and cos Y related? Explain.
An angle of depression is the angle formed by a horizontal line and the line of sight to an
object below the horizontal line.
Horizontal line
Angle of
B
depression
Horizontal line
2. In Example 2, suppose that the angle the wire makes with the ground is 50°.
What is the height of the flagpole to the nearest foot?
20 ft
Examples.
3. Find the value of x. Round your answer to the nearest degree.
a.
b.
10
6.5
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
50
10
x
x
27
68
41
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
15 feet
Angle of
elevation
A
1 Identifying Angles of Elevation and Depression Describe ⬔1 and ⬔2
as they relate to the situation shown.
1
2
3
4
One side of the angle of depression is a horizontal line. ⬔1 is the angle of
depression
from the
airplane
to the
building
.
One side of the angle of elevation is a horizontal line. ⬔2 is the angle of
elevation
154
Geometry Lesson 8-4
Name_____________________________________ Class____________________________ Date ________________
Use the
?
200
tan 35⬚
ratio.
Theodolite 5 ft 35
200 ft
Use a calculator.
b. ⬔4
140
feet tall.
ft 140
The building is about
5
ft, or
145
ft tall.
All rights reserved.
The angle of elevation from the person on the ground to the building
To find the height of the building, add the height of the theodolite, which is
5
3 Aviation An airplane flying 3500 ft above the ground begins a 2 descent
to land at the airport. How many miles from the airport is the airplane when
it starts its descent? (Note: The angle is not drawn to scale.)
2
x
3,500 ft
You
x
2
3500
5280
The airplane is about
156
19
3,500
x
Use the
sine
ratio.
3500
Solve for x.
sin 2⬚
100287 .9792
18.993935
Use a calculator.
Divide by
5280
to convert
feet to miles.
miles from the airport when it starts its descent.
Geometry Lesson 8-5
All-In-One Answers Version A
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
sin 2⬚
2. You sight a rock climber on a cliff at a 32 angle of elevation. The horizontal
ground distance to the cliff is 1000 ft. Find the line-of-sight distance to the
rock climber.
Rock
climber
Ground
100287.9792
155
The angle of depression from the building to the person on the ground
All rights reserved.
x Geometry Lesson 8-5
Solve for x.
140 .041508
35
200
tangent
200
x .
1. Describe each angle as it relates to the situation in Example 1.
a. ⬔3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
tan 35 airplane
Quick Check.
x
x
to the
Name_____________________________________ Class____________________________ Date ________________
Top of
building
with a 5-ft tall theodolite. The angle of elevation to the top of the building is
35. How tall is the building? Use a diagram to represent the situation.
building
Daily Notetaking Guide
Daily Notetaking Guide
2 Surveying A surveyor stands 200 ft from a building to measure its height
from the
32
1000 ft
about 1179 ft
3. An airplane pilot sees a life raft at a 26 angle of depression. The airplane’s
altitude is 3 km. What is the airplane’s surface distance d from the raft?
about 6.8 km
Daily Notetaking Guide
Geometry Lesson 8-5
Geometry
157
39
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 8-6
y
1 Describing a Vector Describe OM as an ordered pair.
NAEP 2005 Strand: Geometry
Describe vectors
Solve problems that involve vector
addition
2
Examples.
Vectors
Lesson Objectives
1
Name_____________________________________ Class____________________________ Date ________________
Give coordinates to the nearest tenth.
Use the sine and cosine ratios to find the values of x and y.
Topic: Position and Direction
Local Standards: ____________________________________
Vocabulary and Key Concepts.
Use sine and cosine.
80
x 80 ( cos 40
Adding Vectors
x
All rights reserved.
.
A vector is any quantity with magnitude (size) and direction.
arrow
A vector can be represented with an
All rights reserved.
kx1 1 x2, y1 1 y2 l
For a x1, y1 and c x2, y2, a c x
x
40 O
x
cos 40 .
)
Solve for the variable.
61.28355545
Use a calculator.
Because point M is in the
third
y
80
y 80( sin 40
y
80
M
)
51.42300878
quadrant, both coordinates are
. To the nearest tenth, OM negative
y
sin 40 k261.3, 251.4l
.
2 Describing a Vector Direction A boat sailed 12 mi east and 9 mi south.
The trip can be described by the vector k12, 29l. Use distance and
direction to describe this vector a second way.
The magnitude of a vector is its size, or length.
N
Draw a diagram for the situation.
The initial point of a vector is the point at which it starts.
the Distance Formula
To find the distance sailed, use
magnitude
The
initial point
of vector ON is the distance from its
terminal point
O to its
k5, 2l
notation for the vector is
N. The ordered pair
.
y
4
2
4 2 O
N
2
4x
d ( 12 0
)2 (
d
144
)2
9 0
81
W
12 mi
(0, 0)
E
9 mi
(12, 9)
Simplify.
d "225
d
.
Distance Formula
S
Simplify.
15
All rights reserved.
A resultant vector is the sum of other vectors.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The terminal point of a vector is the point at which it ends.
Take the square root.
To find the direction the boat sails, find the angle that the vector forms with
the x-axis.
9
tan x 0.75
Use the tangent ratio.
12
x
⫺1
tan1
(0.75)
The boat sailed
Find the angle whose tangent is 0.75.
36.869898
0.75
15
miles at about
Use a calculator.
37 south of
east
.
2
4
Geometry Lesson 8-6
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
3 Adding Vectors Vectors v 4, 3 and w 4, 3 are shown at the right.
Write the sum of the two vectors as an ordered pair. Then draw s,
the sum of v and w.
To find the first coordinate of s, add the
first
2
To find the second coordinate of s, add the second coordinates of
2
v and w.
4
8x
4
6
(4, 3)
Vocabulary.
,
3 (3)
l Add the coordinates.
4
Simplify.
2
Draw vector v using the origin as the initial point. Draw vector w
using the terminal point of v, (4, 3), as the initial point. Draw the
resultant vector s from the
the
terminal point
initial point
of v to
of w.
A transformation of a geometric figure is a change in its position, shape, or size.
v
2
w
4
s8 x
6
All rights reserved.
44
y
All rights reserved.
k
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties; Position and Direction
Local Standards: ____________________________________
s k4, 3l k4, 3l
k 8 , 0 l
Translations
Lesson Objectives
1 Identify isometries
2 Find translation images of figures
(4, 3)
coordinates of
v and w.
Lesson 9-1
2
159
Name_____________________________________ Class____________________________ Date ________________
y
4
Geometry Lesson 8-6
In a transformation, the preimage is the original image before changes are made.
In a transformation, the image is the resulting figure after changes are made.
2
An isometry is a transformation in which the preimage and the image are congruent.
4
A translation (slide) is a transformation that maps all points the same distance and in the
Quick Check.
same direction.
y
51
k221.6, 46.2l
65
O
x
2. A small airplane lands 246 mi east and 76 mi north of the point from which it
took off. Describe the magnitude and the direction of its flight vector.
about 257 mi at 17 N of E
3. Write the sum of the two vectors k2, 3l and k24, 22l as an ordered pair.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A composition of transformations is a combination of two or more transformations.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. Describe the vector at the right as an ordered pair. Give the coordinates to
the nearest tenth.
Examples.
1 Identifying Isometries Does the transformation appear to be an
isometry?
Image
Preimage
The image appears to be the same as the preimage, but
Because the figures appear to be
congruent
turned
.
, the transformation
appears to be an isometry.
k22, 1l
160
40
Geometry Lesson 8-6
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 9-1
161
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
158
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
b. List all pairs of corresponding sides.
2 Naming Images and Corresponding Parts In the diagram,
䉭XYZ is an image of 䉭ABC.
a. Name the images of ⬔B and ⬔C.
C Z
NI and SU; ID and UP ; ND and SP
Y
B
Because corresponding vertices of the preimage and the
image are listed in the same order, the image of ⬔B is ⬔ Y ,
A
X
3. Find the image of 䉭ABC for the translation (x 2, y 1).
and the image of ⬔C is ⬔ Z .
A(3, 3), B(1, 1), C(2, 0)
b. List all pairs of corresponding sides.
Because corresponding sides of the preimage and the image are
XZ , BC and
YZ .
3 Finding a Translation Image Find the image of 䉭ABC
y
under the translation (x, y) → (x 2, y 3).
A(1, 2) → A( 1 2 , 2 3)
A
Use the rule (x, y ) → (x 2, y 3)
B
B(1, 0) → B( 1 2, 0 3 )
All rights reserved.
XY , AC and
All rights reserved.
listed in the same order, the following pairs are corresponding sides:
AB and
x
A⬘(2, 0)
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Image
Yes; the figures appear to be
congruent by a flip and a
slide.
2. In the diagram, NID S SUP.
a. Name the images of ⬔I and point D.
U
lU, P
N
S
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The rule is (x, y) S
Quick Check.
All rights reserved.
, the horizontal change is 2 ( ⫺4
P
(x ⫹ 6, y ⫺ 1)
y
B
), or
6 ,
A
C
4 2
A
2
x
C
2
and the vertical change is 0 1 , or ⫺1 .
and C( 2 , 4 ).
Yes; the figures appear to be
congruent by a flip.
4
2
to describe the translation. Using A(4, 1) and its image
The image of 䉭ABC is 䉭ABC with A( 1 , 1 ), B( 3 , 3 ),
1. Does the transformation appear to be an isometry? Explain.
a.
b.
Preimage
Preimage
Image
B
describe the translation 䉭ABC S 䉭ABC.
You can use any point on 䉭ABC and its image on 䉭ABC
C
C(0, 1) → C( 0 2 , 1 3 )
Examples.
4 Writing a Rule to Describe a Translation Write a rule to
4
.
5 Adding Translations Tritt rides his bicycle 3 blocks north and 5 blocks east of a
pharmacy to deliver a prescription. Then he rides 4 blocks south and 8 blocks
west to make a second delivery. How many blocks is he now from the pharmacy?
The rule (x, y) → (x ⫹ 5, y ⫹ 3) represents a ride of 3 blocks north and 5 blocks east.
The rule (x, y) → (x ⫺ 8, y ⫺ 4) represents a ride of 4 blocks south and 8 blocks
west.
Tritt’s position after the second delivery is the composition of the two
translations.
( 0 , 0 ) translates to ( 0 5 , 0 3 ) or ( 5 , 3 ). Then, ( 5 , 3 )
translates to ( 5 ⫺8 , 3 ⫺4 ) or ( ⫺3 , ⫺1 ).
Tritt is 1 block
south
and 3 blocks
of the
west
pharmacy.
I
D
Geometry Lesson 9-1
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 9-1
163
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Quick Check.
Lesson 9-2
4. Use the rule (x, y) S (x 7, y 1) to find the translation image of 䉭LMN.
Graph the image 䉭LMN.
y
6
Reflections
Lesson Objectives
1 Find reflection images of figures
2
2
M
2
4
L
4
6
A reflection in line r is a transformation such that if a point A is on line r, then
x
N
M
6
All rights reserved.
2
N
Vocabulary.
All rights reserved.
6 4
L
NAEP 2005 Strand: Geometry
Topics: Transformation of Shapes and Preservation of
Properties
Local Standards: ____________________________________
4
the image of A is
itself
B is the point such that r is the
, and if a point B is not on line r, then its image
perpendicular bisector
of BBr.
r
A A
B
Line of
5. Refer to Example 5. Tritt now makes a third delivery. Starting from where
the second delivery was, he goes 5 blocks north and 1 block west. How
many blocks is Tritt from the pharmacy?
reflection
B
He is 4 blocks north and 4 blocks west of the pharmacy.
Example.
164
Geometry Lesson 9-1
All-In-One Answers Version A
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1 Finding Reflection Images If point Q(1, 2) is reflected across
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
162
line x 1, what are the coordinates of its reflection image?
y x=1
Q
Q is 2 units to the left of the reflection line, so its image Q is
2 units to the right of the reflection line. The reflection line is the
x
2
2
perpendicular bisector of QQr if Q is at (3, 2) .
Quick Check.
1. What are the coordinates of the image of Q if the reflection
line is y 1?
(1, 4)
Daily Notetaking Guide
Geometry Lesson 9-2
Geometry
165
41
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
Lesson 9-3
2 Drawing Reflection Images 䉭XYZ has vertices X(0, 3), Y(2, 0),
y
and Z(4, 2). Draw 䉭XYZ and its reflection image in the line x 4.
First locate vertices X, Y, and Z and draw 䉭XYZ in a coordinate
plane.
x4
‚
4
X
2
O
2
of XXr, YYr, and ZZr.
Y
6
Topic: Transformation of Shapes and Preservation of
Properties
Local Standards: ____________________________________
8x
W
, PD and
P
PDr and PW also form congruent angles with line ᐉ
All rights reserved.
A rotation of x about a point R is a transformation for which the following
are true:
PDr form congruent angles with line ᐉ.
艎
(that is, R itself
• The image of R is
• For any point V, RV RV and m⬔VRV x
The center of a regular polygon is a point that is
vertices
from the polygon’s
V
R
R
).
R
x
.
V
equidistant
.
angles are congruent.
vertical
Therefore, PD and PW form congruent angles with line ᐉ
by the
Draw and identify rotation images of
figures
‚
D
Show that PD and PW form congruent angles with line ᐉ.
because
NAEP 2005 Strand: Geometry
Vocabulary.
3 Congruent Angles D is the reflection of D across ᐉ.
isometry
1
2
Draw the reflection image XYZ.
Because a reflection is an
Y
‚
All rights reserved.
perpendicular bisector
X
Z Z
Locate points X, Y, and Z such that the line of reflection x 4
is the
Rotations
Lesson Objective
Examples.
Property.
Transitive
1 Drawing a Rotation Image Draw the image of 䉭LOB under a 60 rotation
D
about C.
Step 1 Use a protractor to draw a 60 angle at vertex C with one side CO.
A
4 C C
B
B
8 4
4
8
12 x
4
8
3. In Example 3, what kind of triangle is 䉭DPD? Imagine the image of point W
reflected across line ᐉ. What can you say about 䉭WPW and 䉭DPD?
isosceles; they are similar by SAS
Step 3 Locate L⬘ and B⬘ in a similar manner. Then draw 䉭L⬘O⬘B⬘.
C
O
O
L
L
L
B
B
C
O
O
L
Name_____________________________________ Class____________________________ Date________________
A
into six equilateral triangles.
a. Name the image of B for a 240 rotation about M.
B
Because 360 6 60 , each central angle of ABCDEF measures
60° . A 240 counterclockwise rotation about center M moves point B
across
triangles. The image of point B is point D .
four
L
167
Geometry Lesson 9-3
3 Finding an Angle of Rotation A regular 12-sided polygon can be formed
by stacking congruent square sheets of paper rotated about the same center
on top of each other. Find the angle of rotation about M that maps W to B.
Consecutive vertices of the three squares form the outline of a regular
12-sided polygon.
M
D
360 12 b. Name the image of M for a 60° rotation about F.
䉭AMF is equilateral, so ⬔AFM has measure 180 3 B
L
.Examples.
C
E
B
O
L
Name_____________________________________ Class____________________________ Date ________________
B
F
B
O
L
C
O
Daily Notetaking Guide
Daily Notetaking Guide
2 Identifying a Rotation Image Regular hexagon ABCDEF is divided
B
C
O
L
Geometry Lesson 9-2
C
O
60°
B
166
C
O
30 , so each vertex of the polygon is a
30⬚
B
W
M
rotation
K
about point M.
60 . A 60
60
You must rotate counterclockwise through 7 vertices to map point W to
image of M under a 60 rotation about point F is point A .
Quick Check.
All rights reserved.
All rights reserved.
rotation of 䉭AMF about point F would superimpose FM on FA, so the
1. Draw the image of LOB for a 50 rotation about point B. Label the vertices
of the image.
point B, so the angle of rotation is 7 ?
30 210
.
4 Compositions of Rotations Describe the image of quadrilateral XYZW
for a composition of a 145 rotation and then a 215 rotation, both about
point X.
Two rotations of 145 and 215 about the same point make a total rotation of
145 215, or
360
. Because this forms a
about point X, the image is
the preimage XYZW
complete rotation
.
O
Quick Check.
O
L
2. Regular pentagon PENTA is divided into 5 congruent triangles. Name the
image of T for a 144 rotation about point X.
P
A
E
E
X
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
B B
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
L
3. In the figure from Example 3, find the angle of rotation about M that maps
B to K.
270°
4. Draw the image of the kite for a composition rotation of two 90 rotations
about point K.
K
T
168
42
Geometry Lesson 9-3
Geometry
N
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 9-3
169
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
Step 2 Use a compass to construct COr CO.
All rights reserved.
2. 䉭ABC has vertices A(3, 4), B(0, 1), and C(2, 3). Draw 䉭ABC and its
reflection image in the line x 3.
y
8
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Quick Check.
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 9-4
Examples.
Symmetry
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
1 Identifying Lines of Symmetry Draw all lines of symmetry for the
isosceles trapezoid.
NAEP 2005 Strand: Geometry
Identify the type of symmetry in
a figure
Topic: Transformation of Shapes and Preservation of
Properties
Draw any lines that divide the isosceles trapezoid so that
half of the figure is a mirror image of the other half.
Local Standards: ____________________________________
Vocabulary.
Line symmetry is the same as reflectional symmetry.
All rights reserved.
A figure has reflectional symmetry if there is a reflection that maps the figure onto itself.
All rights reserved.
A figure has symmetry if there is an isometry that maps the figure onto itself.
There is
one
line of symmetry.
2 Identifying Rotational Symmetry Judging from appearance, do the
letters V and H have rotational symmetry? If so, give an angle of rotation.
The letter V does not have rotational symmetry because it must be rotated
before it is its own image.
360
Line of
Symmetry
The letter H is its own image after one half-turn, so it has rotational
symmetry with a
3 Finding Symmetry A nut holds a bolt in place. Some nuts have square
reflectional
)
A figure has rotational symmetry if it is its own image for some rotation of 180⬚ or less.
A figure has point symmetry if it has 180⬚ rotational symmetry.
The figure is its own image after one half-turn, so it has
rotational symmetry with a
The figure also has
point
180
angle of rotation.
symmetry.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
(
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The heart-shaped figure has
symmetry.
line
angle of rotation.
180
faces, like the top view shown below. Tell whether the nut has rotational
symmetry and/or reflectional symmetry. Draw all lines of symmetry.
The nut has a square outline with a circular opening. The square and circle
are concentric.
The nut is its own image after one quarter-turn, so it has
symmetry.
rotational
90
The nut has 4 lines of symmetry.
170
Geometry Lesson 9-4
Daily Notetaking Guide
Daily Notetaking Guide
171
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Quick Check.
Lesson 9-5
1. Draw a rectangle and all of its lines of symmetry.
Dilations
Lesson Objective
1 Locate dilation images of figures
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
Local Standards: ____________________________________
All rights reserved.
All rights reserved.
Vocabulary.
A dilation is a transformation with center C and scale factor n for which
the following are true:
itself
• The image of C is
)
(that is, C C ).
• For any point R, R is on CR and CR n C
CR .
R
R
1
3
Q
2. a. Judging from appearance, tell whether the figure at the right has
rotational symmetry. If so, give the angle of rotation.
dilation
RrQr is the image of RQ under a
with center C and scale factor 3 .
Q
yes; 180°
An enlargement is a dilation with a scale factor greater than 1.
b. Does the figure have point symmetry?
3. Tell whether the umbrella has rotational symmetry about a line and/or
reflectional symmetry.
The umbrella has both rotational and reflectional symmetry.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
yes
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Geometry Lesson 9-4
A reduction is a dilation with a scale factor less than 1.
Examples.
1 Finding a Scale Factor Circle A with 3-cm diameter and center C is a
dilation of concentric circle B with 8-cm diameter. Describe the dilation.
The circles are concentric, so the dilation has center C .
Because the diameter of the dilation image is smaller, the dilation is a
reduction
Scale factor:
.
diameter of dilation image
diameter of preimage 5
3
8
The dilation is a reduction with center C and scale factor
172
Geometry Lesson 9-4
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
3
8
.
Geometry Lesson 9-5
Geometry
173
43
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
2 Scale Drawings The scale factor on a museum’s floor plan is 1 : 200. The
Quick Check.
length and width of one wing on the drawing are 8 in. and 6 in. Find the
actual dimensions of the wing in feet and inches.
1. Quadrilateral JKLM is a dilation image of quadrilateral JKLM.
Describe the dilation.
1
The floor plan is a reduction of the actual dimensions by a scale factor of 200
.
Multiply each dimension on the drawing by 200 to find the actual
dimensions. Then write the dimensions in feet and inches.
in. 6 in. 200 1200
in. The museum wing measures
133
ft,
100
ft
133 ft, 4 in.
by
.
100 ft
3 Graphing Dilation Images 䉭ABC has vertices A(2, 3), B(0, 4), and
C(6, 12). What are the coordinates of the image of 䉭ABC for a dilation
with center (0, 0) and scale factor 0.75?
The scale factor is 0.75 , so use
B(0, 4) → B(
the rule (x, y ) → (0.75x, 0.75y) .
C(6, 12) → C(
, 0.75 0.75 6
4 )
K
x
K
4
2
L
L
The dilation is a reduction with center (0, 0) and scale factor 1
2.
2. The height of a tractor-trailer truck is 4.2 m. The scale factor for a model
A(2, 3) → A(0.75 ? (2) , 0.75 (3) )
0.75 0
O
M
M
3
in.
4
All rights reserved.
1600
All rights reserved.
8 in. 200 y J
J
2
1 . Find the height of the model to the nearest centimeter.
truck is 54
8 cm
, 0.75 (12) )
, B
(4.5, ⫺9)
, and C
(0, 3)
.
3. Find the image of 䉭PZG for a dilation with center (0, 0) and scale factor 12.
Draw the reduction on the grid and give the coordinates of the image’s
vertices.
y
2
1
Z Z
2
P
1P 2 x
1
1
G
2
Geometry Lesson 9-5
G
(1, 0)
Vertices: P
174
a2 1 , 1 b
2 4
, Z
, and G
a 1 , 21b
2
Daily Notetaking Guide
Daily Notetaking Guide
All rights reserved.
(⫺1.5, ⫺2.25)
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The vertices of the reduction image of 䉭ABC are
Geometry Lesson 9-5
175
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 9-6
Examples.
1 Composition of Reflections in Intersecting Lines The letter D is
Compositions of Reflections
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation
of Properties
reflected in line x and then in line y. Describe the resulting rotation.
Find the image of D through a reflection across line x.
Find the image of the reflection through another reflection across line y.
Local Standards: ____________________________________
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objectives
1 Use a composition of reflections
2 Identify glide reflections
D
D
Vocabulary and Key Concepts.
43
A
D
y
reflections
.
Theorem 9-2
A composition of reflections across two parallel lines is a
.
translation
All rights reserved.
A translation or rotation is a composition of two
All rights reserved.
Theorem 9-1
x
The composition of two reflections across intersecting lines is a
The center of rotation is
is
twice the angle formed by the intersecting lines
86 clockwise, or
Theorem 9-3
is rotated
A composition of reflections across two intersecting lines is a
rotation at point A .
rotation
.
rotation
.
the point where the lines intersect , and the angle
274
. So, the letter D
counterclockwise, with the center of
2 Finding a Glide Reflection Image 䉭ABC has vertices A(4, 5), B(6, 2),
composition of at most three
.
reflections
Theorem 9-5: Isometry Classification Theorem
There are only four isometries. They are the following:
Reflection
Translation
R
R
R R
R
Glide reflection
Rotation
R
R
R
R
A glide reflection is the composition of a glide (translation) and a reflection across a line
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
In a plane, one of two congruent figures can be mapped onto the other by a
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 9-4: Fundamental Theorem of Isometries
and C(0, 0). Find the image of 䉭ABC for a glide reflection where the
translation is (x, y) → (x, y 2) and the reflection line is x 1.
y
A
A
A
B
6
4
2
C
4 2 C
x 1
(4, 5) S (4 0 , 5 2 ), or ( 4 , 7 )
C
2
First, translate 䉭ABC by (x, y) → (x, y 2).
B
B
4
6
(6,2) S (6 0 , 2 2 ), or ( 6 , 4 )
x
(0, 0) S (0 0 , 0 2 ), or ( 0 , 2 )
Then, reflect the translated image across the line x 1 .
2
The glide reflection image 䉭ABC has vertices A
B
(⫺4, 4)
, and C
(2, 2)
(6, 7)
,
.
parallel to the direction of translation.
176
44
Geometry Lesson 9-6
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 9-6
177
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Quick Check.
Lesson 9-7
1. a. Reflect the letter R across a and then b. Describe the resulting rotation.
a
R
1
b
R
Answers may vary. The result is a clockwise rotation about point c
through an angle of 2mlacb.
Tessellations
Lesson Objectives
2
NAEP 2005 Strand: Geometry
Identify transformation in tessellations,
and figures that will tessellate
Identify symmetries in tessellations
Topic: Geometry
Local Standards: ____________________________________
R
c
R
Answers may vary. The result is a translation twice the distance
between ᐉ and m.
R
All rights reserved.
R
b. Use parallel lines ᐉ and m. Draw R between ᐉ and m. Find the image of
R for a reflection across line ᐉ and then across line M. Describe the
resulting translation.
All rights reserved.
Vocabulary and Key Concepts.
Theorem 9-6
Every triangle tessellates.
Theorem 9-7
Every quadrilateral tessellates.
A tessellation, or tiling, is a repeating pattern of figures that completely covers a plane,
艎
m
without gaps or overlaps.
Translational symmetry is the type of symmetry for which there is a translation that maps a
2. a. Find the image of 䉭TEX under a glide reflection
where the translation is (x, y) → (x 1, y) and the
reflection line is y 2. Draw the translation first,
then the reflection.
Glide reflectional symmetry is the type of symmetry for which there is a glide reflection that
E E 4
T
x
2
2
yes; Order of steps does not matter in a glide
reflection.
X
T
maps a figure onto itself.
2
X X
6 4 2 O
4
6
y 2
4
6
E
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
T
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
b. Would the result of part (a) be the same if you reflected
䉭TEX first, and then translated it? Explain.
figure onto itself.
y
Examples.
1 Determining Figures That Will Tessellate Determine whether a regular
15-gon tessellates a plane. Explain.
Because the figures in a tessellation do not overlap or leave gaps, the sum of
the measures of the angles around any vertex must be
a
a
a
(n
180
2
180
(
15
15
for n.
Simplify.
is not
a factor of 360, a regular 15-gon
tessellate a plane.
Geometry Lesson 9-7
179
and one adjoining
will completely cover the plane without gaps or
overlaps. Use arrows to show a translation.
3 Identifying Symmetries in Tessellations List the
Starting at any vertex, the tessellation can be mapped
onto itself using a
has
point
180⬚
rotation, so the tessellation
symmetry centered at any vertex.
All rights reserved.
All rights reserved.
symmetries in the tessellation.
translation
The tessellation also has
translational
b.
symmetry, as can be seen by sliding any triangle
Quick Check.
1. Explain why you can tessellate a plane with an equilateral triangle.
The interior angles of an equilateral triangle measure 60⬚. Because 60
divides evenly into 360, it will tessellate.
180
Geometry Lesson 9-7
All-In-One Answers Version A
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
onto a copy of itself along any of the lines.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Substitute
15
2. Identify a transformation and outline the smallest repeating figure in each
tessellation below.
a.
repeating figures and a transformation in the tessellation.
octagon
)
Name_____________________________________ Class____________________________ Date ________________
2 Identifying the Transformation in a Tessellation Identify the
square
Use the formula for the measure of an angle of a regular polygon.
2
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
A repeated combination of an
)
156
will not
Geometry Lesson 9-6
. Check
n
Because 156
178
360
to see whether the measure of an angle of a regular 15-gon is a factor of 360.
Answers may vary. Sample: rotation
3. List the symmetries in the tessellation.
line symmetry, rotational symmetry,
glide reflectional symmetry,
translational symmetry
Daily Notetaking Guide
Geometry Lesson 9-7
Geometry
181
45
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 10-1
1 Finding a Missing Dimension A parallelogram has 9-in. and 18-in. sides.
NAEP 2005 Strand: Measurement
The height corresponding to the 9-in. base is 15 in. Find the height
corresponding to the 18-in. base.
First find the area of the parallelogram using the 9-in. base and its
corresponding 15-in height.
Topic: Measuring Physical Attributes
Local Standards: ____________________________________
A bh
Vocabulary and Key Concepts.
its
and
base
A
h
.
height
b
bh
Theorem 10-2: Area of a Parallelogram
height
and the corresponding
A
Area of a parallelogram
A 9
(
A
135
A
bh
18
135
18
7.5
Theorem 10-3: Area of a Triangle
and the corresponding
base
A
h
.
height
b
1bh
2
A base of a parallelogram is any of its sides.
Altitude
Base
The altitude of a parallelogram corresponding to a given base is
the segment perpendicular to the line containing that base
drawn from the side opposite the base.
The height of a parallelogram is the length of its altitude.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
a
for h.
in.2
135
Area of a parallelogram
h
135
Substitute
h
Divide each side by
h
Simplify.
2 Finding the Area of a Triangle Find the area of 䉭XYZ.
A 5 12bh
A 5 12
(
A5
triangle
Area of a
30
)(
13
)
195
30
Substitute
X
for b and
13
䉭XYZ has area
195
cm2.
Y
Geometry Lesson 10-1
1. A parallelogram has sides 15 cm and 18 cm. The height corresponding to a
15-cm base is 9 cm. Find the height corresponding to an 18-cm base.
7.5 cm
18 cm
15 cm
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Lesson 10-2
3 Structural Design The front of a garage is a square 15 ft on each side
(
15
)(
10.6
)
79.5
15 ft
Find the force of the wind against the front of the garage.
Use the formula for force.
)(
80
)2
7795.2
7800
Substitute
304.5
for A and
the bases.
h
1h(b 1 b )
1
2
2
A
for v.
Round to the nearest hundred.
7800
The height of a trapezoid is the perpendicular
lb against the front
Quick Check.
2. Find the area of the triangle.
5 cm
30 cm2
13 cm
12 cm
3. Critical Thinking Suppose the bases of the square and triangle in Example 3
are doubled to 30 ft, but the height of each figure stays the same. How is the
force of the wind against the building affected?
The force is doubled.
Leg
Geometry
Leg
Base
Examples.
1 Applying the Area of a Trapezoid A car window is
20 in.
shaped like the trapezoid shown. Find the area of the
window.
18 in.
36 in.
A 5 12h(b1 1 b2)
(
18
)(
20
36
504
The area of the car window is
46
b1
h
b2
A5
Daily Notetaking Guide
Base
distance h between the bases.
A 5 12
Geometry Lesson 10-1
d2
d1
1d d
2 1 2
A
Simplify.
An 80 mi/h wind exerts a force of about
of the garage.
184
b2
Theorem 10-5: Area of a Rhombus or a Kite
diagonals.
80
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
304.5
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
F
b1
The area of a trapezoid is half the product of the height and the sum of
The area of a rhombus or a kite is half the product of the lengths of its
F 0.004Av 2
(
Local Standards: ____________________________________
Theorem 10-4: Area of a Trapezoid
15 ft
ft 2
The total area of the front of the garage is
79.5
304.5
225
ft 2.
F 0.004
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Vocabulary and Key Concepts.
10.6 ft
ft 2
225
All rights reserved.
12
All rights reserved.
Area of the triangle:
1bh
2
2
183
Areas of Trapezoids, Rhombuses, and Kites
Lesson Objectives
1 Find the area of a trapezoid
2 Find the area of a rhombus or a kite
with a triangular roof above the square. The height of the triangular roof
is 10.6 ft. To the nearest hundred, how much force is exerted by an 80 mi/h
wind blowing directly against the front of the garage? Use the formula
F 0.004Av 2, where F is the force in pounds, A is the area of the surface
in square feet, and v is the velocity of the wind in miles per hour.
Use the area formulas for rectangles and triangles to find the area of the
front of the garage.
F
Geometry Lesson 10-1
Name_____________________________________ Class____________________________ Date________________
Example.
15
Z
30 cm
Quick Check.
b
31 cm
13 cm
Simplify.
9 cm
containing that base.
bh
for h.
h
The height of a triangle is the length of the altitude to the line
Area of the square:
for b.
.
A base of a triangle is any of its sides.
182
18
for A and
18
The height corresponding to the 18-in base is 7.5 in.
the product of
half
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The area of a triangle is
15
Simplify.
Use the area to find the height corresponding to the 18-in. base.
b
bh
Substitute 9 for b and
The area of the parallelogram is
h
.
)
15
135
The area of a parallelogram is the product of a
base
All rights reserved.
The area of a rectangle is the product of
All rights reserved.
Theorem 10-1: Area of a Rectangle
All rights reserved.
Find the area of a parallelogram
Find the area of a triangle
Daily Notetaking Guide
)
Area of a
trapezoid
Substitute
18
for h,
20
for b1 , and
36
for b2 .
Simplify.
504
in.2.
Geometry Lesson 10-2
185
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
Examples.
Areas of Parallelograms and Triangles
Lesson Objectives
2
Name_____________________________________ Class____________________________ Date ________________
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
2 Finding Area Using a Right Triangle Find the area of
A
Name_____________________________________ Class____________________________ Date ________________
B
11 ft
Examples.
trapezoid ABCD. First draw an altitude from vertex B to DC
that divides trapezoid ABCD into a rectangle and a right
triangle. The altitude will meet DC at point X.
3 Finding the Area of a Kite Find the area of kite XYZW.
13 ft
DX 11
ft and XC 16 ft
D
X
A 5 12
C
By the Pythagorean Theorem, BX 2 XC 2 BC 2,
52
Taking the square root, BX 12
(
(
A5
)(
b2
)
11
16
.
ft. You may remember
triple
that 5, 12, 13 is a Pythagorean
b1
A 5 12 h
A 5 12 12
144
.
Use the trapezoid area formula.
)
12
Substitute
162
for h,
11
16
for b1 , and
for b2 .
Simplify.
A 5 12
( 6 )( 5 )
Substitute 6 for d1 and 5 for d2.
A5
15
Simplify.
4 cm
W
15
cm 2.
R
S
To find the area, you need to know the lengths of both diagonals.
Draw diagonal SU, and label the intersection of the diagonals point X.
䉭SXT is a
right
24
triangle because the diagonals of a rhombus
The diagonals of a rhombus bisect each other, so TX ft
13 ft
X
are perpendicular.
U
12
T
ft.
You can use the Pythagorean triple 5, 12, 13 or the Pythagorean Theorem
Quick Check.
to conclude that SX 1. Find the area of a trapezoid with height 7 cm and bases 12 cm and 15 cm.
SU 94.5 cm 2
A 5 12
2. In Example 2, suppose BX and BC change so that m⬔C 60 while bases
and angles A and D are unchanged. Find the area of trapezoid ABCD.
67.5 !3 or 116.9 ft 2
A 5 12
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
Z
3 cm
4 Finding the Area of a Rhombus Find the area of rhombus RSTU.
ft 2.
162
The area of trapezoid ABCD is
Use the formula for the area of a kite.
The area of kite XYZW is
All rights reserved.
13 2
All rights reserved.
so BX 2 d2
d1
3 cm
X
XZ 5 d1 5 3 1 3 5 6 and YW 5 d2 5 1 1 4 5 5
ft 11 ft 5 ft.
16
1 cm Y
Find the lengths of the diagonals of kite XYZW.
Because opposite sides of rectangle ABXD are congruent,
10
d1
(
A5
5
ft.
ft because the diagonals of a rhombus bisect each other.
d2
24
)(
Area of a rhombus
)
10
24
Substitute
120
for d1 and
10
for d2 .
Simplify.
The area of rhombus RSTU is
ft 2.
120
Quick Check.
3. Find the area of a kite with diagonals that are 12 in. and 9 in. long.
54 in 2
4. Critical Thinking In Example 4, explain how you can use a Pythagorean
triple to conclude that XU 5 ft.
5 2 ⫹ 12 2 ⫽ 13 2
Geometry Lesson 10-2
Daily Notetaking Guide
Daily Notetaking Guide
187
Geometry Lesson 10-2
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 10-3
2 Finding the Area of a Regular Polygon A library is in the shape
Areas of Regular Polygons
Lesson Objective
1 Find the area of a regular polygon
Local Standards: ____________________________________
a2 a
p
1 ap
2
All rights reserved.
perimeter.
All rights reserved.
The area of a regular polygon is half the product of the apothem and the
A
(
9.0
a2 Theorem 10-6: Area of a Regular Polygon
a<
18.0
Radius
To the nearest 10 ft 2, the area is
Divide 360 by the number of sides.
The apothem bisects the vertex angle of
m⬔1
isosceles
the
1
2
3
triangle formed by
the radii.
m⬔2 1
2
(
60
)
30
90
30
m⬔3 180 (
m⬔1 , m⬔2 188
60
Substitute
)
60
30 , and m⬔3 60
for m⬔1.
180
.
)(
)
144
for s, and simplify.
18.0
Daily Notetaking Guide
Area of a
regular
Substitute
21.7
polygon
144
for a and
for p.
Simplify.
1560
ft 2.
with apothem 8 cm. Leave your answer in simplest radical form.
This equilateral triangle shows two radii forming an angle that measures
s
360 120. Because the radii and a side form an isosceles triangle,
3
60⬚
angles.
the apothem bisects the 120 angle, forming two
60
8 cm 60
You can use a 30-60-90 triangle to find half the length of a side.
1s 5
2
1s 5
2
"3
s 5 16
p 5 ns
p5
( 3 )(
A 5 12 ap
A 5 12 8
A5
longer leg "3 ? shorter leg
?a
"3 ? 8
192
s
Substitute 8 for a.
"3
Multiply each side by 2 .
Find the perimeter.
"3
16
)
48
"3
Substitute 3 for n and
16 !3
for s, and simplify.
48 !3
for p.
Area of a regular polygon
48
"3
"3
)
The area of the equilateral triangle is
60
All-In-One Answers Version A
Substitute 8 for n and
3 Applying Theorem 10-6 Find the area of an equilateral triangle
( )(
The sum of the measures of the angles of a triangle is
Geometry Lesson 10-3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
radii drawn. Find the measure of each numbered angle.
2
144
Center
1 Finding Angle Measures This regular hexagon has an apothem and
m⬔2 )
A 12 ap
21.7
A 12
1562.4
A
Examples.
1
471.25
21.7
Step 3 Find the area A.
Apothem
60
9.0 ft
Solve for a.
Find the perimeter, where n the number of sides
of a regular polygon.
(
(
9.0 ft
Pythagorean Theorem
p ns
The apothem of a regular polygon is the perpendicular distance from the center to a side.
6
)2
23.5
552.25
Step 2 Find the perimeter p.
p (8)
The radius of a regular polygon is the distance from the center to a vertex.
)2 (
81
a2 The center of a regular polygon is the center of the circumscribed circle.
360
23.5 ft
Step 1 Find the apothem a.
Vocabulary and Key Concepts.
m⬔1 18.0 ft
of a regular octagon. Each side is 18.0 ft. The radius of the octagon is
23.5 ft. Find the area of the library to the nearest 10 ft 2.
Consecutive radii form an isosceles triangle, so an apothem bisects
the side of the octagon.
To apply the area formula A 12ap, you need to find a and p.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
186
Daily Notetaking Guide
Substitute 8 for a and
Simplify.
192"3
cm 2.
Geometry Lesson 10-3
Geometry
189
47
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Quick Check.
Lesson 10-4
1. At the right, a portion of a regular octagon has radii and an apothem drawn.
Find the measure of each numbered angle.
ml1 5 45; ml2 5 22.5; ml3 5 67.5
Perimeters and Areas of Similar Figures
Lesson Objective
1
1
NAEP 2005 Strand: Measurement and Number Properties
and Operations
Find the perimeters and areas of
similar figures
Topics: Systems of Measurement; Ratios and Proportional
Reasoning
2
3
Local Standards: ____________________________________
All rights reserved.
All rights reserved.
Key Concepts.
Theorem 10-7: Perimeters and Areas of Similar Figures
If the similarity ratio of two similar figures is ba , then
232 cm 2
a
b
(1) the ratio of their perimeters is
(2) the ratio of their areas is
2. Find the area of a regular pentagon with 11.6-cm sides and an 8-cm
apothem.
a2
b2
and
.
Examples.
1 Finding Ratios in Similar Figures The triangles at the right are
4
similar. Find the ratio (larger to smaller) of their perimeters and of
their areas.
384"3 ft 2
the shortest side of the right-hand triangle has length
5
4
the perimeters is
7.5
.
.
, and the ratio of the areas is
52
42
, or
25
16
Because the ratio of the lengths of the corresponding sides of the regular
octagons is 83, the ratio of their areas is
64
320
A
82
32
, or
64
9
.
Write a proportion.
2880
A
45
Cross-Product
Use the
Divide each side by
45
Property.
64 .
ft 2.
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
.
corresponding sides of two regular octagons is 83. The area of the larger
octagon is 320 ft 2. Find the area of the smaller octagon.
All regular octagons are similar.
The area of the smaller octagon is
Daily Notetaking Guide
5
6.25
2 Finding Areas Using Similar Figures The ratio of the length of the
9
Geometry Lesson 10-3
5
6
5
, and
By the Perimeters and Areas of Similar Figures Theorem, the ratio of
64 A 190
5
4
From larger to smaller, the similarity ratio is
4
All rights reserved.
3. The side of a regular hexagon is 16 ft. Find the area of the hexagon.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The shortest side of the left-hand triangle has length
Geometry Lesson 10-4
191
Name_____________________________________ Class____________________________ Date ________________
fields. Each dimension of the larger field is 3 12 times the dimension of the
Example.
smaller field. Seeding the smaller field costs $8. How much money does
seeding the larger field cost?
4 Finding Similarity and Perimeter Ratios The areas of two similar
pentagons are 32 in.2 and 72 in.2. What is their similarity ratio? What is the
ratio of their perimeters?
Find the similarity ratio a : b.
The similarity ratio of the fields is 3.5 : 1, so the ratio of the areas of the
12.25 to 1
.
fields is (3.5) 2 : (1) 2, or
a2
Because seeding the smaller field costs $8, seeding 12.25 times as much
12.25
land costs
$8 .
.
a2
Quick Check.
1. Two similar polygons have corresponding sides in the ratio 5 : 7.
a. Find the ratio of their perimeters.
b. Find the ratio of their areas.
5:7
All rights reserved.
$98
b2
All rights reserved.
Seeding the larger field costs
b2
a
32
72
16
b
4
6
Take the square root.
3
Perimeters and Areas of Similar Figures
Theorem,
3. The similarity ratio of the dimensions of two similar pieces of window glass
is 3 : 5. The smaller piece costs $2.50. What should be the cost of the larger
piece?
$6.94
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Quick Check.
54 in.2
Geometry
2
the ratio of the perimeters is also 2 : 3 .
The area of the larger parallelogram is 96 in.2. Find the area of the smaller
parallelogram.
48
Simplify.
36
By the
Geometry Lesson 10-4
The ratio of the areas is a 2 : b 2 .
The similarity ratio is 2 : 3 .
25 : 49
2. The corresponding sides of two similar parallelograms are in the ratio 34.
192
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3 Using Similarity Ratios Benita plants the same crop in two rectangular
4. The areas of two similar rectangles are 1875 ft 2 and 135 ft 2. Find the ratio of
their perimeters.
5!5 : 3
Daily Notetaking Guide
Geometry Lesson 10-4
193
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 10-5
A
NAEP 2005 Strand: Measurement
180(sin 36)
Substitute for a and p.
(cos 36) (sin 36)
Simplify.
770.355778
The area is about
Local Standards: ____________________________________
18(cos 36)
1620
A
Topic: Measuring Physical Attributes
Find the area of a regular polygon
using trigonometry
Find the area of a triangle using
trigonometry
2
A 12 Trigonometry and Area
Lesson Objectives
1
Name_____________________________________ Class____________________________ Date ________________
Use a calculator.
770
ft2.
2 Surveying A triangular park has two sides that measure 200 ft and 300 ft
and form a 65 angle. Find the area of the park to the nearest hundred
square feet.
Key Concepts.
Use Theorem 10-8: The area of a triangle is
B
and
the lengths of two sides
c
.
the sine of the included angle
A
1 bc(sin A)
2
Area of 䉭ABC a
C
b
All rights reserved.
The area of a triangle is one half the product of
All rights reserved.
Theorem 10-8: Area of a Triangle Given SAS
300 ft
of the included
sine
angle.
Area 12 side length side length sine of included angle
200
300
sin 65
Area 12 Area sin 65
30,000
65
200 ft
Theorem 9-1
Substitute.
Substitute.
27189.23361
Use a calculator.
ft2.
27,200
The area of the park is approximately
Examples.
the
one half
product of the lengths of two sides and the
1 Finding Area The radius of a garden in the shape of a regular pentagon
A⫽1
2ap
All rights reserved.
formula
C
.
18 ft
360
Because a pentagon has five sides, m⬔ACB 5
72 .
A
M
B
CA and CB are radii, so CA CB . Therefore, 䉭ACM 䉭BCM by
, so m⬔ACM 12m⬔ACB the HL Theorem
Use the
ratio to find a.
cosine
cos 36 a
18
a 18(cos 36)
36 .
sine
Use the
ratio to find AM.
36 AM
18
sin
Use the ratio.
AM Solve.
18(sin 36)
Use AM to find p. Because 䉭ACM 䉭BCM, AB 2 ? AM . Because
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Find the perimeter p and apothem a, and then find the area using the
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
is 18 feet. Find the area of the garden.
Quick Check.
1. Find the area of a regular octagon with a perimeter of 80 in. Give the area
to the nearest tenth.
482.8 in.2
2. Two sides of a triangular building plot are 120 ft and 85 ft long. They include
an angle of 85. Find the area of the building plot to the nearest square foot.
5081 ft 2
the pentagon is regular, p 5 ? AB .
So p 5(2 ? AM ) 10
? AM 10
?
18(sin 36)
180(sin 36) .
Finally, substitute into the area formula A 12ap.
194
Geometry Lesson 10-5
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Circles and Arcs
Lesson Objectives
1 Find the measures of central angles
and arcs
2 Find circumference and arc length
R
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
T
Local Standards: ____________________________________
TRS is a
C
B
measures of the two arcs.
1
0
0
mABC mAB 1 mBC
A
Theorem 10-9: Circumference of a Circle
C
pd
or C d
and the
circumference of the circle
r
O
.
RS is a
T
P
minor arc
.
mRS m⬔RPS
RTS is a
S
P
major arc
mTRS 180
A semicircle is half of a
A minor arc is smaller than
mRTS 360 mRS
A major arc is greater
circle.
a semicircle.
than a semicircle.
.
Adjacent arcs are arcs of the same circle that have exactly one point in common.
Concentric circles are circles that lie in the same plane and have the same center.
Congruent arcs are arcs that have the same measure and are in the same circle or in
congruent circles.
A
Example.
r
.
O
360
0
length of AB semicircle
R
S
T
C
Theorem 10-10: Arc Length
The length of an arc of a circle is the product of the ratio
measure of the arc
R
S
P
Arc length is a fraction of a circle’s circumference.
.
2pr
0
1
1 Finding the Measures of Arcs Find mXY and mDXM in circle C.
B
0
mAB ? 2pr
360
A circle is the set of all points equidistant from a given point called the center.
A center of a circle is the point from which all points are equidistant.
A radius is a segment that has one endpoint at the center and the other endpoint on the
circle.
Congruent circles have congruent radii.
A diameter is a segment that contains the center of a circle and has both endpoints on the
circle.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
p times the diameter
The circumference of a circle is
All rights reserved.
Postulate 10-1: Arc Addition Postulate
The measure of the arc formed by two adjacent arcs is the sum of the
All rights reserved.
Vocabulary and Key Concepts.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson 10-6
195
Geometry Lesson 10-5
0
0
0
mXY mXD mDY
0
0
mXY m/ XCD mDY
0
mXY 56 40
0
mXY 96
1
0
1
mDXM mDX mXWM
1
mDXM 56 180
1
236
mDXM Arc Addition Postulate
M
The measure of a minor arc is the measure
of its corresponding central angle.
Substitute.
Y
40
Simplify.
D
56
C
W
Arc Addition Postulate
X
Substitute.
Simplify.
Quick Check.
1
0
1
1. Use the diagram in Example 1. Find m⬔YCD, mYMW , mMW, and mXMY.
40; 180; 96; 264
A central angle is an angle whose vertex is the center of the circle.
Circumference of a circle is the distance around the circle.
Pi (π) is the ratio of the circumference of a circle to its diameter.
196
Geometry Lesson 10-6
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 10-6
Geometry
197
49
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Lesson 10-7
Examples.
2 Applying Circumference A circular swimming pool 16 feet in
1
POOL
The pool and the fence are concentric circles. The diameter of the pool
4 ft
is 16 ft, so the diameter of the fence is 16 4 4 24 ft. Use
the formula for the circumference of a circle to find the length of fencing
material needed.
24
)
75.36
About
76
Use 3.14 to approximate π.
Simplify.
1
3 Finding Arc Length Find the length of ADB in circle M in terms of π.
0
Because mAB 150,
1
mADB . Arc Addition Postulate
360
150
210
1
1
pr
length of ADB mADB
Arc Length Postulate
360 ? 2
210
2π
360
(
)
18
A
150
Quick Check.
2. The diameter of a bicycle wheel is 22 in. To the nearest whole number, how
many revolutions does the wheel make when the bicycle travels 100 ft?
17 revolutions
1.3p m
joining its endpoints.
Sector
1 Applying the Area of a Circle A circular archery target has a 2-ft diameter.
First find the areas of the archery target and the red bull’s-eye.
The radius of the archery target is 12 2 ft 1 ft 12 in.
(
)
The area of the archery target is πr 2 π
area of
red region
π
100
πr 2
135
424.11501
6 m 100
2
B
10
The area of sector ACB is
π
π
10π
m 2.
Quick Check.
A
120 24 ft
O
0
mAB
360
sector .
Substitute.
π
192
24
2 h
hypotenuse 2 ? shorter leg
12
h
Divide each side by 2 .
AB !3 ?
2
24
䉭AOB has base
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
)r
24
? 576π Simplify.
You can use a 30-60-90 triangle to find the
height h of 䉭AOB and the length of AB.
AB © Pearson Education, Inc., publishing as Pearson Prentice Hall.
? π(
360
1
Use the formula for area of a
2
Step 2 Find the area of 䉭AOB.
about 41 in.2
181.5 cm 2
2
? πr
120
3
1. How much more pizza is in a 14-in.-diameter pizza than in a 12-in. pizza?
2. Critical Thinking A circle has a diameter of 20 cm. What is the area of a sector
bounded by a 208 major arc? Round your answer to the nearest tenth.
199
B
area of sector AOB All rights reserved.
5
36
Simplify.
Step 1 Find the area of sector AOB.
A
)
All rights reserved.
424
π
in.2
the shaded segment. Round your answer to the nearest tenth.
5
18
in.2
9p
Geometry Lesson 10-7
360
in.
3 Finding the Area of a Segment of a Circle Find the area of
C
π( 6
3
area of
yellow region
π
9
in.2
144p
in. Example.
Leave your answer in terms of π.
360
)2 )
12
(6
The area of the red region is πr 2 π ( 3 )2 The radius of the red bull’s-eye region is 12
Name_____________________________________ Class____________________________ Date ________________
2 Finding the Area of a Sector of a Circle Find the area of sector ACB.
area of sector ACB (
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
of a circle
It is yellow except for a red bull’s-eye at the center with a 6-in. diameter.
Find the area of the yellow region to the nearest whole number.
The area of the yellow region is about
mAB
Segment
of a circle
Examples.
144
Daily Notetaking Guide
B
A segment of a circle is the part bounded by an arc and the segment
area of
archery target 3. Find the length of a semicircle with radius 1.3 m in terms of π.
Geometry Lesson 10-6
r
O
0
mAB ? pr2
360
intercepted arc.
Substitute.
A
360
A sector of a circle is a region bounded by two radii and their
D
21 π
1
The length of ADB is 21π cm.
198
measure of the arc
.
area of the circle.
Area of sector AOB 18 cm
M
A
r
O
The area of a sector of a circle is the product of the ratio
B
.
pr 2
Theorem 10-12: Area of a Sector of a Circle
and the
ft of fencing material is needed.
the product of p and the square of the radius
The area of a circle is
All rights reserved.
(
Substitute.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C<
Theorem 10-11: Area of a Circle
All rights reserved.
C < 3.14
Vocabulary and Key Concepts.
Formula for the circumference of a circle
24
Topic: Measuring Physical Attributes
4 ft
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Cπ
Find the areas of circles, sectors, and
segments of circles
Local Standards: ____________________________________
16 ft
All rights reserved.
pd
NAEP 2005 Strand: Measurement
12
"3
12"3
A
(
24"3
144
12"3
)(
24
12
ft.
60 60
12
ft
ft
30
B
ft
O
"3 ? shorter leg
longer leg Multiply each side by 2 .
ft A 5 12 bh
A 5 12
123
ft
123
A
12
)
"3
12"3
ft, or
24"3
Area of a
triangle
Substitute
24"3
ft, and height
.
12
for b and
for h.
Simplify.
Step 3 Subtract the area of 䉭AOB from the area of sector AOB to find the
area of the segment of the circle.
area of segment area of segment<
192
π 353.77047
144
"3
Use a calculator.
To the nearest tenth, the area of the shaded segment is
353.8
ft 2.
Quick Check.
3. A circle has a radius of 12 cm. Find the area of the smaller segment of the
circle determined by a 60 arc. Round your answer to the nearest tenth.
13.0 cm 2
200
50
Geometry Lesson 10-7
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 10-7
201
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
diameter will be enclosed in a circular fence 4 ft from the pool. What
length of fencing material is needed? Round your answer to the next
whole number.
C
Areas of Circles and Sectors
Lesson Objective
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 10-8
Example.
Geometric Probability
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
2 Finding Probability Using Area A circle is inscribed in a square target
NAEP 2005 Strand: Data Analysis and Probability
Use segment and area models to find
the probabilities of events
Local Standards: ____________________________________
Find the area of the square.
A s2 Vocabulary.
20 2
cm 2
400
Find the area of the circle. Because the square has sides of length 20 cm,
the circle’s diameter is 20 cm, so its radius is 10 cm.
Example.
1 Finding Probability Using Segments A gnat lands at random on the
All rights reserved.
All rights reserved.
Geometric probability is a model in which you let points represent outcomes.
edge of the ruler below. Find the probability that the gnat lands on a point
between 2 and 10.
A πr 2 π
(
10
)2 4
5
6
7
8
9
10
The length of the segment between 2 and 10 is
12
All rights reserved.
The length of the ruler is
10
(
A 400 100π
) cm 2
Use areas to calculate the probability that a dart landing randomly in the
square does not land within the circle. Use a calculator. Round to the
nearest thousandth.
area between square and circle
P(between square and circle) area of square
2
8
.
.
length of
favorable
segment
length of
entire
segment
8
23
12
Quick Check.
1. A point on AB is selected at random. What is the probability that it is
a point on CD?
A
0
C
1
2
3
4
5
6
7
D
B
8
9 10
1
π
4
0.215
The probability that a dart landing randomly in the square does not land
21.5
within the circle is about
%.
Quick Check.
2. Use the diagram in Example 2. If you change the radius of the circle as
indicated, what then is the probability of hitting outside the circle?
a. Divide the radius by 2.
about 80.4%
about 96.9%
Geometry Lesson 10-8
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Lesson 11-1
Example.
so that it lands entirely within the outer region of the circle at right. Find the
probability that this happens with a quarter of radius 15
32 in. Assume that the
quarter is equally likely to land anywhere completely inside the large circle.
15
The center of a quarter with a radius of 15
32 in. must land at least 32 in.
beyond the boundary of the inner circle in order to lie entirely outside the
inner circle. Because the inner circle has a radius of 9 in., the quarter must
in., or
9 15
32
9 in.
A
5 p°
15
Euler’s Formula
9
32
2
¢ <
281.66648
in.2
15 in. must land at least
Similarly, the center of a quarter with a radius of 32
15 in. within the outer circle. Because the outer circle has a radius of 12 in.,
32
15 in., or
the quarter must land inside the circle whose radius is 12 in. 32
11 17
32
Local Standards: ____________________________________
Vocabulary and Key Concepts.
in.
Find the area of the circle with a radius of 915
32 in.
pr2
NAEP 2005 Strand: Geometry
Topic: Dimension and Shape
12 in.
All rights reserved.
15
32
Space Figures and Cross Sections
Lesson Objective
1 Recognize polyhedra and their parts
2 Visualize cross sections of space
figures
All rights reserved.
land outside the circle whose radius is 9 in. in.
The numbers of faces (F), vertices (V), and edges (E) of a polyhedron
are related by the formula
F⫹V⫽E⫹2
.
A polyhedron is a three-dimensional figure whose surfaces are polygons.
A face is a flat surface of a polyhedron in the shape of a polygon.
Find the area of the circle with a radius of 1117
32 in.
32
¢ <
417.73672
in.2
Use the area of the outer region to find the probability that the quarter
lands entirely within the outer region of the circle.
417.73672 281.66648
P(outer region) area of outer circle area of large circle
417.73672
P(outer region) 136.07024
Rectangular
prism
Faces
0.32573
417.73672
The probability that the quarter lands entirely within the outer region of
32.6
0.326
the circle is about
or
%.
Quick Check.
3. Critical Thinking Use Example 3. Suppose you toss 100 quarters. Would you
expect to win a prize? Explain.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
11
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A pr2 5 p °
17
203
Geometry Lesson 10-8
Name_____________________________________ Class____________________________ Date ________________
3 Applying Geometric Probability To win a prize, you must toss a quarter
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
100π
b. Divide the radius by 5.
2
5
202
400
P(landing between 2 and 10)
400
11
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2
cm 2
100π
Find the area of the region between the square and the circle.
1
20 cm
with 20-cm sides. Find the probability that a dart landing randomly within
the square does not land within the circle.
Topic: Probability
An edge is a segment that is formed by the intersection of two
faces.
Edge
Vertex
A vertex is a point where three or more edges intersect.
A cross section is the intersection of a solid and plane.
Examples
Yes; theoretically you should win 32.6 times out of 100.
204
Geometry Lesson 10-8
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 11-1
Geometry
205
51
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Examples.
Example.
1 Identifying Vertices, Edges, and Faces How many vertices, edges,
and faces are there in the polyhedron shown? Give a list of each.
There are
four
There are
six
There are
four
A, B, C, and D
vertices:
the solid in Example 1.
A
Count the regions: F 4
.
AB, BC, CD, DA, AC, and BD
edges:
3 Verify Euler’s Formula for the two-dimensional net of
B
.
Count the vertices: V 6
faces: kABC, kABD, kACD, and kBCD .
C
Count the segments: E 9
D
4 6 9 1
FVE2
Euler’s Formula
6 8 E 2
Substitute the number of faces and vertices.
E
12
All rights reserved.
edges on a solid with 6 faces and 8 vertices.
Simplify.
A solid with 6 faces and 8 vertices has
All rights reserved.
2 Using Euler’s Formula Use Euler’s Formula to find the number of
Quick Check.
3. The figure at the right is a trapezoidal prism.
a. verify Euler’s Formula F V E 2 for the prism.
edges.
12
6 8 12 2
Quick Check.
1. List the vertices, edges, and faces of the polyhedron.
T
U
V
2. Use Euler’s Formula to find the number of edges on a polyhedron
with eight triangular faces.
12 edges
Geometry Lesson 11-1
Name_____________________________________ Class____________________________ Date________________
6 14 19 1
207
Geometry Lesson 11-1
Name_____________________________________ Class____________________________ Date ________________
Surface Areas of Prisms and Cylinders
Lesson Objectives
1 Find the surface area of a prism
2 Find the surface area of a cylinder
c. Verify Euler’s Formula F V E 1 for your
two-dimensional net.
Daily Notetaking Guide
Daily Notetaking Guide
Lesson 11-2
Possible answer:
A prism is a polyhedron with exactly two congruent, parallel
prism
Pentagonal
faces.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
The bases of a polyhedron are are the parallel faces.
Local Standards: ____________________________________
Bases
Lateral faces are the faces on a prism that are not bases.
Lateral
edges
An altitude is a perpendicular segment that joins the planes
Vocabulary and Key Concepts.
of the bases.
perimeter of the base
height
and the
All rights reserved.
The lateral area of a right prism is the product of the
.
L.A. ph
lateral area
The surface area of a right prism is the sum of the
areas of the two bases
the
and
.
All rights reserved.
Theorem 11-1: Lateral and Surface Area of a Prism
a bh
c
d Base
a b
h
h
h
Lateral Area
Height
Lateral Area ph
Surface Area L.A. 2B
Area of a base
Theorem 11-2: Lateral and Surface Area of a Cylinder
The lateral area of a right cylinder is the product of the
circumference of the base
height of the cylinder
and the
.
L.A. 2πrh, or L.A. πdh
The surface area of a right cylinder is the sum of the
areas of the two circular bases
lateral area
.
S.A. L.A. 2B, or S.A. 2πrh 2πr 2
r
h
Lateral
2πr
52
prism
Triangular
faces.
d
d Base
c
208
Bases
The lateral area of a prism is the sum of the areas of the lateral
two bases.
c
Area
Geometry Lesson 11-2
Geometry
h
Surface
Area
r
A right prism is a prism whose lateral faces are rectangular regions
and a lateral edge is an altitude.
prisms
right
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
c
d
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Perimeter of base
abcd
a b c d
h
and the
Lateral
faces
The surface area of a prism or a cylinder is the sum of the lateral area and the areas of the
S. A. L.A. 2B
Perimeter
The height is the length of the altitude.
h
An oblique prism is a prism whose lateral faces are not
perpendicular to the bases.
h
Bases
A cylinder is a three-dimensional figure with two congruent circular
bases that lie in parallel planes.
In a cylinder, the bases are parallel circles.
h
right
cylinders
The lateral area of a cylinder is the area of the curved surface.
A right cylinder is one where the segment joining the centers of the bases is an altitude.
h
An oblique cylinder is one in which the segment joining the centers
r
Area of a base
B π r2
Daily Notetaking Guide
prism
oblique
of the bases is not perpendicular to the bases.
oblique
Daily Notetaking Guide
cylinder
Geometry Lesson 11-2
209
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
206
b. Draw a net for the prism.
All rights reserved.
S
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
R, S, T, U, V; RS, RU, RT, VS, VU, VT, SU, UT, TS;
kRSU, kRUT, kRTS, kVSU, kVUT, kVTS
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
R
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
2 Finding Surface Area of Cylinders A company sells cornmeal and barley
in cylindrical containers. The diameter of the base of the 6-in.-high cornmeal
container is 4 in. The diameter of the base of the 4-in.-high barley container
is 6 in. Which container has the greater surface area?
1 Using Formulas to Find Surface Area Find the surface area of a 10-cmhigh right prism with triangular bases having 18-cm edges. Round to the
nearest whole number.
L.A. ⫽ ph
Use the formula
Find the surface area of each container. Remember that r d2.
to find the lateral area and
Cornmeal Container
S.A. ⫽ L.A. ⫹ 2B
to find the surface area of the
prism. The area B of the base is 12 ap, where a is the apothem and p is the
perimeter
.
the formula
Substitute the formulas for lateral
area of a cylinder and area of a circle.
( )( 6 ) 2π ( 2 2)
All rights reserved.
All rights reserved.
5
24
p 1 8
5
32
p
S.A. 5 L.A. 1 2
Use the formula for surface area.
r2
2π 2
cm.
54
B
p
5 2prh 1 2
cm
18
18
cm
The triangle has sides of length 18 cm, so p 3 ? 18 cm, or
Barley Container
S.A. 5 L.A. 1 2
5 2 p
2π
Substitute for r and h.
( 3 )( 4 ) 2π ( 3 2)
Simplify.
5
24
p 1
Combine like terms.
5
42
p
p
Because 42π in.2 ⬎ 32π in.2, the
surface area.
B
h 2pr2
r
p
18
container has the greater
barley
Use the diagram of the base.
60 a
All rights reserved.
1. Use formulas to find the lateral area and surface area of the prism.
Divide both sides by
"3
9
a
"3
?
"3
"3 ?
B 12 ? 3
54
9 !3
"3
"3
81
81 !3
)(
54
540 10
) 2(
!3
81
162 !3
cm2.
Substitute
ph
for L.A.
) Substitute
54
for p,
10
81 !3
for h and
for B.
400p cm2
3. The company in Example 2 wants to make a label to cover the cornmeal
container. The label will cover the container all the way around, but will not
cover any part of the top or bottom. What is the area of the label to the
nearest tenth of a square inch?
75.4 in.2
Use a calculator.
821 cm 2
Rounded to the nearest whole number, the surface area is
210
for p.
2. Find the surface area of a cylinder with height 10 cm and radius 10 cm in
terms of π.
Simplify.
820.59223
<
54
for a and
Use the formula for surface area.
2B
ph
Rationalize the denominator.
Simplify.
S.A. 5 L.A. 1 2B
(
"3
3
81 !3
Substitute
Geometry Lesson 11-2
.
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Lesson 11-3
Geometry Lesson 11-2
A right cone is a cone in which the altitude is perpendicular to the
Surface Areas of Pyramids and Cones
Lesson Objectives
1 Find the surface area of a pyramid
2 Find the surface area of a cone
211
Name_____________________________________ Class____________________________ Date ________________
Slant height
plane of the base.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Altitude
The slant height of a cone is the distance from the vertex to the edge
of the base.
Local Standards: ____________________________________
/
h
r
The lateral area of a cone is the area of the curved lateral surface.
Base
Vocabulary and Key Concepts.
The surface area of a cone the sum of the lateral area and the area
ᐉ
.
B
The surface area of a regular pyramid is the sum of the
lateral area
S.A. L.A. area of the base
and the
Examples.
pyramid with base edges 7.5 ft and slant height 12 ft.
12 ft
B
circumference of the base
L.A. 12 ? 2πr ? ᐉ, or L.A. and the
ᐉ
slant height
.
r
B
prᐉ
The surface area of a right cone is the sum of the lateral area and the area of the base.
B
A regular pyramid is a pyramid whose base is a regular polygon and whose lateral faces are
congruent isosceles triangles.
The altitude of a pyramid or a cone is the perpendicular segment from the vertex to the plane
of the base.
The height of a pyramid or a cone is the length of
the altitude.
Lateral
Vertex
edge
Lateral face
Altitude
The slant height of a regular pyramid is the length
of the altitude of a lateral face.
Base
Base
The lateral area of a pyramid is is the sum of the
edge
Regular Pyramid
7.5 ft
The perimeter p of the square base is 4 You are given ᐍ 12
find the lateral area.
1
L.A. 5
L.A. 5 12
(
30
)(
)
Substitute.
Simplify.
Because the base is a square with side length 7.5 ft,
B s2 7.5 2 56.25
S.A. 5 L.A. 1 B
The surface area of a pyramid is is the sum of the lateral area and the area of the base.
S.A. 5
180
1
236.25
.
Use the formula for surface area of a pyramid.
56.25
Substitute.
Simplify.
The surface area of the square pyramid is
All-In-One Answers Version A
Daily Notetaking Guide
ft, so you can
Find the area of the square base.
S.A. 5
Geometry Lesson 11-3
ft.
30
30
Use the formula for lateral area of a pyramid.
12
180
ft, or
7.5
ft and you found that p ᐍ
p
2
L.A. 5
areas of the congruent lateral faces.
212
cone
1 Finding Surface Area of a Pyramid Find the surface area of a square
.
Theorem 11-4: Lateral and Surface Area of a Cone
The lateral area of a right cone is half the product of the
S.A. L.A. All rights reserved.
slant height
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
L.A. and the
1 pᐉ
2
All rights reserved.
Theorem 11-3: Lateral and Surface Area of a Regular Pyramid
The lateral area of a regular pyramid is half the product of the
perimeter of the base
Right
of the base.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
12 m
6m
3
The area of each base of the prism is
432 m 2 ; about 619 m 2
"3 .
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
9
B 5 12ap
Quick Check.
!3 ? shorter leg longer leg
9
a
B
.
9 cm
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
a ? "3
apothem
Use 30°-60°-90° triangles to find the
30
9 cm
Daily Notetaking Guide
236.25
ft2.
Geometry Lesson 11-3
Geometry
213
53
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
2 Finding Lateral Area of a Cone Leandre uses paper cones to cover her
Lesson 11-4
plants in the early spring. The diameter of each cone is 1 ft, and its height is
1.5 ft. How much paper is in the cone? Round your answer to the nearest
tenth.
1
2
to find the lateral area of the cone.
0.5
The cone’s diameter is 1 ft, so its radius r is
2
2
1.5
0.25 Theorem 11-5: Cavalieri’s Principle
If two space figures have the same height and the same cross sectional area
"2.5
2.25
Find the lateral area.
L.A. 5 p
(
L.A. 5 p
r
ᐍ
0.5
)
Use the formula for lateral area of a cone.
"2.5
2.4836471
L.A. <
0.5
Substitute
"2.5
for r and
All rights reserved.
0.5
for ᐍ.
height of the prism
height of the cylinder
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2. Find the lateral area of a cone with radius 15 in. and height 20 in. to the
nearest square inch.
1178 in.2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
V Bh, or V 55 m 2
Geometry Lesson 11-3
Name_____________________________________ Class____________________________ Date ________________
2 Finding Volume of a Triangular Prism Find the volume of the prism.
A composite space figure is a three-dimensional figure that is the combination of two or more
simpler figures.
Examples.
1 Finding Volume of a Cylinder Find the volume of the cylinder.
Leave your answer in terms of π.
The formula for the volume of a cylinder is
The diagram shows h and d, but you must find r.
altitude
and the other leg is the
r 5 12d 5 8
V5 p
r2
h
5p?
82
? 9
576
p
Use the formula for the volume of a cylinder.
2
20
841
(
400
)(
21
The area B of the base is 21bh 12 20
area of the base to find the volume of the prism.
V B
h
V
210
V
8400
)
Simplify.
40
215
Geometry Lesson 11-4
10 m
21
. Use the
Simplify.
8400
ft3.
5m
441
Substitute.
The volume of the triangular prism is
576π
150 m 3
Use the formula for the volume of a prism.
?
16 ft
Substitute.
6m
210
9 ft
20 m
All rights reserved.
2
29
.
1. Find the volume of the triangular prism.
40 m
. Use the
Pythagorean Theorem to calculate the length of the other leg.
V pr 2h
Quick Check.
29 m
where one leg
All rights reserved.
base
is the
h
r
B
Name_____________________________________ Class____________________________ Date ________________
with triangular bases. The
right triangle
area of the base
.
Daily Notetaking Guide
Daily Notetaking Guide
right triangular prism
B
Volume is the space that a figure occupies.
5
base of the triangular prism is a
h
pr 2h
The volume of the cylinder is
The prism is a
area of a base
Bh
and the
1. Find the surface area of a square pyramid with base edges 5 m and slant
height 3 m.
.
.
Theorem 11-7: Volume of a Cylinder
The volume of a cylinder is the product of the
Quick Check.
214
volume
Theorem 11-6: Volume of a Prism
The volume of a prism is the product of the
V
ft2.
2.5
at every level, then they have the same
and the
Use a calculator.
The lateral area of the cone is about
Local Standards: ____________________________________
Vocabulary and Key Concepts.
All rights reserved.
Topic: Measuring Physical Attributes
ft.
The altitude of the cone, radius of the base, and slant height form a
right
Pythagorean Theorem
triangle. Use the
to find the slant height ᐍ.
ᐉ
Find the volume of a prism
Find the volume of a cylinder
2. The cylinder shown is oblique.
a. Find its volume in terms of π.
256p m3
16 m
m3.
8m
3 Finding Volume of a Composite Figure Find the volume
25
4 cm
Bh .
( 14
Volume of Prism II Bh ( 6
Volume of Prism III Bh ( 6
1400
Volume of Prism I Bh 1400
?
) ? 25 4 ) ? 25
4 ) ? 25 4
?
?
600
The volume of the composite space figure is
216
54
10 cm
6 cm
4 cm
Each prism’s volume can be found using the formula V Sum of the volumes 4 cm
6 cm
6 cm
6 cm
6 cm
4
4 cm
Geometry Lesson 11-4
Geometry
2600
600
600
600
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
cm
4 cm
cm
cm
25
14
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
14 cm
to find the volume.
6 cm
rectangular prisms
10 cm
You can use three
cm
of the composite space figure.
All rights reserved.
prᐍ
NAEP 2005 Strand: Measurement
b. Find its volume to the nearest tenth of a cubic meter.
804.2 m 3
3. Find the volume of the composite space figure.
2 in.
12 in.3
4 in.
2 in.
4 in.
2600
1 in.
cm3.
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 11-4
217
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Use the formula L.A. Volumes of Prisms and Cylinders
Lesson Objectives
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 11-5
2 Finding Volume of a Pyramid Find the volume of a
Volumes of Pyramids and Cones
square pyramid with base edges 16 m and slant height 17 m.
Lesson Objectives
1
Find the volume of a pyramid
Find the volume of the cone
2
17 m
NAEP 2005 Strand: Measurement
The altitude of a right square pyramid intersects the base
at the center of the square. Because each side of the square
base is 16 m, the leg of the right triangle along the base is
8 m, as shown.
Topic: Measuring Physical Attributes
Local Standards: ____________________________________
h
16 m
Key Concepts.
16 m
Theorem 11-8: Volume of a Pyramid
.
h
B
Theorem 11-9: Volume of a Cone
All rights reserved.
height of the pyramid
and the
1Bh
3
All rights reserved.
area of the base
V
17 m
h
The volume of a pyramid is one third the product of the
8m
Step 1 Find the height of the pyramid.
The volume of a cone is one third the product of the
and the
area of the base
V 13Bh, or V .
height of the cone
h
1pr 2h
3
172 5 82 1 h2
Use the
289 5 64 1 h2
Simplify.
5 h2
225
r
h5
B
Pythagorean Theorem
64
Subtract
15
.
from each side.
Find the square root of each side.
Examples.
All rights reserved.
1 Finding Volume of a Pyramid Find the volume of a square pyramid with
base edges 15 cm and height 22 cm.
Because the base is a square, B ?
15
225
.
1
V5
13
15
B
h
Use the formula for volume of a pyramid.
3
(
V5
225
)(
22
)
1650
Substitute
225
for B and
22
for h.
Simplify.
The volume of the square pyramid is
1650
cm3.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Step 2 Find the volume of the pyramid.
1
V5
B
(
5 13
5
16
?
15
Substitute.
1280
Simplify.
1280
m 3.
of the oblique cone in terms of π.
r5
V5
1
d 5 3 in.
2
11 in.
1
p
3
r2
33
h
11
)
p
Use the formula for the volume of a cone.
Substitute 3 for r and
11
6 in.
for h.
Simplify.
The volume of the cone is
33p
in.3.
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 11-5
219
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Quick Check.
Lesson 11-6
1. Find the volume of a square pyramid with base edges 24 m and slant height 13 m.
Surface Areas and Volumes of Spheres
Lesson Objective
1 Find the surface area and volume
of a sphere
960 m 3
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Local Standards: ____________________________________
Vocabulary and Key Concepts.
All rights reserved.
12 m
6m
All rights reserved.
Theorem 11-10: Surface Area of a Sphere
2. Find the volume of each cone in terms of π and also rounded as indicated.
a. to the nearest cubic meter
The surface area of a sphere is four times the product of p and
the
square
S.A. of the
radius
r
of the sphere.
4pr 2
Theorem 11-11: Volume of a Sphere
The volume of a sphere is four thirds the product of p and
144p m3; 452 m3
the
cube
S.A. of the
radius
r
of the sphere.
4pr 3
3
A sphere is the set of all points in space equidistant from a
42 mm
21 mm
6174p mm3; 19,396 mm3
3. A small child’s teepee is 6 ft tall and 7 ft in diameter. Find the volume of the
teepee to the nearest cubic foot.
77 ft 3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
b. to the nearest cubic millimeter
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
)
3 Finding Volume of an Oblique Cone Find the volume
5
Geometry Lesson 11-5
16
The volume of the square pyramid is
5 13 p( 3 2)(
218
Use the formula for
the volume of a pyramid.
h
3
center
diameter
given point.
The center of a sphere is the given point from which all points
r
on the sphere are equidistant.
The radius of a sphere is a segment that has one endpoint at
d
the center and the other endpoint on the sphere.
The diameter of a sphere is a segment passing through the
center with endpoints on the sphere.
circumference
sphere
A great circle is the intersection of a plane and a sphere
containing the center of the sphere.
A great circle divides a sphere into two congruent
Hemispheres
Great
hemispheres.
circle
The circumference of a sphere is the circumference of its
great circle.
220
Geometry Lesson 11-5
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 11-6
Geometry
221
55
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
Example.
1 Finding Surface Area The circumference of a rubber ball is 13 cm.
3 Using Volume to Find Surface Area The volume of a sphere is 1 in.3.
Find its surface area to the nearest tenth.
Calculate its surface area to the nearest whole number.
Step 1 First, find the radius.
13
13
2p
p
r
5 2pr
Substitute
5r
Solve for r.
13
for C.
3
5
r2
5 4p ?
(
Use the formula for surface area of a sphere.
)
13
13
2
2p
Substitute
for r.
53.794371
Use a calculator.
cm2.
Substitute r p
30
2
15
.
Simplify.
The volume of the sphere is
cm3.
4500p
Quick Check.
1. Find the surface area of a sphere with d 14 in. Give your answer in two
ways, in terms of π and rounded to the nearest square inch.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
15
4500
Use the formula for volume of a sphere.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
r3
3
m
30 c
5
p
r
cube root
Find the
of each side.
4p
4.83597587
Use r, S.A. 4π r 2, and
a calculator.
in.2.
4.8
Geometry Lesson 11-6
100 in.2
3. Find the volume to the nearest cubic inch of a sphere with diameter 60 in.
113,097 in.3
4. The volume of a sphere is 4200 ft 3. Find the surface area to the nearest tenth.
1258.9 ft 2
196p in.2, 616 in.2
222
Solve for r 3.
2. Find the surface area of a spherical melon with circumference 18 in. Round
your answer to the nearest ten square inches.
Leave your answer in terms of π.
5 43p ?
Substitute.
Quick Check.
54
2 Finding Volume Find the volume of the sphere.
4
Use the formula for volume of a sphere.
r3
S. A. To the nearest whole number, the surface area of the rubber ball is
3
r3
p
To the nearest tenth, the surface area of the sphere is
Simplify.
p
<
3
3
2p
169
5
V5
All rights reserved.
All rights reserved.
4p
p
3
3
1 4
3πr
Step 2 Use the radius to find the surface area.
S.A. 5 4
4
V5
Use the formula for circumference.
All rights reserved.
C5 2
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 11-6
223
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 11-7
2 Finding the Similarity Ratio Find the similarity ratio of two similar
Areas and Volumes of Similar Solids
cylinders with surface areas of 98π ft 2 and 2π ft 2.
Use the ratio of the surface areas to find the similarity ratios.
NAEP 2005 Strand: Measurement
Topic: Systems of Measurement
a2 5
b2
Vocabulary and Key Concepts.
(2) the ratio of their volumes is a3 : b3 .
All rights reserved.
All rights reserved.
Theorem 11-12: Areas and Volumes of Similar Solids
If the similarity ratio of two similar solids is a : b, then
(1) the ratio of their corresponding areas is a2 : b2 , and
98p
a2 5
b2
Local Standards: ____________________________________
a
b5
Simplify.
1
7
8 in.
9 in.
3 in.
Both figures have the same shape. Check that the ratios of the
corresponding dimensions are equal.
The ratio of the radii is
The cones are
224
56
3
9
, and the ratio of the heights is
not similar
Geometry Lesson 11-7
Geometry
because
8
3
9 ⬆ 26
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
26 in.
.
48
a3 5
3
b
a
b5
The ratio of the volumes is
162
:
b3
.
Simplify.
27
2
Take the
cube root
of each side.
3
Step 2 Use the similarity ratio to find the surface area S1 of the smaller pyramid.
22
S1
S2 5
S1
135
The ratio of the surface areas is
32
S1
S2 5
4
4
Substitute
9
the
Daily Notetaking Guide
a2
:
b2
.
Simplify.
9
4
9
60
?
135
larger
135
for S2, the surface area of
pyramid.
Solve for S 1 .
Simplify.
The surface area of the smaller pyramid is
Daily Notetaking Guide
a3
8
S1 5
.
of each side.
3 Using a Similarity Ratio Two similar square pyramids have volumes of
S1 5
8
26
square root
Take the
1
The similarity ratio is 7 : 1 .
a3 5
3
b
similarity ratio.
.
Step 1 Find the similarity ratio.
The similarity ratio of two similar solids is the ratio of their corresponding linear dimensions.
Examples.
b2
:
2p
49
48 cm 3 and 162 cm 3. The surface area of the larger pyramid is 135 cm 2.
Find the surface area of the smaller pyramid.
Similar solids have the same shape and all of their corresponding parts are proportional.
1 Identifying Similar Solids Are the two solids similar? If so, give the
a2
The ratio of the surface areas is
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objective
1 Find relationships between the ratios
of the areas and volumes of similar
solids
60
cm2.
Geometry Lesson 11-7
225
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date ________________
Quick Check.
Lesson 12-1
1. Are the two cylinders similar? If so, give the similarity ratio.
Tangent Lines
Lesson Objectives
yes; 6 : 5
1
10
12
6
2
5
NAEP 2005 Strand: Geometry
Use the relationship between a radius
and a tangent
Use the relationship between two
tangents from one point
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
Vocabulary and Key Concepts.
2. Find the similarity ratio of two similar prisms with surface areas 144 m 2 and 324 m 2.
All rights reserved.
All rights reserved.
Theorem 12-1
A
If a line is tangent to a circle, then the line is perpendicular to the radius drawn
P
to the point of tangency.
* )
B
O
AB ' OP
Theorem 12-2
2:3
A
P
If a line in the plane of a circle is perpendicular to a radius at its endpoint
the line is tangent to the circle
on the circle, then
* )
.
AB is tangent to 䉺 O.
O
B
A
B
3. The volumes of two similar solids are 128 m 3 and 250 m 3. The surface area
of the larger solid is 250 m 2. What is the surface area of the smaller solid?
160
m2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
Theorem 12-3
The two segments tangent to a circle from a point outside the circle
congruent
are
.
O
AB > CB
C
A tangent to a circle is a line, segment, or ray in the plane of the circle that intersects the
circle in exactly one point.
The point of tangency is the point where a circle and a tangent intersect.
A
tangent
B
point of tangency
A triangle is inscribed in a circle if all vertices of the triangle lie on the circle.
A triangle is circumscribed about a circle if each side of the triangle is tangent
to the circle.
226
Geometry Lesson 11-7
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Examples.
180
Triangle Angle-Sum Theorem
x
180
Substitute.
x
180
Simplify.
x
68
Solve.
Y
15 cm
X
O
ODWZ. Because OZ is a radius of 䉺 O, OZ 3 cm.
supplement
of a
right
Because the radius of 䉺 P is 7 cm, PD 7⫺3⫽4
OD2 PD2 OP2
8 ft
R
angle,
triangle.
cm.
Simplify.
15.524175
Use a calculator to find the square root.
The distance between the centers of the pulleys is about
15.5
cm.
Quick Check.
1. ED is tangent to 䉺 O. Find the value of x.
D
E
x
52
O
38
Geometry Lesson 12-1
All-In-One Answers Version A
tangent
Y
S
Daily Notetaking Guide
Z
6 ft
11 ft
C
X
T
7 ft
U
W
UX XR
YS RY
8 ft
SZ ZT
6 ft
11
ft
By Theorem 12-3, two segments tangent to a circle from
a point outside the circle are
.
congruent
WU TW 7 ft
p XY YZ
ZW WX
XR RY
YS 11
64
SZ
Definition of perimeter p
ZT TW WU UX
8 8 6 6 7 7 The perimeter is
228
is not
, PA
Find the perimeter of XYZW.
W
Substitute.
OP2
Simplify.
PO 2 ⫽ PA 2 ⫹ OA 2
4 Circles Inscribed in Polygons 䉺 C is inscribed in quadrilateral XYZW.
Pythagorean Theorem
4 2 OP2
Substitute.
194
Because
A
.
Is 䉭OAP a right triangle?
13 2 5 2
cm.
⬔ODP is also a right angle, and 䉭OPD is a
OP 12 2 ⱨ
144
P
13
P
15 cm
Because opposite sides of a rectangle have the same measure,
241
PA 2 ⫹ OA 2
12
to 䉺 O at A.
D
Z
right
angle, 䉭OAP must be a right triangle, and
PO 2 ⱨ PA2 OA2
7 cm
3 cm
Draw OP. Then draw OD parallel to ZW to form rectangle
Because ⬔ODP is the
5
PO 2 22°
two circular pulleys, as shown. Find the distance
between the centers of the pulleys. Round your
answer to the nearest tenth.
15
O
For PA to be tangent to 䉺 O at A, ⬔A must be a
B
2 Applying Tangent Lines A belt fits tightly around
DW 3 cm and OD 䉺 O such that PO 12, and point A is on 䉺 O such
that PA 13. Is PA tangent to 䉺 O at A? Explain.
right
All rights reserved.
22
112
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
angle.
All rights reserved.
m⬔A m⬔B m⬔C right
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
x°
Because BA is tangent to 䉺 C, ⬔A must be a
Use the Triangle Angle-Sum Theorem to find x.
2
3 Finding a Tangent 䉺 O has radius 5. Point P is outside
C
value of x.
15
227
Name_____________________________________ Class____________________________ Date ________________
Examples.
1 Finding Angle Measures BA is tangent to 䉺 C at point A. Find the
90
Geometry Lesson 12-1
Daily Notetaking Guide
11
Segment Addition Postulate
Substitute.
Simplify.
64
ft.
Geometry Lesson 12-1
Geometry
229
57
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Quick Check.
Lesson 12-2
2. A belt fits tightly around two circular pulleys, as shown. Find the distance
between the centers of the pulleys.
1
35 in.
2
14 in.
Chords and Arcs
Lesson Objectives
8 in.
NAEP 2005 Strand: Geometry
Use congruent chords, arcs, and
central angles
Recognize properties of lines through
the center of a circle
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
about 35.5 in.
All rights reserved.
All rights reserved.
Vocabulary and Key Concepts.
Theorem 12-4
Within a circle or in congruent circles
(1) Congruent central angles have
(2) Congruent chords have
(3) Congruent arcs have
congruent
congruent
congruent
chords.
arcs.
central angles.
Theorem 12-5
Within a circle or in congruent circles
(1) Chords equidistant from the center are
4. 䉺 O is inscribed in 䉭PQR. 䉭PQR has a perimeter of 88 cm. Find QY.
15 cm X
P
O
12 cm
Z
17 cm
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
No; 4 2 ⫹ 7 2 ⫽ 8 2
Q
Y
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
(2) Congruent chords are
equidistant
.
congruent
from the center.
Theorem 12-6
In a circle, a diameter that is perpendicular to a chord bisects the
chord
arcs
and its
.
All rights reserved.
3. In Example 3, if OA 4, AP 7, and OP 8, is PA tangent to 䉺 O at A?
Explain.
Theorem 12-7
In a circle, a diameter that bisects a chord (that is not a diameter) is
perpendicular
to the chord.
Theorem 12-8
In a circle, the perpendicular bisector of a chord contains the
of the circle.
center
A chord is a segment whose endpoints are on a circle.
R
P
Q
O
Geometry Lesson 12-1
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
Quick Check.
A
1 Using Theorem 12-4 In the diagram, radius OX bisects ⬔AOB.
1. If you are given that BC DF in the circles, what can you conclude?
What can you conclude?
congruent
chords.
congruent
arcs.
congruent
O
central angles have
X
O
QS 7 7
Substitute.
A
Q
QS 14
Simplify.
AB QS
Chords that are equidistant from the center of a circle
are congruent.
AB 14
Substitute
7
14
The distance from the center of 䉺 O to PQ is measured along a
perpendicular line.
2
(
16
P
r
15 in.
O
Substitute.
17
Find the square root of each side.
17
18
16
x
36
3. Use the circle at the right.
M
Substitute.
Simplify.
OM 2
The radius of 䉺 O is
58
8
Use the Pythagorean Theorem.
r
232
)
r 2 289
PM 2
18
S
A diameter that is perpendicular
to a chord bisects the chord.
PQ
r 2 8 2 15 2
OP 2
7
16
Q
16 in.
The distance from O to PQ is 15 in., and PQ 16 in.
Find the radius of 䉺 O.
PM R
2. Find the value of x in the circle at the right.
for QS.
3 Using Diameters and Chords P and Q are points on 䉺 O.
1
4
C
4
All rights reserved.
B
Segment Addition Postulate
2
F
B
QS QR RS
PM C
chords have
2 Using Theorem 12-5 Find AB.
1
D
P
䉺O 䉺P
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
because
BX
congruent
0
0
AX BX because
B
0 0
lO O lP; BC O DF
by the definition of an angle bisector.
All rights reserved.
AX lBOX
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
⬔AOX 231
Geometry Lesson 12-2
a. Find the length of the chord to the nearest unit.
6.8
about 11
4
x
b. Find the distance from the midpoint of the chord to the midpoint of its
minor arc.
2.8
in.
Geometry Lesson 12-2
Geometry
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 12-2
233
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
230
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 12-3
Topic: Relationships Among Geometric Figures
x
Local Standards: ____________________________________
x
x
B
1 mAC
2
C
All rights reserved.
half the measure of its intercepted arc.
m⬔B All rights reserved.
Theorem 12-9: Inscribed Angle Theorem
A
Theorem 12-10
The measure of an angle formed by a tangent and a chord is
B
half the measure of the intercepted arc.
1 1
2mBDC
m⬔C x
An inscribed angle has a vertex that is on a
A
Intercepted arc
circle and sides that are chords of the circle.
.
B
C
Inscribed angle
An intercepted arc is an arc with endpoints on
the sides of an inscribed angle and its other
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
supplementary
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
.
angle.
3. The opposite angles of a quadrilateral inscribed in a circle are
DE
(m 0
2
0
m EF
)
points in the interior of the angle.
x°
1
(
80
70
)
75
1
mEFG
1
y
1
2
1
2
y
(
0
m EF
Inscribed Angle Theorem
(
70
0
m FG
120
)
Arc Addition Postulate
)
Substitute.
95
Simplify.
2 Using Corollaries to Find Angle Measures Find the values of a and b.
By Corollary 2 to the Inscribed Angle Theorem, an angle inscribed in a
semicircle is a right angle, so a B
䉺 O is
180
. Therefore, the angle whose intercepted arc has
measure b must have measure 180 Theorem 12-10
0
0
0
mRT m URT m UR
m⬔BRT 1
2
(
180
126
27
)
Substitute
and
126
180
126°
A
Use the properties of tangents to find m⬔TRS.
m⬔BRS 90
m⬔BRS m⬔
BRT
63
m⬔TRS
27
perpendicular
to
the radius of a circle at its point of tangency.
90
m⬔
m⬔TRS
TRS
58
a°
.
twice
the measure of the inscribed
235
Geometry Lesson 12-3
Angle Measures
and Segment Lengths
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
Vocabulary and Key Concepts.
U
Simplify.
A tangent is
or
S
for mURT
for mUR.
All rights reserved.
m⬔BRT N
Arc Addition Postulate
32
the measure of the intercepted
116 .
Lesson Objectives
1 Find the measures of angles formed
by chords, secants, and tangents
2 Find the lengths of segments
associated with circles
R
0
mRT
All rights reserved.
2
)
90
half
Lesson 12-4
Angle Addition Postulate
Substitute.
Theorem 12-11
The measure of an angle formed by two lines that
(1) intersect inside a circle is half the sum of the measures of the
y°
1
m⬔1 1 (x 1 y)
2
x°
difference of the measures of the
Simplify.
x°
intercepted arcs.
(2) intersect outside a circle is half the
y° 1
y° 1
x°
y° 1
x°
intercepted arcs.
m⬔1 Quick Check.
1 (x 2 y)
2
105
2. For the diagram at the right, find the measure of each numbered angle.
ml1 5 90, ml2 5 110, ml3 5 90, ml4 5 70
4
1 60°
2
80°
3
3. In Example 3, describe two ways to find m⬔NRS using Theorem 12-10.
mlNRS 5 mlNRU 1 mlURS 5 1
2mRU 1 27 5 63 1 27 5 90
0
1
m⬔NRS 1
2m RUS 2(180) 90
236
Geometry Lesson 12-3
All-In-One Answers Version A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. In Example 1, find m⬔DEF.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
O
Name_____________________________________ Class____________________________ Date ________________
T
RB is tangent to 䉺 A at point R. Find m⬔BRT and m⬔TRS.
32°
b°
Daily Notetaking Guide
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date________________
m⬔BRT 90 .
The sum of the measures of the three angles of the triangle inscribed in
angle, so b 2( 58
3 *Using
) Theorem 12-10 RS and TU are diameters of circle A.
G
Simplify.
Because the inscribed angle has
Geometry Lesson 12-3
90°
Substitute.
arc, the intercepted arc has
234
D y°
Arc Addition Postulate
2
y
congruent
1
0
1
Because EFG is the intercepted arc of ⬔D, you need to find mFG in order
1
to find mEFG . The arc measure of a circle is 360, so
0
mFG 360 70 80 90 120 .
y
Corollaries to the Inscribed Angle Theorem
right
C
2
C
C
2. An angle inscribed in a semicircle is a
Inscribed Angle Theorem
B
D
D
1. Two inscribed angles that intercept the same arc are
F
80°
2
Vocabulary and Key Concepts.
The measure of an inscribed angle is
1
mDEF
1
70°
E
1 Using the Inscribed Angle Theorem Find the values of x and y.
NAEP 2005 Strand: Geometry
Find the measure of an inscribed
angle
Find the measure of an angle formed
by a tangent and a chord
2
Examples.
Inscribed Angles
Lesson Objectives
1
Name_____________________________________ Class____________________________ Date ________________
Theorem 12-12
For a given point and circle, the product of the lengths of the two segments from
the point to the circle is constant along any line through the point and circle.
I.
a c
b
P
d
abcd
III.
II.
x
w
z
y
t
P
(w x)w (y z)y
A secant is a line that intersects a circle at two points.
(y z)y Daily Notetaking Guide
P
t2
secant
A
O
Daily Notetaking Guide
y
z
B
Geometry Lesson 12-4
Geometry
237
59
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Examples.
1 Finding Arc Measures An advertising agency wants a frontal photo of a “flying
Quick Check.
A
X
T
Theorem 12-11(2)
2
[(
12 (
12
72
72
x
A
)
)
360 ⫺ x
x
]
Substitute.
2x
360
Simplify.
x
110°
70°
72°
All rights reserved.
m⬔T 250
Y
All rights reserved.
1
0
(mXAY mXY )
1
1. Find the value of w.
w°
saucer” ride at an amusement park. The photographer stands at the vertex of
the angle formed by tangents to the “flying saucer.” What is the measure of the
arc that will be in the photograph? In the 0
diagram, the photographer
stands at
1
)
)
point T. TX and TY intercept minor arc XY and major arc XAY.
0
Let mXY x.
1
360
Then mXAY x.
72
180
72
180
Add x to both sides.
x
108
Solve for x.
2. Critical Thinking To photograph a 160 arc of the “flying saucer” in Example 1,
should you move toward or away from the object? What angle should the
tangents form?
away; 20°
Distributive Property.
3. Find the value of x to the nearest tenth.
20
x
arc will be in the advertising agency’s photo.
108⬚
14
13.8
16
200 ft
the center of the circle. Because the radius is 125 ft, the
125 250 ft. The length of
2
diameter is
B
x
the other segment along the diameter is
250 ft 50
ft, or 200 ft.
xx
50
x2 x
Theorem 12-12(1)
200
10,000
Multiply.
100
Solve for x.
200
The shortest distance from point A to point B is 2x or
238
Center
b. Critical Thinking Complete the circle with the given center. Then use
Theorem 12-12 to find the value of x. Show that your answers to parts
(a) and (b) are equal.
16
in.; (8 2 4"3)
8 1 4"3
Name_____________________________________ Class____________________________ Date________________
(x 4 )2 (y 1 )2 25
(x 4 )2 (y 1 )2 ( 5 )2
c
c
h
Key Concepts.
y
Theorem 12-13
2
8 6 4 2 O
Examples.
0 )2 (x 8
)
2
y2
(
"5
)2
5
Substitute
(
"5 for r.
8, 0
) for (h, k) and
Simplify.
a circle with center (5, 8) that passes through the point (15, 13).
First find the radius.
Distance
Use the
15
5 8
#(
)2 ( 13
)2
2
)
20
21
#(
( )2
Substitute
(
5, 8
Formula to find r.
) for (h, k) and (
15, 13
) for (x, y).
Simplify.
441
#400
#841
29
Then find the standard equation of the circle with center (5, 8) and radius
(x h )2 (y k )2 r 2
(x 5 )2 (y 8 )2 (x 5 ) 2 (y 8 ) 2 240
60
(
29
841
Geometry Lesson 12-5
Geometry
6
8
x
4 Applying the Equation of a Circle A diagram locates a radio tower at
2 Using the Center and a Point on a Circle Write the standard equation of
2(
h )
k x y )2
r #(
4
4
Standard form
29
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
)] 2 (y 2
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
8
y
4
1 Writing the Equation of a Circle Write the standard equation of a circle
with center (8, 0) and radius "5.
(
r
and the radius is 5 .
(⫺4, 1)
6
All rights reserved.
x
O
All rights reserved.
(h, k)
Relate the equation to the standard form
(x h)2 (y k)2 r 2.
c
k
The center is
(x, y)
r
(h, k)
and radius r is (x h) 2 (y k) 2 r 2.
[x 239
circle with equation (x 4) 2 (y 1) 2 25. Then graph the circle.
Local Standards: ____________________________________
(x h )2 (y k )2 r 2
Geometry Lesson 12-4
3 Graphing a Circle Given Its Equation Find the center and radius of the
NAEP 2005 Strand: Geometry
Topic: Position and Direction
The standard form of an equation of a circle with center
16
1.07
8 1 4"3
Name_____________________________________ Class____________________________ Date ________________
Circles in the
Coordinate Plane
Lesson 12-5
5
Daily Notetaking Guide
Daily Notetaking Guide
Arc
x
8 in.
(8 2 4"3) in.
ft.
Geometry Lesson 12-4
Lesson Objectives
1 Write an equation of a circle
2 Find the center and radius of a circle
4. The basis of a design of a rotor for a Wankel engine is an equilateral
triangle. Each side of the triangle is a chord to an arc of a circle. The
opposite vertex of the triangle is the center of the arc. In the diagram
at the right, each side of the equilateral triangle is 8 in. long.
a. Use what you know about equilateral triangles and find the value of x.
All rights reserved.
The perpendicular bisector of the chord AB contains
x
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
50 ft
point A to point B along the arc of a circle with a
radius of 125 ft. Find the shortest distance from
point A to point B.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2 Finding Segment Lengths A tram travels from
(6, 12) on a coordinate grid where each unit represents 1 mi. The radio
signal’s range is 80 mi. Find an equation that describes the position and
range of the tower.
(6, ⫺12)
The center of a circular range is at
(x h)2 (y k)2 r 2
(
x 6
(x )
2
[y (
) (
6 2 y
12
12
)]
2
)2 , and the radius is
80 .
Use standard form.
80
2
6400
Substitute.
This is an equation for position
and range of the tower.
Quick Check.
1. Write the standard equation of each circle.
a. center (3, 5); radius 6
(x ⫺ 3) 2 ⫹ (y ⫺ 5) 2 ⫽ 36
b. center (2, 1); radius "2
(x ⫹ 2) 2 ⫹ (y ⫹ 1) 2 ⫽ 2
.
Standard form
)2
Substitute
(
5, 8
) for (h, k) and
29
for r.
Simplify.
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 12-5
241
All-In-One Answers Version A
Geometry: All-In-One Answers Version A (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 12-6
2. Write the standard equation of the circle with center (2, 3) that passes
through the point (1, 1).
Locus: A Set of Points
Lesson Objective
(x ⫺ 2) 2 ⫹ (y ⫺ 3) 2 ⫽ 13
1
NAEP 2005 Strand: Geometry
Draw and describe a locus
Topic: Dimension and Shape
Local Standards: ____________________________________
3. Find the center and radius of the circle with equation
(x 2) 2 (y 3) 2 100. Then graph the circle.
y
Vocabulary.
16
center: (2, 3); radius: 10
x
4
8
All rights reserved.
O
(2, 3)
All rights reserved.
A locus is a set of points, all of which meet a stated condition.
8
Examples.
1 Describing a Locus in a Plane Draw and describe the locus of points in a
4. When you make a call on a cellular phone,
a tower receives the call. In the diagram,
the centers of circles O, A, and B are
locations of cellular telephone towers.
Write an equation that describes the
position and range of Tower O.
plane that are 1.5 cm from 䉺 C with radius 1.5 cm.
Use a compass to draw 䉺 C with radius 1.5 cm.
y
24
All rights reserved.
8
16
20
24
x
8
cm
4
4
5
1.
8 4 O
C
cm
4
5
1.
A
8
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
B
16
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
20
x 2 ⫹ y 2 ⫽ 144
The locus of points in the interior of 䉺 C that are 1.5
the center C.
cm from 䉺 C is
The locus of points exterior to 䉺 C that are 1.5 cm from 䉺 C is a circle with
center C and radius 3 cm.
The locus of points in a plane that are 1.5 cm from a point on 䉺 C with
radius 1.5 cm is point C and a circle with radius 3 cm and center C.
Geometry Lesson 12-5
Daily Notetaking Guide
Daily Notetaking Guide
Geometry Lesson 12-6
243
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
2 Drawing a Locus for Two Conditions Point P is 10 cm from point Q.
Quick Check.
Describe the locus of points in a plane that are 6 cm from point P and 8 cm
from point Q.
*
)
1. Draw and describe the locus: In a plane, the points 2 cm from line XY.
* )
* )
Two lines parallel to XY , each 2 cm from XY .
6 cm
2 cm
8 cm
P
Q
10 cm
10
X
8 cm
center P and radius 6 cm.
The set of points in a plane that are 8 cm from point Q is a circle with
All rights reserved.
The set of points in a plane that are 6 cm from point P is a circle with
2. Draw the locus: In a plane, the points equidistant from two points X and Y
and 2 cm from the midpoint of XY.
center Q and radius 8 cm.
Because the sum (14 cm) of the radii is greater than
overlap by
4
10
cm, the circles
cm on PQ. The circles intersect at two points, so the locus
is the two points where the circles
intersect
X
.
4 cm from a plane M.
M
44 cm
艎
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3 Describing a Locus in Space Describe the locus of points in space that are
44 cm
Y
2 cm
All rights reserved.
6 cm
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
242
}
}
2 cm
Y
2 cm
3. Draw and describe each locus.
a. In a plane, the points that are equidistant from two parallel lines.
A line midway between the two parallel lines,
parallel to and equidistant from each.
b. In space, the points that are equidistant from two parallel planes.
A plane midway between the two parallel planes,
parallel to and equidistant from each.
Imagine plane M as a horizontal plane. Because distance is measured along
a perpendicular segment from a point to a plane, the locus of points in
space that are 4 cm from a plane M are a plane 4 cm above and parallel
to plane M and another plane 4 cm below and parallel to plane M.
A
244
Geometry Lesson 12-6
All-In-One Answers Version A
Daily Notetaking Guide
Daily Notetaking Guide
B
Geometry Lesson 12-6
Geometry
245
61
Geometry: All-In-One Answers Version B (continued)
Guided Problem Solving 1-2
Chapter 1
1. They represent three-dimensional objects on a twodimensional surface. 2. nine 3. See the figure in 4, below.
Practice 1-1
1. 47, 53 2. 42, 54 3. -64, 128 4. Sample: 2 or 3
5. 6 or 8 6. Y or A 7. any hexagon 8. a 168.75° angle
9. 34 10. Sample: The farther out you go, the closer the
4.
ratio gets to a number that is approximately 0.618.
11. 0, 1, 1, 2, 3, 5, 8, 13
Guided Problem Solving 1-1
All rights reserved.
1. The pattern is easier to visualize. 2. The graph will go up.
3. Use Years for the horizontal axis. 4. Use Number of
Stations for the vertical axis. 5. increasing 6. greater
7. Yes. Since the number of stations increases steadily from
1950 to 2000, we can be confident that the number of stations
in 2010 will be greater than in 2000. 8. Patterns are necessary
to reach a conclusion through inductive reasoning. 9. (any
list of numbers without a pattern would apply) 2,435; 16,439;
16,454; 3,765; 210,564
5. Yes. It is similar to the foundation drawing, except there are
no numbers. 6. no
7.
Practice 1-2
1.
Top
Fro
nt
Rig
ht
nt
h
Rig
1
1
1
1
1
1
Top
Front
Right
1
1
1
Front
1. Collinear points lie on the same line. 2. Answers may vary.
(Some people might note that the y-coordinate of two of the
points is the same so that the third point must have the same
y-coordinate to be collinear. Since it does not, the points are
not collinear.) 3. horizontal 4. No. 5. No. 6. All points
must have the same y-coordinate, -3. 7. No. 8. a1, 21 b
2
Practice 1-4
1. true 2. false 3. false 4. false 5. JK, HG 6. any
* )
3
Front
Top
5. A, C, D 6. C 7. D 8. B 9. A
L1
3
Guided Problem Solving 1-3
Front
4.
2
1. AC 2. any two of the following: ABD, DBC, CBE,
ABE, ECD, ADE, ACE, ACD 3. * yes) 4. no 5. yes
6. yes
* )7. yes 8. yes 9. G 10. LM * 11.) the empty set
12. KP 13. Sample: plane ABD 14. AB 15. no
16. yes 17. the empty set 18. no 19. yes 20. yes
t
Right
3.
4
Right
Practice
* ) 1-3
Right
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2.
Fro
Front
8. Yes.
9.
All-In-One Answers Version B
Right
* )* )
* )*
)
three of the following pairs: EF and JH ; EF and GK ; HG
* )* ) * )* ) * )* ) * )* )
and JE ; HG and FK ; JK and EH ; JK and FG ; EJ and
* )* ) * )* ) * )* ) * )* ) * )
FG ; EH and FK ; JE and KG ; EH and KG ; JH and KF ;
* ) * )
JH and GF 7. planes A and B 8. planes A and C
Geometry
41
Geometry: All-In-One Answers Version B
)
)
)
)
)
9. Sample: EG 10. EF and ED or EG and ED
) )
11. FE , FD 12. yes
13. Sample:
S
(continued)
Practice 1-6
1. any three of the following: &O, &MOP, &POM, &1
2. &AOB 3. &EOC 4. &DOC 5. 51 6. 90
7. 141 8. 68 9. &ABD, &DBE, &EBF, &DBF, &FBC
10. &ABF, &DBC 11. &ABE, &EBC
T
Guided Problem Solving 1-6
U
1. Angle Addition Postulate 2. supplementary angles
3. m&RQS + m&TQS = 180 4. (2x + 4) + (6x + 20) = 180
5. x = 19.5 6. m&RQS = 43; m&TQS = 137 7. The sum
of the angle measures should be 180; m&RQS + m&TQS =
43 + 137 = 180. 8a. x = 11 8b. m&AOB = 17;
m&COB = 73
14. Sample:
R
Practice 1-7
W
All rights reserved.
Q
1.
P
X
B
Y A
2.
Guided Problem Solving 1-4
X
1. Opposite rays are two collinear rays with the same
endpoint. 2. a line 3–4. See graph in Exercise 5 answer.
5. Answers may vary. Sample: (0,0) (Answers will be
Y
coordinates (x, y), wherey = 32x, x < 2.)
y
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3.
6
B(4,6)
4
X
A(2,3)
2
Y
x
2
2
2
C(0,0)
4
X
6
4.
6. yes 7. L(4, 2)
Practice 1-5
1. 4 2. 12 3. 20 4. 6 5. 22 6. -3, 4 7. no 8. -2
9. 11 10. 29 11. 29
45°
90°
Guided Problem Solving 1-5
1. AD DC 2. AD = DC 3. Segment Addition Postulate.
4. Since AD = DC, AC = 2(AD). 5. AC = 2(12) = 24
6. y = 15 7. DC = AD = 12 8. Answers may vary.
9. ED = 11, DB = 11, EB = 22
42
Geometry
All-In-One Answers Version B
L1
Geometry: All-In-One Answers Version B (continued)
Practice 1-8
5.
1.–5.
2
y
6
A
1
C
4
A
2
1
2
6
6.
D
4
2
6 x
4
2
E
B
4
2 ? 2
2
All rights reserved.
2
6
C
7.
X
6. 12 7. 13 8. (5, 5) 9. (-2 12 , 6) 10. (-0.3, 3.4)
11. (5, -2) 12. yes; AB = BC = CD = DA = 6
13. "401 20.025
14.
y
S
6
Q
8.
4
2
P
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2 ? X
R
2
4
6
8x
15. 24.7 16. (3.5, 3)
W
Guided Problem Solving 1-8
9. true 10. false 11. false 12. true
Guided Problem Solving 1-7
1. &DBC > &ABC 2. complementary angles
3. &CBD 4. m&CBD = m&CBA = 41
5. m&ABD = m&CBA + m&CBD = 41 + 41 = 82
6. m&ABE + m&CBA = 90
m&ABE + 41 = 90
m&ABE = 49
7. m&DBF = m&ABE = 49 8. Answers may vary. Sample:
The sum of the measures of the complementary angles should
be 90 and the sum of the measures of the supplementary
angles should be 180. 9. m&CBD= 21, m&FBD = 69,
m&CBA =21, and m&EBA = 69
L1
All-In-One Answers Version B
1. Distance Formula 2. The distance d between two points
A(x1,y1) to B(x2,y2) is d 5 Í (x2 2 x1) 2 1 (y 2 2 y1) 2
3. No; the differences are opposites but the squares of the
differences are the same.
4. XY 5 Í (5 2 (26)) 2 1 (22 2 9) 2 5. To the nearest
tenth, XY = 15.6 units. 6. To the nearest tenth, XZ =
12.0 units. 7. Z is closer to X. 8. The results are the same;
e.g., XY 5 Í (26 2 5) 2 1 (9 2 (22)) 2 5 Í 242 , or
about 15.6 units, as before.
9. YZ 5 Í (17 2 (26)) 2 1 (23 2 9) 2 5 Í 673 ; to the
nearest tenth, YZ = 25.9 units. To the nearest tenth,
XY+YZ+XZ = 53.5 units.
Geometry
43
Geometry: All-In-One Answers Version B
Practice 1-9
(continued)
1B: Reading Comprehension
* ) * )*
1. 792 in.2 2. 2.4 m2 3. 16p 4. 7.8p 5. 26 cm; 42 cm2
6. 29 in.; 42 in.2 7. 40 m; 99 m2 8. 26; 22 9. 30; 44
10. 156.25p 11. 10,000p 12. 36 13. 26; 13
) *
)
1. Answer may vary. Sample: AB 6 CD , EF 6 GH ,
Guided Problem Solving 1-9
JK > LM, JL > KM, m/AJF + m/FJK = 180°,
) * )
/HKB > /KMD,DN ' CD 2. Points A, M, and S
* ) * )
* )
are collinear. 3. AB , HI , and LN intersect at point M.
4. a
1. six 2. It is a two-dimensional pattern you can fold to form
a three-dimensional object. 3. rectangles
1C: Reading/Writing Math Symbols
4.
1. Line BC is parallel to line MN. 2. Line CD
3. Line segment GH 4. Ray AB 5. The length of
6
4
4
4
4
segment XY is greater than the length of segment ST.
6. MN = XY 7. GH = 2(KL) 8. ST ' UV
9. plane ABC || plane XYZ 10. AB 6 DE
6
8
4
4
4
1. parallel planes 2. Segment Addition Postulate
3. supplementary angles 4. opposite rays 5. isometric
drawing 6. perpendicular lines 7. foundation drawing
8. right angle 9. congruent sides
6
1E: Vocabulary Check
6
2
Net: A two-dimensional pattern that you can fold to form a
three-dimensional figure.
Conjecture: A conclusion reached using inductive reasoning.
Collinear points: Points that lie on the same line.
Midpoint: A point that divides a line segment into two
congruent segments.
Postulate: An accepted statement of fact.
2
6. They are equal. 7. 208 in. 8. Answers will
vary. Sample: 2(4 6) + 2(4 8) + 2(6 8); the results are the
same, 208 in.2 9. 6(72) = 294 in.2
5. 208 in.
1A: Graphic Organizer
1. Tools of Geometry 2. Answers may vary. Sample: patterns
and inductive reasoning; measuring segments and angles; basic
constructions; and the coordinate plane 3. Check students’
work.
44
Geometry
1F: Vocabulary Review Puzzle
C
O
L
L
I
N
E
A
R
P
P
D
C
L
Z
J
P
S
Y
J
E
Z
C
C
C
E
O
R
H
L
O
E
G
E
L
K
C
A
F
R
T
S
V
I
L
A
B
B
N
L
I
L
A
E
A
O
R
E
U
N
N
X
W
L
T
Z
S
L
B
G
N
W
I
N
N
Y
L
T
I
Y
Y
U
R
G
T
A
C
N
E
L
F
Q
G
G
D
P
C
O
S
C
S
B
R
R
K
A
C
I
E
N
I
L
R
D
M
E
P
I
D
O
U
A
X
T
F
N
R
V
P
V
E
U
A
A
J
D
V
K
C
P
Y
H
U
E
D
I
C
V
D
N
E
H
X
N
I
I
T
X
S
G
M
S
L
X
S
E
G
M
E
N
T
E
O
M
I
A
V
I
X
K
F
L
P
L
S
D
K
V
T
P
R
C
O
F
J
R
N
E
H
B
L
E
V
D
V
J
S
S
R
R
E
N
L
W
F
Q
G
N
I
N
O
S
A
E
R
E
V
I
T
C
U
D
N
I
R
O
T
C
E
S
I
B
L
P
J
A
E
N
R
U
K
R
S
T
R
A
I
G
H
T
A
N
G
L
E
W
U
Q
A
P
E
L
G
N
A
E
T
U
C
A
W
K
M
F
B
O
Z
N
T
U
B
A
R
N
K
I
W
T
K
Q
R
U
I
V
C
F
All-In-One Answers Version B
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4
All rights reserved.
1D: Visual Vocabulary Practice
8
L1
Geometry: All-In-One Answers Version B
Chapter 2
Practice 2-1
1. Sample: It is 12:00 noon on a rainy day. 2. Sample: 6
3. If you are strong, then you drink milk. 4. If a rectangle is
a square, then it has four sides the same length. 5. If x = 26,
then x - 4 = 22; true. 6. If m is positive, then m2 is
positive; true. 7. If lines are parallel, then their slopes are
equal; true. 8. Hypothesis: If you like to shop; conclusion:
Visit Pigeon Forge outlets in Tennessee. 9. If you visit Pigeon
Forge outlets, then you like to shop.
10. Drinking Sustain
makes you train harder and run faster. 11. If you drink
Sustain, then you will train harder and run faster. 12. If you
All rights reserved.
train harder and run faster, then you drink Sustain.
Guided Problem Solving 2-1
1. Hypothesis: x is an integer divisible by 3. 2. Conclusion:
x2 is an integer divisible by 3. 3. Yes, it is true. Since 3 is a factor
of x, it must be a factor of x ? x = x2. 4. If x2 is an integer
divisible by 3 then x is an integer divisible by 3. 5. The converse
is false. Counterexamples may vary. Let x2 = 3. Then x 5 Í 3 ,
which is not an integer and is not divisible by 3. 6. No. The
conditional is true, so there is no such counterexample. 7. No.
By definition, a general statement is false if a counterexample
can be provided. 8. If 5x + 3 = 23, then x = 4. The original
statement and the converse are both true.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 2-2
1. Two angles have the same measure if and only if they are
congruent. 2. The converse, “If ∆n∆ = 17, then n = 17,” is not
true. 3. If a whole number is a multiple of 5, then its last digit
is either 0 or 5. If a whole number has a last digit of 0 or 5,
then it is a multiple of 5. 4. If two lines are perpendicular,
then the lines form four right angles. If two lines form four
right angles, then the lines are perpendicular. 5. Sample:
Other vehicles, such as trucks, fit this description. 6. Sample:
Baseball also fits this definition. 7. Sample: Pleasing,
smooth, and rigid all are too vague. 8. yes 9. no 10. yes
Guided Problem Solving 2-2
1. A good definition is clearly understood, precise, and
reversible. 2. &3 and &4, &5 and &6 3. No 4. They are
not supplementary. 5. A linear pair has a common vertex,
shares a common side, and is supplementary. 6. yes 7. yes;
yes 8. linear pairs: &1 and &2, &3 and &4; not linear pairs:
&1 and &3, &1 and &4, &2 and &3, &2 and &4
Practice 2-3
1. Football practice is canceled for Monday. 2. DEF is a
right triangle. 3. If two lines are not parallel, then they
intersect at a point. 4. If you vacation at the beach, then you
like Florida. 5. Tamika lives in Nebraska. 6. not possible
7. It is not freezing outside. 8. Shannon lives in the smallest
state in the United States.
L1
All-In-One Answers Version B
(continued)
Guided Problem Solving 2-3
1. conditional; hypothesis 2. Yes 3. Beth will go.
4. Anita, Beth, Aisha, Ramon 5. No; only two students went.
6. Beth, Aisha, Ramon; no—only two went. 7. Aisha, Ramon
8. The answer is reasonable. It is not possible for another pair
to go to the concert. 9. Ramon
Practice 2-4
1. UT = MN 2. y = 51 3. JL 4. Addition;
Subtraction Property of Equality; Multiplication Property of
Equality; Division Property of Equality 5. Substitution
6. Substitution 7. Symmetric Property of Congruence
8. Definition of Complementary Angles; 90, Substitution; 3x,
Simplify; 3x, 84, Subtraction Property of Equality; 28, Division
Property of Equality
Guided Problem Solving 2-4
1. Angle Addition Postulate 2. Substitution Property of
Equality; Simplify; Addition Property of Equality; Division
Property of Equality 3. 40 4. yes; yes 5. 13; 13
Practice 2-5
1. 30 2. 15 3. 6 4. m&A = 135; m&B = 45
5. m&A = 10; m&B = 80 6. m&PMO = 55;
m&PMQ = 125; m&QMN = 55
7. m&BWC = m&CWD, m&AWB + m&BWC = 180;
m&CWD + m&DWA = 180; m&AWB = m&AWD
Guided Problem Solving 2-5
1. 90 2. See graph in Exercise 5 answer. 3. on the positive
y-axis 4. Answers may vary. B can be any point on the
positive y-axis. Sample: B(0, 3).
5.
y
B(0,3)
A(1,3)
X(4,0)
x
O(0,0)
C(3,1)
6. They are adjacent complementary angles. 7. Answers may
vary. C can be any point on the line y 5 21 x, x . 0.
3
Sample: C(3, -1). 8. a right angle 9. Yes; their sum
corresponds to the right angle formed by the positive x-axis
and the positive y-axis. 10. Answers may vary. D can be any
point on the negative x-axis, sample: D(-4, 0)
Geometry
45
Geometry: All-In-One Answers Version B
(continued)
2A: Graphic Organizer
2D: Visual Vocabulary Practice
1. Reasoning and Proof 2. Answers may vary. Sample:
1. Law of Detachment 2. hypothesis 3. Distributive
Property 4. Reflexive Property 5. Law of Syllogism
6. biconditional 7. conclusion 8. good definition
9. Symmetric Property
conditional statements; writing biconditionals; converses; and
using the Law of Detachment 3. Check students’ work.
2B: Reading Comprehension
2E: Vocabulary Check
1. 42 degrees 2. 38 degrees 3. b
Truth value: “True” or “false” according to whether the
2C: Reading/Writing Math Symbols
statement is true or false, respectively
Hypothesis: The part that follows if in an if-then statement.
Biconditional: The combination of a conditional statement
1. Segment MN is congruent to segment PQ. 2. If p, then q.
3. The length of MN is equal to the length of PQ.
4. Angle XQV is congruent to angle RDC. 5. If q, then p.
6. The measure of angle XQV is equal to the measure of
angle RDC. 7. p if and only if q. 8. a S b 9. AB = MN
10. m/XYZ = m/RPS 11. b S a 12. AB > MN
13. a 4 b 14. /XYZ /RPS
2F: Vocabulary Review Puzzle
P
O
S
T
U
L
A
T
2
V
4
E
P
R
5
7
C
O
G
C
O
N
S
T
D
I
J
C
C
L
P
A
R
C
O
L
I
N
A
A
J
T
N
E
E
U
T
G
C
E
A
L
T
M
N
E
U
I
G
S
R
C
T
I
O
11
N
T
D
12
H
Y
P
O
L
O
T
H
E
A
L
L
E
L
N
E
E
U
T
I
N
R
N
D
8
A
O
I
O
6
T
A
9
N
10
A
C
O
R
E
S
I
S
S
I
N
T
Chapter 3
Practice 3-1
1. corresponding angles 2. alternate interior angles 3. sameside interior angles 4. 1 and 5, 2 and 6, 3 and 8,
4 and 7 5. 4 and 6, 3 and 5 6. 4 and 5, 3
and 6 7. m1 = 100, alternate interior angles;
m2 = 100, corresponding angles or vertical angles
8. m1 = 135, corresponding angles; m2 = 135, vertical
angles 9. x = 103; 77°, 103° 10. x = 30; 85°, 85°
46
Geometry
Guided Problem Solving 3-1
1. The top and bottom sides are parallel, and the left and right
sides are parallel. 2. The two diagonals are transversals, and
also each side of the parallelogram is a transversal for the two
sides adjacent to it. 3. Corresponding angles, interior and
exterior angles are formed. 4. v, w and x; By the Alternate
Interior Angles Theorem, v = 42, w = 25 and x = 76.
5. Answers may vary. Possible answer: By the Same-Side
Interior Angles Theorem, (w + 42) + (y + 76) = 180. Since
w = 25, y = 37. (The two y’s are equal by Theorem 3-1.)
6. w = 25, y = 37, v = 42, x = 76; yes 7. v = 42, w = 35,
x = 57, y = 46
All-In-One Answers Version B
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
3
All rights reserved.
and its converse; it contains the words “if and only if.”
Conclusion: The part of an if-then statement that follows
then.
Conditional: An if-then statement.
Geometry: All-In-One Answers Version B (continued)
Practice 3-2
Practice 3-5
1. l and m, Converse of Same-Side Interior Angles Theorem
2. none 3. BC and AD , Converse of Same-Side Interior
Angles Theorem 4. BH and CI, Converse of Corresponding
Angles Postulate 5. 43 6. 90 7. 38 8. 100
1. x = 120; y = 60 2. n = 5173 3. a = 108; b = 72
4. 109 5. 133 6. 129 7. 30 8. 150 9. 6 10. 5
11. BEDC 12. FAE 13. FAE and BAE
14. ABCDE
Guided Problem Solving 3-2
Guided Problem Solving 3-5
1. and m 2. transversals 3. x 4. the angles measuring
19xº and 17xº 5. 180º 6. 17xº 7. 180 - 19x = 17x or
19x+17x = 180 8. x = 5 9. With x = 5, 19x = 95 and
17x = 85. 10. x = 6
1. A theater stage, consisting of a large platform surrounding
a smaller platform. The shapes in the bottom part of the figure
may represent a ramp for actors to enter and exit. 2. The
measures of angles 1 and 2 3. 8; octagon 4. (8 - 2)180 =
1080 degrees 5. 135 6. 45 7. Yes, angle 1 is an obtuse angle
and angle 2 is an acute angle. 8. trapezoids 9. 360
All rights reserved.
Practice 3-3
1. True. Every avenue will be parallel to Founders Avenue, and
therefore every avenue will be perpendicular to Center City
Boulevard, and therefore every avenue will be perpendicular
to any street that is parallel to Center City Boulevard.
2. True. The fact that one intersection is perpendicular, plus the
fact that every street belongs to one of two groups of parallel
streets, is enough to guarantee that all intersections are
perpendicular. 3. Not necessarily true. If there are more than
three avenues and more than three boulevards, there will be
some blocks bordered by neither Center City Boulevard nor
Founders Avenue. 4. a ⊥ e 5. a || e 6. a || e 7. a || e
8. If the number of ⊥ statements is even, then 1 || n. If it is
odd, then 1 ⊥ n.
Practice 3-6
1. y = 13 x - 7 2. y = -2x + 12 3. y = 45 x - 2
4. y = 4x - 13
5.
y
y 2x
6.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y
6
4 y 5x 4
2
1. supplementary angles 2. right angle 3. Any one of the
following: Postulate 3-1, or Theorem 3-1, 3-2, 3-3 or 3-4 4. 90
5. It is congruent; Postulate 3-1 6. 90 7. a ⊥ c 8. It is true
for any line parallel to b. 9. Yes. The point is that a
2 4 6 x
–6–4–2
–4
–6
y
7.
6
4
2
Practice 3-4
1. 125 2. 143 3. 129 4. 136 5. x = 35; y = 145; z = 25
6. v = 118; w = 37; t = 62 7. 50 8. 88 9. m1 = 33;
m2 = 52 10. right scalene 11. obtuse isosceles
12. equiangular equilateral
2 4 6 x
–6–4–2
–2
–4
–6
Guided Problem Solving 3-3
transversal cannot be perpendicular to just one of two parallel
lines. It has to be perpendicular to both, or else to neither.
6
4
y 4x 1
2 4 6 x
–6–4–2
–2
–6
8. y = x + 4 9. y = 12 x - 3 10. y = -21 x -
1
2
Guided Problem Solving 3-4
11. y = -6x + 45 12 y = -11; x = 2 13. y = 2; x = 0
1. three 2. 180 3. right triangle 4. z = 90; Because it is
given in the figure that BD ⊥ AC. 5. Theorem 3-12, the
Triangle Angle-Sum Theorem 6. x = 38 7. y = 36
8. #ABD is a 36-54-90 right triangle. #BCD is a 38-52-90
right triangle. 9. 74 10. #ABC is a 52-54-74 acute triangle.
11. Yes, all three are acute angles, with &ABC visibly larger
than &A and &C. 12. &BCD
14.
L1
All-In-One Answers Version B
y
(–2, 0) 6
4
2
–6 –4 –2 2 4 6 x
–2
–4
–6 (0, –1)
Geometry
47
Geometry: All-In-One Answers Version B
Guided Problem Solving 3-7
y
6 (0, 6)
4
2
y
K
2 4 6 x
–6–4–2
–2
–4
–6
16.
1.
(6, 0)
H
4
4
6
4
2 (4, 0)
y
B
6. Product = - 85 -1. Sides GH and GK are not
perpendicular. 7. #GHK has no pair of perpendicular sides.
It is not a right triangle. 8. No 9. &HGK; approximately 80
10. Slope of LM = 72 and slope of LN = - 27 . The product
of the slopes is -1, so LM and LN are perpendicular.
2
2
4
2. a right angle 3. m1 m2 = -1 4. sides GH and GK
5. Slope of GH = 35 ; slope of GK = 83
Guided Problem Solving 3-6
O A
O
4
2 4 6 x
–6–4–2
–2
–4 (0, –4)
–6
1.
x
G
y
2
x
C
All rights reserved.
15.
(continued)
Practice 3-8
1.
2.
2
Q
å
å
AB = 5 ; slope of BC = - 5 . The absolute values of the
2
2
slopes are the same, but one slope is positive and the other
is negative. 5. Point-slope form: y - 0 = 5 (x - 0);
2
slope-intercept form: y = 5 x 6. Point-slope form:
2
5
y - 5 = - (x - 2) or y - 0 = - 5 (x - 4);
2
2
å
slope-intercept form: y = - 5 x + 10 7. Of line AB : (0, 0);
2
å
of line BC : (0, 10) 8. #ABC appears to be an isosceles
triangle, which is consistent with a horizontal base and two
remiaining sides having slopes of equal magnitude and opposite
sign. 9. Slope = 0; y = 0; y-intercept = (0, 0) just as for line
å
AB (they intersect on the y-axis).
Practice 3-7
1. neither; 3 2 13, 3 ? 13 2 -1 2. perpendicular;
2
1
2
2 ? -2 = -1 3. parallel; -3 = -3 4. perpendicular;
T
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y 2y
2. m 5 x22 2 x11 3. y - y1 = m(x-x1) 4. Slope of
3.
K
4. Sample:
b
a
y = 2 is a horizontal line, x = 0 is a vertical line
5. perpendicular; -1 ? 1 = -1 6. neither; 12 2 -53,
9
9
5
1
2 ? –3 2 -1 7. neither; 2 2 4, 2 ? 4 2 -1 8. parallel;
2
1
1
2 = 2 9. y = 3 x 10. y = 2x - 4
48
Geometry
All-In-One Answers Version B
L1
Geometry: All-In-One Answers Version B (continued)
7. They appear to be both congruent and parallel. 8. The
answers to Step 7 are confirmed. 9. yes; a parallelogram
5.
3A: Graphic Organizer
b
b
b
b
1. Parallel and Perpendicular Lines 2. Answers may vary.
Sample: properties of parallel lines; finding the measures of
angles in triangles; classifying polygons; and graphing lines
3. Check students’ work.
3B: Reading Comprehension
1. 14 spaces 2. 6 spaces 3. 60° 4. corresponding
5. $7000; $480 6. the width of the stalls, 10 ft 7. b
3C: Reading/Writing Math Symbols
6.
å å
All rights reserved.
1. m # n 2. m/1 + m/2 = 180 3. AB CD
4. m/MNP + m/MNQ = 90 5. /3 /EFD
6. Line 1 is parallel to line 2. 7. The measure of angle ABC
is equal to the measure of angle XYZ.
8. Line AB is perpendicular to line DF. 9. Angle ABC and
angle ABD are complementary. 10. Angle 2 is a right angle,
or the measure of angle 2 is 90°. 11. Sample answer:
å å
CB GD, m/BAF = m/GFA
a
c
3D: Visual Vocabulary Practice/High-Use
Academic Words
Guided Problem Solving 3-8
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. a line segment of length c 2. Construct a quadrilateral
with one pair of parallel sides of length c, and then examine
the other pair. 3. The procedure is given on p. 181 of the text.
1. property 2. conclusion 3. describe 4. formula
5. measure 6. approximate 7. compare 8. contradiction
9. pattern
3E: Vocabulary Check
m
Transversal: A line that intersects two coplanar lines in two
points.
l
Alternate interior angles: Nonadjacent interior angles that
lie on opposite sides of the transversal.
4. Adjust the compass to exactly span line segment c, end to
end. Then tighten down the compass adjustment as necessary.
5.
m
l
6.
m
l
L1
All-In-One Answers Version B
Same-side interior angles: Interior angles that lie on the
same side of a transversal between two lines.
Corresponding angles: Angles that lie on the same side of a
transversal between two lines, in corresponding positions.
Flow proof: A convincing argument that uses deductive
reasoning, in which arrows show the logical connections
between the statements.
3F: Vocabulary Review
1. C 2. E 3. D 4. B 5. A 6. F 7. K 8. H 9. L 10. G
11. I 12. J
Geometry
49
Geometry: All-In-One Answers Version B
1. m1 = 110; m2 = 120 2. m3 = 90; m4 = 135
3. CA JS , AT SD , CT JD 4. C J,
A S, T D
5. Yes; GHJ IHJ by
Theorem 4-1 and by the Reflexive Property of . Therefore,
GHJ IHJ by the definition of triangles.
6. No; QSR TSV because vertical angles are
congruent, and QRS TVS by Theorem 4-1, but none
of the sides are necessarily congruent. 7a. Given
7b. Vertical angles are . 7c. Theorem 4-1 7d. Given
7e. Definition of triangles
Guided Problem Solving 4-1
1. right triangles 2. m&A = 45, m&B = m&L = 90,
and AB = 4 in. 3. #ABC is congruent to #KLM means
corresponding sides and angles are congruent. 4. x and t
5. 45 6. m&K = m&M = 45 7. 3x = 45 8. x = 15 9. 4
10. 2t = 4 11. t = 2 12. The angle measures indicate that
the two triangles are isosceles right triangles. This matches the
appearance of the figure. 13. m&M = 60
Practice 4-4
1. BD is a common side, so ADB CDB by SAS, and
A C by CPCTC. 2. FH is a common side, so
FHE HFG by ASA, and HE FG by CPCTC.
3. QS is a common side, so QTS SRQ by AAS.
QST SQR by CPCTC. 4. ZAY and CAB are
vertical angles, so ABC AYZ by ASA. ZA AC by
CPCTC. 5. JKH and LKM are vertical angles, so
HJK MLK by AAS, and JK KL by CPCTC.
6. PR is a common side, so PNR RQP by SSS, and
N Q by CPCTC. 7. First, show that ACB and
ECD are vertical angles. Then, show ABC EDC by
ASA. Last, show A E by CPCTC.
Guided Problem Solving 4-4
1. ISOS and SP bisects &ISO. 2. Prove whatever additional
facts can be proven about #ISP and #OSP, based on the
given information. 3. IS > SO. 4. &ISP &PSO 5. SP
6. #ISP #OSP by Postulate 4-2, the Side-AngleSide(SAS)Postulate 7. It does not matter. The Side-AngleSide Postulate applies whether or not they are collinear.
8. It does follow, because #ISP #OSP and because IP and
PQ are corresponding parts.
1. A compass with a fixed setting was used to draw two
circular arcs, both centered at point P but crossing in
different locations, which were labeled A and B. The compass
was used again, with a wider setting, to draw two intersecting
circular arcs, one centered at A and one at B. The point at
å
which the new arcs intersected was labeled C. Finally, line CP
was drawn. 2. Find equal lengths or distances and explain
å
why CP is perpendicular to . 3. #ACP and #BCP
4. AP = PB and AC = BC. 5. #APC > #BPC, by
Postulate 4-1, the Side-Side-Side (SSS) Postulate
6. &APC > &BPC by CPCTC 7. Since &APC > &BPC,
m&APC = m&BPC and m&APC + m&BPC = 180, it
follows that m&APC = m&BPC = 90. 8. from the
definition of perpendicular and the fact that m&APC =
m&BPC = 90 9. The distances do not matter, so long as
AP = BP and AC = BC. That is what is required in order that
#APC > #BPC. 10. Draw a line, and use the construction
technique of the problem to construct a second line
perpendicular to the first. Then do the same thing again to
construct a third line perpendicular to the second line. The
first and third lines will be parallel, by Theorem 3-10.
Practice 4-3
Practice 4-5
Practice 4-2
1. ADB CDB by SAS 2. not possible
3. TUS XWV by SSS 4. not possible
5. DEC GHF by SAS 6. PRN PRQ
by SSS 7. C 8. AB and BC 9. A and B
10. AC 11a. Given 11b. Reflexive Property of
Congruence 11c. SAS Postulate
Guided Problem Solving 4-2
1. not possible 2. ASA Postulate 3. AAS Theorem
4. ASA Postulate 5. not possible 6. AAS Theorem
7. Statements
Reasons
1. K M, KL ML
1. Given
2. JLK PLM
2. Vertical s are .
3. JKL PML
3. ASA Postulate
8. BC EF 9. KHJ HKG or KJH HGK
Guided Problem Solving 4-3
1. Corresponding angles and alternate interior angles.
2. &EAB and &DBC. 3. &EBA and &DCB. 4. &EAB
50
Geometry
All rights reserved.
Practice 4-1
and &DBC 5. &EAB > &DBC, AE > BD, and
&E > &D. 6. #AEB > #BDC by Postulate 4-3, the AngleSide-Angle (ASA) Postulate 7. Yes; Theorem 4-2, the
Angle-Angle-Side (AAS) Theorem; &EBA > &DCB.
8. No, because now there is no way to demonstrate a second
pair of congruent sides, nor a second pair of congruent angles.
1. x = 35; y = 35 2. x = 80; y = 90 3. t = 150
4. x = 55; y = 70; z = 125 5. x = 6 6. z = 120
7. AD ; D F 8. KJ ; KIJ KJI 9. BA ;
ABJ AJB 10. 130 11. 130 12. x = 70; y = 55
Guided Problem Solving 4-5
1. One angle is obtuse. The other two angles are acute and
congruent. 2. Highlight an obtuse isosceles angle and find its
angle measures, then find all the other angle measures
represented in the figure.
All-In-One Answers Version B
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Chapter 4
(continued)
Geometry: All-In-One Answers Version B (continued)
3. Possible answer:
8.
hypotenuse
shorter leg =
2; yes, this matches the figure. 9. The
solution remains the same: x = 3 and y = 2. The reason is
that one is still solving the same two equations, x = y + 1 and
x + 3 = 3y.
Practice 4-7
4. 60 5. 30, because the measure of each base angle is half
the measure of an angle of the equilateral triangle. 6. 120°
1. ZWX YXW; SAS 2. LNP LMO; SAS
3. ADF BGE; SAS
4. F
common side: FG
G G
F
because the sum of the angles of the highlighted triangle must
equal 180°.
7. The other measures are 90° and 150°. Examples:
E
All rights reserved.
5.
common angle: L
L
L
N
K
J
M
6. Sample:
8. Yes; 3 120 = 360. 9. Answers may vary.
Practice 4-6
1. Statements
1. AB # BC , ED # FE
2. B, E are right s.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
H
3. AC FD , AB ED
4. ABC DEF
Reasons
1. Given
2. Perpendicular lines
form right s.
3. Given
4. HL Theorem
MJN and MJK
are right s.
2.
Perpendicular lines
form right s.
MN MK
MJN MJK
Given
HL Theorem
MJ MJ
Reflexive Property of 3. RS VW 4. none 5. mC and mF = 90
6. ST UV or SV UT 7. mA and mX = 90
8. GI # JH
Guided Problem Solving 4-6
1. Two congruent right triangles. Each one has a leg and a
hypotenuse labeled with a variable expression. 2. the values
of x and y for which the triangles are congruent by HL
3. The two shorter legs are congruent. 4. x = y + 1 5. The
hypotenuses are congruent. 6. x + 3 = 3y 7. x = 3; y = 2
L1
All-In-One Answers Version B
Statements
1. AX AY
2. CX # AB, BY # AC
3. mCXA and
mBYA = 90
4. A A
5.
BYA CXA
Reasons
1. Given
2. Given
3. Perpendicular lines
form right s.
4. Reflexive Property
of 5. ASA Postulate
Guided Problem Solving 4-7
1. The figure, a list of parallel and perpendicular pairs of sides,
and one known angle measure, namely m&A = 56
2. nine 3. They are congruent and have equal measures.
4. m&A = m&1 = m&2 = 56 5. m&4 = 90
6. m&3 = 34 7. m&DCE = 56 8. m&5 = 22
9. m&FCG = 90 10. m&6 = 34 11. m&7 = 34,
m&8 = 68, and m&9 = 112 12. m&9 = 56 + 56 = 112
13. m&FIC = 180 -(m&2 + m&3) = 90;
m&DHC = m&4 = 90; m&FJC = 180 - m&9 = 68;
m&BIG = m&FIC = 90
4A: Graphic Organizer
1. Congruent Triangles 2. Answers may vary. Sample:
congruent figures; triangle congruence by SSS, SAS, ASA,
and AAS; proving parts of triangles congruent; the Isosceles
Triangle Theorem 3. Check students’ work.
4B: Reading Comprehension
1. Yes. Using the Isosceles Triangle Theorem, /W /Y.
It is given that WX YX and WU YV. Therefore
nWUX nYVX by SAS. 2. There is not enough
information. You need to know if AC EC, if /A /E, or
if /B /D. 3. a
Geometry
51
(continued)
4C: Reading/Writing Math Symbols
Guided Problem Solving 5-2
1. Angle-Angle-Side 2. triangle XYZ 3. angle PQR
4. line segment BD 5. line ST 6. ray WX 7. hypotenuseleg 8. line 3 9. angle 6 10. Angle-Side-Angle
1.
y
4
1. theorem 2. congruent polygons 3. base angle of an
isosceles triangle 4. CPCTC 5. postulate 6. vertex angle of
an isosceles triangle 7. corollary 8. base of an isosceles
triangle 9. legs of an isosceles triangle
O
corresponding sides congruent and corresponding angles
congruent.
CPCTC: An abbreviation for “corresponding parts of
congruent triangles are congruent.”
4F: Vocabulary Review Puzzle
1. postulate 2. hypotenuse 3. angle 4. vertex 5. side
6. leg 7. perpendicular 8. polygon 9. supplementary
10. parallel 11. corresponding
Chapter 5
Practice 5-1
1a. 8 cm 1b. 16 cm 1c. 14 cm 2a. 9.5 cm 2b. 17.5 cm
2c. 14.5 cm 3. 17 4. 7 5. 42 6. 16.5 7a. 18 7b. 61
8. PR 6 YZ, PQ 6 XZ, XY 6 RQ
Guided Problem Solving 5-1
1. 30 units 2. The three sides of the large triangle are each
bisected by intersections with the two line segments lying in
the interior of the large triangle. 3. the value of x 4. They
are called midsegments. 5. They are parallel, and the side
labeled 30 is half the length of the side labeled x. 6. x = 60
7. Yes; the side labeled x appears to be about twice as long as
the side labeled 30. 8. No. Those lengths are not fixed by the
given information. (The triangle could be vertically stretched
or shrunk without changing the lengths of the labeled sides.)
All one can say is that the midsegment is half as long as the
side it is parallel to.
Practice 5-2
1. WY is the perpendicular bisector of XZ. 2. 4 3. 9
4. right triangle 5. 5 6. 17 7. )isosceles triangle 8. 3.5
9. 21 10. right triangle 11. JP is the bisector of LJN.
12. 9 13. 45 14. 14 15. right isosceles triangle
52
Geometry
4
C
B x
2. See answer to Step 1, above. 3. See answer to Step 1,
above. 4. Plot a point and explain why it lies on the bisector
of the angle at the origin. 5. line : y 5 23 x 1 25 ; line m:
4
2
x = 10 6. C(10, 5) 7. CA = CB = 5; yes 8. Theorem 5-5,
the Converse of the Angle Bisector Theorem 9. m&AOC =
m&BOC < 27 10. Draw , m, and C, then draw OC. Since
OA = OB = 10, it follows that #OAC > #OBC, by HL.
Then CA > CB and &AOC > &BOC by CPCTC.
Practice 5-3
1. (-2, 2) 2. (4, 0) 3. altitude 4. median
5. perpendicular bisector 6. angle bisector
7a. (2, 0) 7b. (-2, -2) 8a. ( 0, 0) 8b. (3, -4)
Guided Problem Solving 5-3
1. the figure and a proof with some parts left blank 2. Fill in
the blanks. 3. AB 4. Theorem 5-2, the Perpendicular
Bisector Theorem 5. BC; XC 6. the Transitive Property of
Equality 7. Perpendicular Bisector. (This converse is
Theorem 5-3.) 8. The point of the proof is to demonstrate
that n runs through point X. It would not be appropriate to
show that fact as already given in the figure. 9. Nothing
essential would change. Point X would lie outside #ABC
å
(below BC ), but the proof woud run just the same.
Practice 5-4
1. I and III 2. I and II 3. The angle measure is not 65.
4. Tina does not have her driver’s license. 5. The figure does
not have eight sides. 6. ABC is congruent to XYZ.
7a. If you do not live in Toronto, then you do not live in
Canada; false. 7b. If you do not live in Canada, then you do
not live in Toronto; true. 8. Assume that mA 2 mB.
9. Assume that LM does not intersect NO. 10. Assume that
it is not sunny outside. 11. Assume that mA 90. This
means that mA + mC 180. This, in turn, means that the
sum of the angles of ABC exceeds 180, which contradicts the
Triangle Angle-Sum Theorem. So the assumption that
mA 90 must be incorrect. Therefore, mA 90.
Guided Problem Solving 5-4
1. Ice is forming on the sidewalk in front of Toni’s house.
2. Use indirect reasoning to show that the temperature of the
sidewalk surface must be 32°F or lower. 3. The temperature
All-In-One Answers Version B
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Angle: Formed by two rays with the same endpoint.
Congruent angles: Angles that have the same measure.
Congruent segments: Segments that have the same length.
Corresponding polygons: Polygons that have
m
A
4D: Visual Vocabulary Practice
4E: Vocabulary Check
<
All rights reserved.
Geometry: All-In-One Answers Version B
Geometry: All-In-One Answers Version B (continued)
All rights reserved.
of the sidewalk in front of Toni’s house is greater than 32°F.
4. Water is liquid (ice does not form) above 32°F. 5. There is
no ice forming on the sidewalk in front of Toni’s house.
6. The result from step 5 contradicts the information identified
as given in step 1. 7. The temperature of the sidewalk in
front of Toni’s house is less than or equal to 32°F. 8. If the
temperature is above 32°F, water remains liquid. This is
reliably true. Converse: If water remains liquid, the
temperature is above 32°F. This is not reliably true. Adding salt
will cause water to remain liquid even below 32°F.
9. Suppose two people are each the world’s tallest person. Call
them person A and person B. Then person A would be taller
than everyone else, including B, but by the same token B
would be taller than A. It is a contradiction for two people
each to be taller than the other. So it is impossible for two
people each to be the World’s Tallest Person.
Practice 5-5
1. M, N 2. C, D 3. R, P 4. A, T
5. yes; 4 + 7 8, 7 + 8 4, 8 + 4 7 6. no; 6 +
10 17 7. yes; 4 + 4 4 8. yes; 11 + 12 13,
12 + 13 11, 13 + 11 12 9. no; 18 + 20 40
10. no; 1.2 + 2.6 4.9 11. BO, BL, LO 12. RS, ST, RT
13. D, S, A 14. N, S, J 15. 3 x 11
16. 8 x 26 17. 0 x 10 18. 9 x 31
Guided Problem Solving 5-5
1.
5C: Reading/Writing Math Symbols
1. L 2. F 3. O 4. G 5. A 6. I 7. M 8. H 9. E
10. K 11. B 12. D 13. N 14. J 15. C
5D: Visual Vocabulary Practice
1. median 2. negation 3. circumcenter 4. contrapositive
5. centroid 6. equivalent statements 7. incenter 8. inverse
9. altitude
5E: Vocabulary Check
Midpoint: A point that divides a line segment into two
congruent segments.
Midsegment of a triangle: The segment that joins the
midpoints of two sides of a triangle.
Proof: A convincing argument that uses deductive reasoning.
Coordinate proof: A proof in which a figure is drawn on a
coordinate plane and the formulas for slope, midpoint, and
distance are used to prove properties of the figure.
Distance from a point to a line: The length of the
perpendicular segment from the point to the line.
5F: Vocabulary Review
1. altitude 2. line 3. median 4. negation 5. contrapositive
6. incenter 7. orthocenter 8. slope-intercept
9. exterior 10. obtuse 11. alternate interior 12. centroid
13. equivalent 14. right 15. parallel
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Chapter 6
Practice 6-1
2. The side opposite the larger included angle is greater than
the side opposite the smaller included angle. 3. The angle
opposite the larger side is greater than the angle opposite the
smaller side 4. The opposite sides each have a length of
nearly the sum of the other two side lengths. 5. The opposite
sides are the same length. They are corresponding parts of
triangles that are congruent by SAS.
5A: Graphic Organizer
1. Relationships Within Triangles 2. Answers may vary. Sample:
midsegments of triangles; bisectors in triangles; concurrent lines,
medians, and altitudes; and inverses, contrapositives, and indirect
reasoning 3. Check students’ work.
1. parallelogram 2. rectangle 3. quadrilateral 4. kite,
quadrilateral 5. trapezoid, isosceles trapezoid, quadrilateral
6. square, rectangle, parallelogram, rhombus, quadrilateral
7. x = 7; AB = BD = DC = CA = 11 8. f = 5;
g = 11; FG = GH = HI = IF = 17 9. parallelogram
10. kite
Guided Problem Solving 6-1
1. a labeled figure, which shows an isosceles trapezoid
2. The nonparallel sides are congruent. 3. the measures of the
angles and the lengths of the sides 4. m&G = c
5. c + (4c - 20) = 180 6. 40 7. m&D = m&G = 40,
m&E = m&F = 140 8. a - 4 = 11 9. 15 10. DE = FG =
11, EF = 15, DG = 32 11. 40 + 40 + 140 + 140 = 360 =
(4 - 2)180 12. m&D = m&G = 39, m&E = m&F = 141
Practice 6-2
5B: Reading Comprehension
1. The width of the tar pit is 10 meters. 2. b
L1
All-In-One Answers Version B
1. 15 2. 32 3. 7 4. 12 5. 9 6. 8 7. 100 8. 40; 140; 40
9. 113; 45; 22 10. 115; 15; 50 11. 55; 105; 55 12. 32; 98; 50
13. 16 14. 35 15. 28
Geometry
53
Geometry: All-In-One Answers Version B
x
know AB > AD, but it was necessary to know m&B = 90.
The key fact, which enables the use of Theorem 6-11 in addition
to Theorem 6-3, is that ABCD is a rectangle. It does not matter
whether all four sides are congruent. 9. 40
4. the measures of the angles 5. The angles are
Practice 6-5
supplementary angles, because they are consecutive.
6. x + 9x = 180 7. 18 and 162 8. No; the lengths of the
sides are irrelevant in this problem. 9. 30 and 150
1. 118; 62 2. 59; 121 3. 96; 84 4. 101; 79 5. x = 4
6. x = 1 7. 105.5; 105.5 8. 118; 118 9. 90; 63; 63
10. 107; 107 11. x = 8 12. x = 7
Practice 6-3
Guided Problem Solving 6-5
1. no 2. yes 3. yes 4. yes 5. x = 2; y = 3 6. x = 64;
y = 10 7. x = 8; the figure is a because both pairs of
opposite sides are congruent. 8. x = 25; the figure is a because the congruent opposite sides are 6 by the Converse of
the Alternate Interior Angles Theorem. 9. No; the congruent
opposite sides do not have to be 6. 10. No; the figure could
be a trapezoid. 11. Yes; both pairs of opposite sides are
congruent. 12. Yes; both pairs of opposite sides are 6 by the
converse of the Alternate Interior Angles Theorem. 13. No;
only one pair of opposite angles is congruent. 14. Yes; one
pair of opposite sides is both congruent and 6.
1. Isosceles trapezoid ABCD with AB > DC 2. &B > &C
and &BAD > &D 3. AB > DC is Given. DC > AE
because opposite sides of a parallelogram are congruent
(Theorem 6-1). AB > AE is from the Transitive Property of
Congruency. 4. Isosceles; >; because base angles of an
isosceles triangle are congruent. 5. &1 > &C because
corresponding angles on a transversal of two parallel lines are
congruent. 6. &B > &C by the Transitive Property of
Congruency. 7. &BAD is a same-side interior angle with &B,
and &D is a same-side interior angle with &C. 8. This is not
a problem, because for AD > BC there is a similar proof with
a line segment drawn from B to a point E lying on AD.
9. The two drawn segments can be shown to be congruent,
and then one has two congruent right triangles by the HL
Theorem. &B > &C follows by CPCTC and &BAD > &D
because they are supplements of congruent angles.
Guided Problem Solving 6-3
1. a labeled figure, which shows a quadrilateral that appears to
be a parallelogram 2. The consecutive angles are
supplementary. 3. find values for x and y which make the
quadrilateral a parallelogram 4. m&A + m&D = 180, so
that &A and &D meet the requirements for same-side interior
angles on the transversal of two parallel lines (Theorems 3-2
and 3-6). 5. &B > &D, by Theorem 6-2. 6. 3x + 10 + 5y =
180; 8x + 5 = 5y 7. x = 15, y = 25 8. m&A = m&C = 55
and m&B = m&D = 125, which matches the appearance of
the figure. 9. (3x + 10) + (8x + 5) = 180; yes
Practice 6-4
1a. rhombus 1b. 72; 54; 54; 72 2a. rectangle 2b. 37; 53;
106; 74 3a. rectangle 3b. 60; 30; 60; 30 4a. rhombus
4b. 22; 68; 68; 90 5. Possible; opposite angles are congruent in
a parallelogram. 6. Impossible; if the diagonals are
perpendicular, then the parallelogram should be a rhombus,
but the sides are not of equal length. 7. x = 7; HJ = 7;
IK = 7 8. x = 6; HJ = 25; IK = 25 9a. 90; 90; 29; 29
9b. 288 cm2 10a. 38; 90; 90; 38 10b. 260 m2
Guided Problem Solving 6-4
1. A labeled figure, which shows a parallelogram. One angle is a
right angle, and two adjacent sides are congruent. Algebraic
expressions are given for the lengths of three line segments.
2. diagonals 3. Find the values of x and y. 4. It is a square.
Theorem 6-1 and the fact that AB > AD imply that all four
sides are congruent. Theorems 3-11 and 6-2 plus the fact that
m&B = 90 imply that all four angles are right angles.
54
Geometry
Practice 6-6
1. (1.5a, 2b); a 2. (0.5a, 0); a 3. (0.5a, b); " a2 1 4b2
2b
4. 0 5. 1 6. -12 7. 2b
3a 8. -3a 9. E(a, 3b); I(4a, 0)
10. D(4a, b); I(3a, 0) 11. (-4a, b) 12. (-b, 0)
Guided Problem Solving 6-6
1. a rhombus with coordinates given for two vertices
2. Arhombus is a parallelogram with four congruent sides.
3. the coordinates of the other two vertices 4. They are the
diagonals. 5. They bisect each other. 6. W(-2r, 0),
Z(0, -2t) 7. No; neither Theorem 6-3 nor any other theorem
or result would apply. 8. Slope of WX = 0 and slope of YZ
is undefined. This confirms Theorem 6-10, which says that the
diagonals of a rhombus are perpendicular.
Practice 6-7
p
p2
p
q 1b. y = mx + b; q = q (p) + b; b = q - q ;
p
p
p2
y = q x + q - q 1c. x = r + p 1d. y = q (r + p) +
2
2
2
rp
rp
p
p
p
q - q ; y = q + q + q - q ; y = q + q; intersection
rp
at (r + p, q + q) 1e. qr 1f. (r, q) 1g. y = mx + b;
2
2
q = qr (r) + b; b = q - rq ; y = rqx + q - rq
2
2
2
rp
1h. y = qr (r + p) + q - rq ; y = rq + q + q - rq ;
1a.
All-In-One Answers Version B
L1
All rights reserved.
1. the ratio of two different angle measures in a parallelogram
2. The consecutive angles are supplementary.
3.
9x
5. congruent; bisect 6. 4x - y + 1 = (2x - 1) + (3y + 5);
2x - 1 = 3y + 5 7. x = 7 1 ; y = 3 8. It was not necessary to
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Guided Problem Solving 6-2
(continued)
Geometry: All-In-One Answers Version B (continued)
rp
rp
Guided Problem Solving 6-7
the given information, only one side and one angle of the two
triangles can be proven to be congruent. Another side or angle
is needed. If it were given that QUSV is a parallelogram, then
the proof could be made. 3. All four sides are congruent.
4. Yes. Since EG > EG by the Reflexive Property,
nEFG > nEHG by SSS. 5. b
1. kite DEFG with DE = EF with the midpoint of each side
identified 2. A kite is a quadrilateral with two pairs of
6C: Reading/Writing Math Symbols
y = q + q; intersection at (r + p, q + q)
rp
1i. (r + p, q + q) 2a. (-2a, 0) 2b. (-a, b)
All rights reserved.
b
2c. (-3a
2 , 2 ) 2d.
b
a
adjacent sides congruent and no opposite sides congruent.
3. The midpoints are the vertices of a rectangle.
4. D(-2b, 2c), G(0, 0) 5. L(b, a + c), M(b, c), N(-b, c),
K(-b, a + c) 6. Slope of KL = slope of NM = 0, slopes
of KN and LM are undefined 7. Opposite sides are
parallel; it is a rectangle. 8. Adjacent sides are perpendicular.
9. right angles 10. Answers will vary. Example: a = 3, b = 2,
c = 2 yields the points D(-4, 4), E(0, 6), F(4, 4), G(0, 0) with
midpoints at (-2, 2), (-2, 5), (2, 5), and (2, 2). Connecting
these midpoints forms a rectangle. 11. Construct DF and
EG. Slope of DF = 0, so DF is horizontal. Slope of EG is
undefined, so EG is vertical.
1. AH, CH, or BH 2. DG, FG, or EG 3. G
4. DE 5. rhombus 6. rectangle 7. square 8. isosceles
trapezoid
6D: Visual Vocabulary Practice/High-Use
Adademic Words
1. solve 2. deduce 3. equivalent 4. indirect 5. equal
6. analysis 7. identify 8. convert 9. common
6E: Visual Vocabulary Check
Consecutive angles: Angles of a polygon that share a
common side.
Kite: A quadrilateral with two pairs of congruent adjacent
sides and no opposite sides congruent.
Parallelogram: A quadrilateral with two pairs of parallel
sides.
Rhombus: A parallelogram with four congruent sides.
Trapezoid: A quadrilateral with exactly one pair of parallel
sides.
6A: Graphic Organizer
1. Quadrilaterals 2. Answers may vary. Sample: classifying
quadrilaterals; properties of parallelograms; proving that a
quadrilateral is a parallelogram; and special parallelograms
3. Check students’ work.
6B: Reading Comprehension
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. QT > SR, QR > ST, QT 6 RS, QR 6 TV
2. No, it cannot be proven that nQTV > nSRU because with
6F: Vocabulary Review Puzzle
1
2
P
A
R
A
L
4
T
L
5
R
E
L
A
P
O
E
G
Z
R
A
O
I
H
7
O
M
8
B
U
S
9
C
3
I
M
N
I
C
D
E
S
N
E
N
T
6
R
O
G
E
R
A
M
R
C
S
E
U
E
N
M
A
T
C
G
All-In-One Answers Version B
I
B
Q
U
A
R
N
L1
C
D
E
10
H
Y
P
O
T
E
N
L
T
E
E
S
R
U
Geometry
S
E
55
Geometry: All-In-One Answers Version B
(continued)
Guided Problem Solving 7-4
Chapter 7
1. #ABC, #ACD, #BCD 2. 1 3. 1; 1 4. 1 5. 1
6. 2; 1 7. 2 8. "2 9. "2 10. yes 11. no (This would
1. 1 : 278 2. 18 ft by 10 ft 3. 18 ft by 16 ft 4. 10 ft by 3 ft
5. true 6. false 7. true 8. false 9. true 10. true
11. 12 12. 12 13. 33 14. 48 15. 6 16. 52 17. 14 : 5
7
18. 12 : 7 19. 38 20. 13
Guided Problem Solving 7-1
42
1
or
1. ratios 2. 42,000,000
3. the denominator
1,000,000
x 5
1
4. 29,000
5. Cross-Product Property 6. 0.029
1,000,000
7. 0.348 8. yes 9. 21.912
require the Pythagorean Theorem.)
Practice 7-5
1. BE 2. BC 3. JD 4. BE 5. 16
3 6. 4 7. x =
y = 4 8. 15
9.
x
=
6;
y
=
6
10.
2 11. 10
4
25
9;
Guided Problem Solving 7-5
1. parallel 2. CE; BD 3. 6; 15 4. 90 5. yes 6. yes 7. The
sides would not be parallel. 8. They are similar triangles.
1. ABC XYZ, with similarity ratio 2 : 1
2. Not similar; corresponding sides are not proportional.
3. Not similar; corresponding angles are not congruent.
4. ABC KMN, with similarity ratio 4 : 7
5. I 6. O 7. NO 8. LO 9. 3.96 ft 10. 3.75 cm
11. 32 12. 53 13. 7 21 14. 4 12 15. 37
Guided Problem Solving 7-2
1. equal 2. 6.14
3. 2.61; 6.14 4. 19.3662; 19.2182 5. no
2.61
6. no 7. 2.3706; 2.3525 8. Since the quotients are not equal,
the ratios are not equal, and the bills are not similar
rectangles. 9. 4.045
Practice 7-3
1. AXB RXQ because vertical angles are . A R (Given). Therefore AXB RXQ by the AA 3
XM
PX
Postulate. 2. Because MP
LW = WA = AL = 4 , MPX LWA by the SSS Theorem. 3. QMP AMB
QM
2
because vertical s are . Then, because AM = PM
BM = 1 ,
QMP AMB by the SAS Theorem. 4. Because
BX
AX = BX and CX = RX, AX
CX = RX . AXB CXR
because vertical angles are . Therefore AXB CXR by
20
48
the SAS Theorem. 5. 15
2 6. 7 7. 3 8. 36 9. 33 ft
All rights reserved.
7A: Graphic Organizer
Practice 7-2
1. Similarity 2. Answers may vary. Sample: ratios and
proportions; similar polygons; proving triangles similar; and
similarity in right triangles 3. Check students’ work.
7B: Reading Comprehension
1. DHB ACB 2. AA similarity postulate
The triangles have two similar angles. 3. 1 : 2 4. 1 : 2
5 HB 6. 250 ft 7. a
5. DB
AB
CB
7C: Reading/Writing Math Symbols
1. no 2. no 3. yes 4. no 5. yes 6. no 7. yes, AAS
8. yes, Hypotenuse-Leg Theorem 9. yes, SAS or ASA or
AAS 10. not possible
7D: Visual Vocabulary Practice
1. Angle-Angle Similarity Postulate 2. golden ratio
3. Side-Side-Side Similarity Theorem 4. geometric mean
5. scale 6. Cross-Product Property 7. golden rectangle
8. simplest radical form 9. Side-Angle-Side Similarity
Theorem
7E: Vocabulary Check
1. no; N/A 2. yes; WT, RS 3. It is a trapezoid. 4. They
are congruent. 5. They are parallel. 6. They are congruent.
7. #RSZ and #TWZ 8. AA~ or Angle-Angle Similarity
Postulate 9. No; there is only one pair of congruent angles.
10. yes; parallelogram, rhombus, rectangle, and square
Similarity ratio: The ratio of lengths of corresponding sides
of similar polygons.
Cross-Product Property: The product of the extremes of a
proportion is equal to the product of the means.
Ratio: A comparison of two quantities by division.
Golden rectangle: A rectangle that can be divided into a
square and a rectangle that is similar to the original rectangle.
Scale: The ratio of any length in a scale drawing to the
corresponding actual length.
Practice 7-4
7F: Vocabulary Review
Guided Problem Solving 7-3
1. 16 2. 8 3. 10"2 4. 6"5 5. h 6. y 7. a 8. c
9. 92 10. x = 6; y = 6"3 11. x = 4 "5 ; y = "55
12. 2"15 in.
56
Geometry
1. L 2. E 3. I 4. A 5. J 6. K 7. C 8. D 9. G 10. O
11. N 12. M 13. H 14. B 15. F
All-In-One Answers Version B
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 7-1
Geometry: All-In-One Answers Version B (continued)
Practice 8-5
Chapter 8
Practice 8-1
1. " 51 2. 2 " 65 3. 2 " 21 4. 18 " 2 5. 46 in.
6. 78 ft 7. 279 cm 8. 19 m 9. acute 10. obtuse 11. right
1a. angle of depression from the plane to the person
1b. angle of elevation from the person to the plane
1c. angle of depression from the person to the sailboat
1d. angle of elevation from the sailboat to the person
2. 116.6 ft 3. 84.8 ft 4. 46.7 ft 5. 31.2 yd
6a.
Guided Problem Solving 8-1
1. the sum of the lengths of the sides 2. Pythagorean Theorem
3. 7 cm 4. 4 cm 3 cm 5. c2 = 42 + 32 6. 5 7. 12 cm
8. perimeter of rectangle = 14 cm; yes 9. Answers will vary;
35
30 ft
All rights reserved.
example: Draw a 4 cm 3 cm grid, copy the given figure,
measure the lengths with a ruler, add them together. 10. 20 cm
6b. 26 ft
Practice 8-2
1. x = 2; y = " 3 2. 8 " 2 3. 14 " 2 4. 2 5. x = 15;
y = 15 " 3 6. 3 " 2 7. 42 cm 8. 10.4 ft, 12 ft 9. a = 4;
b = 3 10. p = 4 " 3; q = 4 " 3; r = 8; s = 4 " 6
Guided Problem Solving 8-2
1. 30°-60°-90° triangle 2. l 3. h 4. Í 3 5. 24 or 8Í 3
!3
6. 2 7. 48 or 16Í 3 8. 28 ft 9. 0.28 min 10. yes
!3
11. 34 ft
1. &e = 1, &d = &4 2. congruent 3. m&e = m&d
4. 7x - 5 = 4(x + 7) 5. 11 6. 72 7. 72 8. yes
9. 44, 44
Practice 8-6
1. 46.0, 46.0 2. 89.2, -80.3 3. 38.6 mi/h; 31.2º north of
east 4. 134.5 m; 42.0º south of west 5. 55º north of east
6. 33º west of north 7a. 1, 5
1. tan E = 34 ; tan F = 34 2. tan E = 52; tan F = 52 3. 12.4
4. 31.0° 5. 7.1 6. 6.4 7. 26.6 8. 71.6 9. 39 10. 72
11. 39 12. 54
A
B
6
4
2 4 6x
642
2
4
6
Guided Problem Solving 8-3
1.
y
7b.
Practice 8-3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Guided Problem Solving 8-5
8a. 1, -1 8b.
20 cm
y
6
4
80 cm
642
2
4
6
2. 180 3. m&A = 2m&X 4. 90 5. base: 40 cm, height:
10 cm 6. 4 7. 4 8. 76 9. 152 10. 28 11. yes 12. 46
Practice 8-4
1. sin P = 2"710 ; cos P =
3
7
= 65
3
5
8
= 17
2. sin P = 45; cos P =
15
3. sin P = "611 ; cos P
4. sin P = 17
; cos P
5. 64 6. 11.0 7. 7.0 8. 7.8 9. 53 10. 6.6 11. 11.0
12. 11.5
4 6x
9. Sample:
N
W
48
E
Guided Problem Solving 8-4
1. The sides are parallel. 2. sine 3. sin 30° 5 w6 4. 3.0
5. yes 6. cosine 7. cos x° 5 34 8. 41 9. Answers may
vary. Sample: cos 60° 36; sin 49° 34 10. 5.2, 2.6
L1
All-In-One Answers Version B
S
Geometry
57
Geometry: All-In-One Answers Version B
(continued)
Guided Problem Solving 8-6
8D: Visual Vocabulary Practice
x ; y
1. Check students’ work. 2. 100
100
1. 30°-60°-90° triangle 2. inverse of tangent 3. congruent
sides 4. tangent 5. Pythagorean Theorem 6. hypotenuse
7. 45°-45°-90° triangle 8. Pythagorean triple 9. obtuse
3. 100 cos 30°; 100 sin 30° 4. 86.6; 50 5. 86.6, 50
6. 86.6, -50 7. 173.2, 0 8. 173; due east 9. yes
10. 100; due east
triangle
8A: Graphic Organizer
8E: Vocabulary Check
1. Right Triangles and Trigonometry 2. Answers may vary.
Obtuse triangle: A triangle with one angle whose measure is
Sample: the Pythagorean Theorem; special right triangles; the
tangent ratio; sine and cosine ratios; angles of elevation and
depression; vectors 3. Check students’ work.
Isosceles triangle: A triangle that has at least two congruent
between 90 and 180.
sides.
Hypotenuse: The side opposite the right angle in a right
Right triangle: A triangle that contains one right angle.
Pythagorean triple: A set of three nonzero whole numbers
1. A 2. J 3. B 4. J 5. B 6. B 7. b
a, b, and c that satisfy the equation a2 + b2 = c2.
8C: Reading/Writing Math Symbols
1. F 2. G 3. D 4. A 5. C 6. B 7. H 8. E
5 10. #ABC , #XYZ 11. m&A < 52°
9. sin21 A 5 12
7
12. tan Z 5 24
58
J
A
T
N
A
T
L
U
S
E
R
Y
G
D
H
P
G
H
S
N
L
P
U
R
F
S
I
M
E
E
M
Q
O
I
I
Y
H
G
E
A
K
U
R
R
E
L
O
F
L
S
B
T
N
T
N
L
O
N
Q
G
M
A
M
M
D
M
T
P
B
K
V
I
X
E
X
X
E
C
S
Q
E
A
L
V
Y
D
W
M
E
T
D
O
T
S
B
U
N
T
P
X
L
T
Z
I
T
O
R
N
I
F
A
D
R
Y
R
L
P
A
H
S
R
P
S
Z
S
E
O
E
E
E
J
I
I
M
T
A
H
Y
M
C
O
N
E
D
C
L
M
Z
C
R
D
E
G
H
E
D
P
I
L
U
T
I
N
E
B
M
J
C
V
O
R
F
P
S
U
T
A
B
A
F
N
V
E
C
T
O
R
Y
O
O
R
I
N
I
V
G
N
T
M
A
D
D
K
E
B
C
O
N
G
R
U
E
N
T
G
H
N
T
B
R
A
A
N
G
L
E
O
F
D
E
P
R
E
S
S
I
O
N
R
A
E
U
A
R
A
X
L
L
Y
T
N
E
J
O
B
M
C
O
O
V
Z
L
P
R
O
P
O
R
T
I
O
N
Geometry
All-In-One Answers Version B
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
8F: Vocabulary Review Puzzle
All rights reserved.
triangle.
8B: Reading Comprehension
L1
Geometry: All-In-One Answers Version B (continued)
8.
Chapter 9
4
A
2
Practice 9-1
1. No; the triangles are not the same size. 2. Yes; the ovals
are the same shape and size. 3a. C and F 3b. CD
and CD, DE and DE, EF and EF, CF and CF
4. (x, y) S (x - 2, y - 4) 5. (x, y) S (x + 4, y - 2)
6. (x, y) S (x + 2, y + 2) 7. W(-2, 2), X(-1, 4),
Y(3, 3), Z(2, 1) 8. J(-5, 0), K(-3, 4), L(-3, -2)
9. (x, y) S (x + 13, y - 13) 10. (x, y) S (x,y)
(x + 3, y + 3)
11a. P(-3, -1) 11b. P(0, 8),
N(-5, 2), Q(2, 3)
6 4 2
2
A
9.
All rights reserved.
1. the four vertices of a preimage and one of the vertices of
the image 2. Graph the image and preimage. 3. C(4, 2) and
C(0, 0) 4. x = 4, y = 2, x + a = 0, y + b = 0 5. a = -4;
b = -2; (x, y) → (x - 4, y - 2) 6. A(-1, 4), B(1, 3),
6 x
2 4
6 x
4 B T
T
A
4
2
y
4
6 4 2
Guided Problem Solving 9-1
T
4
B B
A
10. (-6, 4)
D(-2, 1)
7.
y
Guided Problem Solving 9-2
A
B
A
B
D
1. a point at the origin, and two reflection lines
2. A reflection is an isometry in which a figure and its image
have opposite orientations. 3. the image after two successive
reflections
4.
C
D
2 O C 2
x
y
y=3
4
x
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y
T
B
O(0,0)
4
8. yes 9. (x, y) → (x + 1, y - 3), A’(4, 3), B’(6, 2), D’(3, 0)
Practice 9-2
1. (-3, -2) 2. (-2, -3) 3. (-1, -4)
4. (4, -2) 5. (4, -1) 6. (3, -4)
y
7a.
5.
y=3
y
O'(0,6)
4
Z
x
2
O(0,0)
Y
6 4 2
2 4
W
4
6x
X
6.
y=3
7b.
4
y
O'(0,6)
y
4
X
6 4
Y
2
x
W
Z
4
O(0,0)
6x
4
L1
All-In-One Answers Version B
4
O"(0,6)
7. (0, - 6) 8. Yes. The x-coordinate remains 0 throughout.
9. O(0, 0) and O(0, 6)
Geometry
59
Geometry: All-In-One Answers Version B
Practice 9-3
1. I 2. I 3. I 4. GH 5. G 6. ST
7.
(continued)
5. slope of OB 5 252 ; slope of OC 5 52 ; slope of OD 5 252 ;
the slopes of perpendicular line segments are negative
reciprocals. 6. square 7. yes 8. B(2, 7), C(7, -2),
D(-2, -7)
P
Practice 9-4
1. The helmet has reflectional symmetry. 2. The teapot has
reflectional symmetry. 3. The hat has both rotational and
O
8.
reflectional symmetry.
4.
5.
P P
Q O
Q
N
All rights reserved.
N
6.
line symmetry and 72° rotational
symmetry
9.
Q'
7.
P'
S
R'
P
line symmetry and 90° rotational
symmetry
Q
8.
line symmetry and 45° rotational
symmetry
Guided Problem Solving 9-3
1. The coordinates of point A, and three rotation
transformations. It is assumed that the rotations are
counterclockwise. 2. Parallelogram, rhombus, square
3.
slope of A 5 2
y
5
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
R
9.
line symmetry
10.
A
x
line symmetry and 45° rotational
symmetry
O
11.
4.
y
180° rotational symmetry
B
A
x
O
12.
13.
COOK
C
D
60
Geometry
All-In-One Answers Version B
L1
Geometry: All-In-One Answers Version B (continued)
14.
that the screen image has 16 16 = 256 as large an area as
the figure on the transparency. 9. 2520 ft2
H
O
A
X
Practice 9-6
1. I. D II. C III. B IV. A
2.
B B B
Guided Problem Solving 9-4
m
1. the coordinates of one vertex of a figure that is symmetric
about the y-axis 2. Line symmetry is the type of symmetry
for which there is a reflection that maps a figure onto itself.
3.
C C
Practice 9-5
1.
5
3
2.
1
2
m
C
All rights reserved.
3. the coordinates of another vertex of the figure
4. images (and preimages) 5. reflection across the y-axis
6. (-3, 4) 7. Yes 8. (-6, 7)
3. 2 4. yes 5. no 6. no
m
R
7.
R
8.
A
J
m
J
T
J
A
4.
R
m
R
y
5.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4
A
A
1. A description of a square projected onto a screen by an
overhead projector, including the square’s area and the scale
factor in relation to the square on the transparency. 2. The
scale factor of a dilation is the number that describes the size
change from an original figure to its image. 3. the area of
the square on the transparency 4. smaller; The scale factor
16 > 1, so the dilation is an enlargement. 5. 1 ; 1 6. 1 ;
16 16
256
1
1
Being
as high and
as wide, the square on the
16
16
tranparency has 1 3 1 5 1 times the area. 7. 3 ft2
16
16
256
256
8. The shape does not matter. Regardless of the shape, the
figure is being enlarged by a factor of 16 in two directions, so
All-In-One Answers Version B
O
2
2
6 x
4
T
T
Guided Problem Solving 9-5
L1
2
T
9. P(-12, -12), Q(-6, 0), R(0, -6) 10. P(-2, 1),
Q(-1, 0), R(0, 1)
E
S
O
B
y
6.
B
4
T
2
6 4 2 O
2
E
4
2
4 x
S
7. reflection 8. rotation 9. glide reflection
10. translation
Guided Problem Solving 9-6
1. assorted triangles and a set of coordinate axes
2. a transformation 3. the transformation that maps one
Geometry
61
Geometry: All-In-One Answers Version B
triangle onto another 4. They are congruent, which confirms
that they could be related by one of the isometries listed.
Corresponding vertices: E and P, D and Q, C and M.
5. counterclockwise; clockwise. The triangles have opposite
orientations, so the isometry must be a reflection or glide
reflection. 6. No; there is no reflection line that will work.
7. a glide reflection consisting of a translation (x, y) →
(x + 11, y) followed by a reflection across line y = 0
(the x-axis) 8. Yes, both are horizontal. 9. 180º rotation
with center Q 22 1 , 0 R
2
(continued)
4.
5. Possible answer:
6. Check students’ work.
7. Answers may vary. Sample:
Practice 9-7
All rights reserved.
1.
translational symmetry
2.
3.
line symmetry,
rotational symmetry,
translational symmetry,
glide reflectional symmetry
9A: Graphic Organizer
1. Transformations 2. Answers may vary. Sample: reflections;
translations; compositions of reflections; and symmetry
3. Check students’ work.
9B: Reading Comprehension
1. D 2. E, R, 5, and 20 3. The center of the county is
the center of the square mile bounded by roads 12, 13, K
and L. 4. C 5. 10 miles 6. a
4.
9C: Reading/Writing Math Symbols
rotational symmetry,
translational symmetry
5.–6. Samples:
5.
6.
1. reflection; opposite 2. rotation; same 3. glide reflection;
opposite 4. translation; same 5. rotation; same
6. translation; same
9D: Visual Vocabulary Practice/High-Use
Academic Words
1. theorem 2. symbols 3. reduce 4. application 5. simplify
6. explain 7. characteristic 8. table 9. investigate
7. yes 8. no 9. yes
Guided Problem Solving 9-7
1. two different-sized equilateral triangles and a trapezoid
2. A tessellation is a repeating pattern of figures that
completely covers a plane, without gaps or overlaps.
3. Theorem 9-7 says that every quadrilateral tessellates. If the
three polygons can be joined to form a quadrilateral (using
each polygon at least once), that quadrilateral can be the basis
of a tessellation.
62
Geometry
9E: Vocabulary Check
Transformation: A change in the position, size, or shape of a
geometric figure.
Preimage: The given figure before a transformation is
applied.
Image: The resulting figure after a transformation is applied.
Isometry: A transformation in which an original figure and
its image are congruent.
Composition: A transformation in which a second
transformation is performed on the image of the first
transformation.
All-In-One Answers Version B
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
line symmetry,
rotational symmetry,
translational symmetry,
glide reflectional symmetry
Geometry: All-In-One Answers Version B (continued)
9F: Vocabulary Review Puzzle
1
2
P
O
G
L
Y
H
E
T
R
Y
T
R
Y
D
R
O
N
I
D
3
S
Y
M
M
E
R
4
O
B
5
T
U
S
E
F
S
6
I
S
All rights reserved.
E
D
M
E
C
L
T
R
L
I
O
A
O
13
T
M
L
O
E
10
15
E
T
A
I
T
O
I
A
N
R
A
14
L
N
N
E
8
11
N
S
L
O
C
U
R
C
9
A
T
L
R
T
A
T
I
N
T
S
U
F
16
O
T
I
O
R
12
D
O
I
L
A
T
I
O
N
N
D
N
E
R
17
M
18
S
C
A
V
L
E
T
N
E
R
I
T
O
E
N
X
Chapter 10
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
H
I
O
7
Practice 10-1
1. 8 2. 60 3. 1.92 4. 0.8125 5. 9 6. 1500 7. 8.75
8. 3.44 9. 30 10. 16 11. 12 12. 48 13. 40
Guided Problem Solving 10-1
1. 21 bh 2. Check students’ work; B(7, 7) 3. a right angle
Guided Problem Solving 10-3
1. 10 2. The angles are all congruent and the sides are all
congruent. 3. 360 4. 36 5. bisects 6. 2 7. 18 8. 180
9. 72 10. 180 11. 30; 15; 75
Practice 10-4
1. 4 : 5; 16 : 25 2. 5 : 3; 25 : 9 3. 1 : 2 4. 8 : 3 5.
2
2
6. 2700
7 cm 7. 1 : 2 8. 2 : 5 9. 108 ft
576
5
in.2
4. 7 5. OA 6. 7 7. 24.5 8. a square 9. 49 10. 12
11. 24.5 12. 21 b2
Guided Problem Solving 10-4
Practice 10-2
a < 20mm, b < 30mm and c < 40mm. 4. 40 mm 5 1 mm
5. < 90 mm 6. < 16 mm 7. < 320 mm2 8. 450 9. 8000
10. yes 11. 16,000
å
1. 48 cm2 2. 784 in.2 3. 90 m2 4. 374 ft2 5. 160 cm2
6. 176.25 in.2 7. 42.5 square units 8. 226.2 in.2
1. scale 2. Check students’ work. 3. Answers will vary.
Let a be the side opposite the 30° angle, b be the side opposite
the 50° angle, and c be the side corresponding to 200 yd. Then
200 yd
5 yd
Guided Problem Solving 10-2
Practice 10-5
1. 12 h(b1 1 b2) 2. 3; 4; 2 3. 9 4. no 5. yes 6. 3
1. 174.8 cm2 2. 578 ft2 3. 1250.5 mm2 4. 1131.4 in.2
5. 37.6 m2 6. 30.1 mi2 7. 55.4 km2 8. 357.6 in.2
9. 9.7 mm2 10. 384.0 in.2 11. 83.1 m2 12. 65.0 ft2
13. 119.3 ft2 14. 8 m2
Practice 10-3
1.
3.
4.
6.
x = 7; y = 3.5 " 3 2. x = 30; p = 4 " 3; q = 8
m1 = 45; m2 = 45; m3 = 90; m4 = 90
m5 = 60; m6 = 30; m7 = 120 5. 120 in.2
137 in.2
L1
All-In-One Answers Version B
Geometry
63
Geometry: All-In-One Answers Version B
(continued)
Guided Problem Solving 10-5
10D: Visual Vocabulary Practice
1. subtract 2. 50 3. 12 ap 4. 30 5. 4 6. tan4308 7. 166
8. 116 9. yes 10. 151
1. area of a triangle 2. radius 3. altitude of a parallelogram
4. area of a trapezoid 5. area of a kite or rhombus
6. similarity ratio 7. apothem 8. perimeter (of an octagon)
9. height of a triangle
11
0
1. 16p 2. 7.8p 3. Samples: DFA, ABF 4. Samples: FE ,
0
0 0
FD 5. Samples: FE , ED 6. 32 7. 54 8. 71 9. 50
10. 140 11. 220 12. 130 13. 6p in. 14. 16p cm
10E: Vocabulary Check
Guided Problem Solving 10-6
Apothem: The distance from the center to the side of a
Semicircle: Half a circle.
Adjacent arcs: Arcs on the same circle with exactly one
point in common.
1. 11; 2 2. 11; 3; 33 3. circumference; 2 4. 31 5. 26 6. 41
7. 105 8. 102 9. 3027.15
regular polygon.
Circumference: The distance around a circle.
Concentric circles: Circles that lie on the same plane and
49
1. 49p 2. 21.2 3. 49
6 p 4. 6 p - 21.2 5. 4p 6.
7. 98 p 8. 23 p 9. 6p 10. 25
3 p 11. 3 12. 2
18
5p
Guided Problem Solving 10-7
? pr 2 2. 45 3. 25 4. 30 5. 27 6. a piece from
1. mAB
360
the outer ring 7. yes 8. a piece from the top tier
10F: Vocabulary Review Puzzle
1. adjacent arcs 2. semicircle 3. center 4. arc length
5. base 6. circle 7. central angle 8. congruent arcs
9. diameter 10. major arc 11. concentric circles
12. minor arc 13. Pythagorean triple 14. radius
15. apothem 16. sector 17. height 18. circumference
Highness, there is no royal road to geometry.
Chapter 11
Practice 10-8
3
10
2
5
1. 8.7% 2. 10.9% 3. 15.3% 4.
5.
6. 1 7. 4.0%
8. 37.5% 9. 20% 10. 50% 11. 40% 12. 33 13 %
Practice 11-1
Guided Problem Solving 10-8
1. 8 2. 5 3. 12 4. C 5. D 6. B 7. A 8. triangle
9. rectangle 10. rectangle
1. outcomes 2. event 3. AF 4. FG 5. EG 6. AE;
Check students’ work. 7. 2 or about 67% 8. 2 or about
3
3
67%; yes 9. 3 or 60%
5
10A: Graphic Organizer
1. Area 2. Answers may vary. Sample: areas of
parallelograms and triangles; areas of trapezoids, rhombuses,
and kites; areas of regular polygons; perimeters and areas of
similar figures; trigonometry and area; circles and arcs; areas
of circles and sectors; geometric probability 3. Check
students’ work.
10B: Reading Comprehension
1. E 2. J 3. B 4. G 5. A 6. I 7. D 8. C 9. H 10. F
0
å å
11. a2 + b2 = c2 12. mAC 5 1058 13. MN RS
0 0
14. "225 5 15 15. #JKL 16. XY > PQ
17. ~ABCD 18. 42 = 16 19. b
10C: Reading/Writing Math Symbols
1. 2860 ft2 2. 3331 ft2 3. 86 bundles 4. 2823 squares 5. $9.00
6. $27.00 7. 45 minutes 8. 21.5 hours 9. $645 10. $1419
64
Geometry
All rights reserved.
have the same center.
Practice 10-7
Guided Problem Solving 11-1
1. a two-dimensional network 2. Verify Euler’s Formula for
the network shown. 3. 4 4. 6; 9 5. 4 + 6 = 9 + 1; This is
true. 6. F becomes 3, V stays at 6, E becomes 8. Euler’s
Formula becomes 3 + 6 = 8 + 1, which is still true.
7. 4 + 4 = 6 + 2; This is true.
Practice 11-2
1. 62.8 cm2 2. 226.2 cm2 3. 44.0 ft2 4. 84.8 cm2
5a. 320 m2 5b. 440 m2 6a. 576 in.2 6b. 684 in.2
7a. 48 cm2 7b. 60 cm2 8a. 1500 m2 8b. 1800 m2
9. 94.5p cm2 10. 290p ft2
Guided Problem Solving 11-2
1. a picture of a box, with dimensions 2. the amount of
cardboard the box is made of, in square inches 3. A front
and a back, each 4 in. 7 1 in.; a top and a bottom, each
2
4 in. 1 in.; and one narrow side, 1 in. 7 1 in. 4. Front and
2
2
back: each 30 in. ; top and bottom: each 4 in.2; one narrow
side: 7 1 in.2 5. 75 1 in.2 6. 75 1 in.2 is about half a square
2
2
2
foot (1 ft2 = 144 in.2), which is reasonable. The answer is an
All-In-One Answers Version B
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Practice 10-6
Geometry: All-In-One Answers Version B (continued)
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
approximation because the solution did not consider the
overlap of cardboard at the glued joints, nor does it factor in
the trapezoid-shaped cutout in the front and back sides.
7. 3 21 in.2
Guided Problem Solving 11-6
1. the diameter of a hailstone, along with its density compared
Practice 11-3
to normal ice 2. the hailstone’s weight 3. V 5 4 pr3
3
4. 2.8 in. 5. 91.95 in.3 6. 1.7 lb 7. The density of normal
ice. This was given for comparison purposes only. 8. 0.01 oz
1. 620 cm2 2. 188 ft2 3. 36p cm2 4. 800p m2 5. 109.6 m2
6. 387.7 ft2 7. 118.4 cm2 8. 668.2 ft2
Practice 11-7
Guided Problem Solving 11-3
1. 7 : 9 2. 5 : 8 3. yes; 7 : 4 4. not similar 5. 125 cm3
6. 256 in.3 7. 25 ft2 8. 600 cm2
1. a solid consisting of a cylinder and a cone 2. the surface
area of the solid 3. The top of the solid is a circular base of
the cylinder, A1 = pr2. The side of the solid is the lateral area
of the cylinder, A2 = 2prh. The bottom of the solid is the
lateral area of the cone, A3 = pr. 4. r = 5 ft; h = 6 ft;
Guided Problem Solving 11-7
or a little less than half. This seems reasonable. 8. 628 ft2
1. two different sizes for an image on a balloon, and the
volume of air in the balloon for one of those sizes 2. the
volume of air for the other size 3. The two sizes of the clown
face have a ratio of 4 : 8, or 1 : 2. 4. 1 : 8 5. 108 in.3 8 =
864 in.3 6. Yes, the two clown face heights are lengths,
measured in inches (not in.2 or in.3). 7. 351 cm3
Practice 11-4
11A: Graphic Organizer
1. 1131.0 m3 2. 7.1 in.3 3. 1781.3 in.3 4. 785.4 cm3
5. 120 in.3 6. 35 ft3 7. 1728 ft3 8. 265 in.3 9. 76 ft3
10. 1152 m3
1. Surface Area and Volume 2. Answers may vary. Sample:
space figures and nets; space figures and drawing; surface
areas; and volumes. 3. Check students’ work.
Guided Problem Solving 11-4
11B: Reading Comprehension
1. dimensions for a layer of topsoil, and two options for buying
1. Area = 12 base times height; area of a triangle
2. slope = the difference of the y-coordinates divided by the
/ 5 "52 1 122 5 13 ft 5. Top: A1 = 78.5 ft2; side:
A2 = 188.5 ft2; bottom: 204.2 ft2 6. 471 ft2 7. About 0.43,
the soil 2. which purchasing option is cheaper 3. 1 ft
3
4. 1400 ft3 5. 467 bags; $1167.50 6. 1400 ft3 < 52 yd3
7. $1164 (or slightly less if the topsoil volume is not rounded
up to the next whole number) 8. buying in bulk 9. The two
costs are quite close. It makes sense that buying in bulk would
be slightly cheaper for a good-sized backyard. 10. Buying
top soil by the bag would be less expensive because its would
cost $335 and in bulk would cost at least $345.93.
Practice 11-5
1. 34,992 cm3 2. 400 in.3 3. 10,240 in.3 4. 4800 yd3
5. 314.2 in.3 6. 4955.3 m3 7. 1415.8 in.3 8. 1005.3 m3
9. 20 10. 2
Guided Problem Solving 11-5
1. a description of a plumb bob, with dimensions 2. the plumb
bob’s volume 3. equilateral; 2 cm 4. a = "3 cm; B = 6"3
cm2 5. Prism: V1 = 36"3 cm3; pyramid: V2 = 6"3 cm3;
plum bob: V < 73 cm3 6. 1 ; this seems reasonable 7. 42 cm3
7
Practice 11-6
1. 2463.0 in.2 2. 6,157,521.6 m2 3. 12.6 cm2 4. 153.9 m2
5. 1436.8 mi3 6. 268,082.6 cm3 7. 7238.2 m3 8. 14.1 cm3
9. 5547 cm3 10. 42 cm3
L1
All-In-One Answers Version B
difference of the x-coordinates; slope of the line between
2 points 3. a squared plus b squared equals c squared, the
Pythagorean Theorem; the lengths of the sides of a right
triangle 4. area = pi times radius squared; area of a circle
5. volume = pi times radius squared times height; volume of a
cylinder 6. the sum of the x-coordinates of two points
divided by 2, and the sum of the y-coordinates of two points
divided by 2; midpoint of a line segment with endpoints
(x1, y1) and (x2, y2) 7. circumference = two times pi times
the radius; circumference of a circle 8. volume = one-third
pi times radius squared times height; volume of a cone
9. distance equals the square root of the quantity of the sum
of the square of the difference of the x-coordinates, squared,
plus the difference of the y-coordinates, squared; the distance
between two points (x1, y1) and (x2, y2) 10. area = one-half
the height times the sum of base one plus base two; area of a
trapezoid 11. a
11C: Reading/Writing Math Symbols
1. 527.8 in2 2. 136 ft2
11D: Visual Vocabulary Practice
1. prism 2. cross section 3. sphere 4. cylinder
5. cone 6. pyramid 7. polyhedron 8. altitude 9. base
Geometry
65
Geometry: All-In-One Answers Version B
(continued)
11E: Vocabulary Check
Practice 12-3
Face: A surface of a polyhedron.
Edge: An intersection of two faces of a polyhedron.
Altitude: For a prism or cylinder, a perpendicular segment
1. A and D; B and C 2. ADB and CDB
3. 55 4. x = 45; y = 50; z = 85 5. x = 90; y = 70
6. 180 7. 70 8. x = 120; y = 100; z = 140
9a. mA = 90 9b. mB = 80 9c. mC = 90
9d. mD = 100
11F: Vocabulary Review
1. polyhedron 2. edge 3. vertex 4. cross section 5. prism
6. altitude 7. right prism 8. slant height 9. cone
10. volume 11. sphere 12. hemispheres 13. similar figures
14. hypotenuse
Chapter 12
Guided Problem Solving 12-3
1. C 2. 180 3. a diameter 4. Check students’ work.
5. Check students’ work. 6. Check students’ work.
7. Check students’ work. 8. 90 9. Check students’ work.
Practice 12-4
1. 87 2. 35 3. 120 4. 72 5. x = 58; y = 59; z = 63
6. x = 30; y = 30; z = 120 7. x = 16; y = 52
8. x = 138; y = 111; z = 111 9. x = 30; y = 60
10. 10 11. 4 12. 3.2 13. 6
All rights reserved.
that joins the planes of the bases.
Surface area: For a prism, cylinder, pyramid, or cone, the
sum of the lateral area and the areas of the bases.
Volume: A measure of the space a figure occupies.
Practice 12-1
1. 32 2. 72 3. 15 4. 6 5. "634 6. 4 "3
7. circumscribed 8. inscribed 9. circumscribed
Guided Problem Solving 12-1
1. The area of one is twice the area of the other.
2. inscribed; circumscribed
3.
4.
Guided Problem Solving 12-4
1. tangent 2. 360. 3. 12 4. 85 5. 2 6. 180 7. 95
8. 104; 86; 75 9. 360 10. yes; The sum of the measures of
the arcs should be 360. 11. Opposite angles are not
supplementary.
Practice 12-5
(0, 5)
r
5. 2r 6. 4r2 7. "2r 8.2r 2 9. 4r 2 = 2(2r 2): One area is
double the other area. 10.yes 11. The same: The area of
(0, 0)
one circle is twice the area of the other circle.
Practice 12-2
0
Guided Problem Solving 12-2
1. perpendicular 2. bisects 3. 4; 3 4. right 5. radius 6.
Pythagorean Theorem 7. 5 8. yes 9. 8; 4
Geometry
x
0
congruent chords, AB = BC = CA. Then, because an
equilateral triangle is equiangular, mABC = mBCA =
mCAB.
66
(5, 0)
5
1. r = 13; mAB 134.8 2. r = 3 "5; mAB 53.1
3. 3 4. 4.5 5.
3 6.0
8.5 7. Q T; PR SU
0
8. A J; BC KL 9. Because congruent arcs have
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. C(0, 0), r = 6 2. C(-1, -6), r = 4 3. x2 + y2 = 49
4. (x - 5)2 + (y - 3)2 = 4 5. (x + 5)2 + (y - 4)2 = 14
6. (x + 2)2 + (y + 5)2 = 2 7. x2 + y2 = 4
8. x2 + (y - 3)2 = 16 9. (x - 7)2 + (y + 2)2 = 4
10. x2 + (y + 20)2 = 100
y
11.
y
12.
5
4
3
2
1
3
(3, 5)
1 2 3 4 5
x
All-In-One Answers Version B
L1
Geometry: All-In-One Answers Version B (continued)
Guided Problem Solving 12-5
Guided Problem Solving 12-6
1. (2, 2); 5 2. 43 3. They are perpendicular. 4. - 34 5. 39
4
6. y 5 234 x 1 39
7.
13 8. y 5 3 x 1 39 ; yes
4
4
4
1. a line 2. Check students’ work. 3. They are
perpendicular. 4. - 1 5. 2 6. the midpoint of PQ 7. (3, 2)
2
9. x-intercept: a = 25
; y-intercept: b = 25
3
4
12A: Graphic Organizer
Practice 12-6
1. Circles 2. Answers may vary. Sample: tangent lines; chords
1.
and arcs; inscribed angles; and angle measures and segment
lengths 3. Check students’ work.
12B: Reading Comprehension
T
Answers may vary. Sample answers:
1. AD, CE 2. BF, CD, AD 3. HD 4. AH
0 00
1 1 1
0 0
5. AF , AB , BC 6. CDF, DAB, DAC 7. AC , ED
1
8. /AOC 9. /DCE 10. DEA 11. /DAH
12. /AGB 13. /AHC 14. /DCE 15. AH 16. a
1.5 cm
All rights reserved.
8. -4 9. y = 2x - 4 10. (2, 0); yes 11. y 5 21 x 1 2
2.
A
12C:
Math Symbols
* )Reading/Writing
* )
B
1. BC ' DE 2. EF is tangent to (G. 3. WX > ZX
4. AB ' XY 5. HM = 4
3.
12D: Visual Vocabulary Practice
Y
X
4.
B
A
1. tangent to a circle 2. standard form of an equation of a
circle 3. secant 4. chord 5. circumscribed about
6. intercepted arc 7. inscribed in 8. inscribed angle
9. locus
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
12E: Vocabulary Check
Inscribed in: A circle inside a polygon with the sides of the
E
C
polygon tangent to the circle.
Chord: A segment whose endpoints are on a circle.
Point of tangency: The single point of intersection of a
5. a third line parallel to the two given lines and midway
between them 6. a sphere whose center is the given
point and whose radius is the given distance 7. the points
within, but not on, a circle of radius 1 in. centered
at the given point
tangent line and a circle.
Tangent to a circle: A line, segment, or ray in the plane of a
circle that intersects the circle in exactly one point.
Circumscribed about: circle outside a polygon with the
vertices of the polygon on the circle.
8.
0.75 in.
R
L1
S
0.75 in.
All-In-One Answers Version B
12F: Vocabulary Review
1. D 2. F 3. G 4. A 5. C 6. B 7. E 8. M 9. I
10. L 11. N 12. H 13. K 14. J
Geometry
67
Geometry: All-In-One Answers Version B
Name_____________________________________ Class____________________________ Date________________
Lesson 1-1
Example.
Patterns and Inductive Reasoning
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
2 Using Inductive Reasoning Make a conjecture about the sum of the cubes
NAEP 2005 Strand: Geometry
Use inductive reasoning to make
conjectures
of the first 25 counting numbers.
Find the first few sums. Notice that each sum is a perfect square and that
the perfect squares form a pattern.
Topic: Mathematical Reasoning
Local Standards: ____________________________________
3
1
3
1 3
1 3
1 3
1 Vocabulary.
conjecture
is a conclusion you reach using inductive reasoning.
A counterexample is an example for which the conjecture is incorrect.
All rights reserved.
A
All rights reserved.
Inductive reasoning is reasoning based on patterns you observe.
half
2
2
96
2
48
the sum of the first 25 counting numbers, or (1 ⫹ 2 ⫹ 3 ⫹ . . . ⫹ 25) 2.
2
the preceding term. The next two
24
and 24 2
12
.
Quick Check.
1. Find the next two terms in each sequence.
a. 1, 2, 4, 7, 11, 16, 22,
29 ,
37
b. Monday, Tuesday, Wednesday,
,...
,
Thursday
Friday
,...
c.
,
,...
Answers may vary. Sample:
Geometry Lesson 1-1
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
terms are 48 192
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Each term is
All rights reserved.
2
1
2
(1 2)
2
(1 2 3)
2
(1 2 3 4)
2
(1 2 3 4 5)
the sum of the cubes of the first 25 counting numbers equals the square of
Use the pattern to find the next two terms in the sequence.
384, 192, 96, 48, . . .
L1
Quick Check.
2. Make a conjecture about the sum of the first 35 odd numbers. Use your
calculator to verify your conjecture.
1
1
1 3
4
22
1 3 5
9
32
1 3 5 7
16
42
1 3 5 7 9 25
52
12
The sum of the first 35 odd numbers is 35 2, or 1225.
L1
Daily Notetaking Guide
Geometry Lesson 1-1
3
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 1-2
Example.
Drawings, Nets, and Other Models
2 Drawing a Net Draw a net for the figure with a square base
NAEP 2005 Strand: Geometry
Topic: Dimension and Shape
and four isosceles triangle faces. Label the net with its dimensions.
10
cm
Lesson Objectives
1 Make isometric and orthographic
drawings
2 Draw nets for three-dimensional
figures
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1 1 2
9 3 2
36 6 2
100 10 2
225 15 The sum of the first two cubes equals the square of the sum of the first
two
counting numbers. The sum of the first three cubes equals the
square of the sum of the first three counting numbers. This pattern
continues for the fourth and fifth rows. So a conjecture might be that
Example.
1 Finding and Using a Pattern Find a pattern for the sequence. 384
2
2
3
2
3
3
2 3
3
3
3
2 3 4
3
3
3
3
2 3 4 5
Think of the sides of the square base as hinges, and “unfold” the
figure at these edges to form a net. The base of each of the four
square
isosceles triangle faces is a side of the
. Write in the
known dimensions.
Local Standards: ____________________________________
Vocabulary.
8 cm
orthographic drawing
All rights reserved.
on isometric dot paper.
An
is the top view, front view, and right-side view of a three-
dimensional figure.
A net is a two-dimensional pattern you can fold to form a three-dimensional figure.
All rights reserved.
An isometric drawing of a three-dimensional object shows a corner view of the figure drawn
Example.
8 cm
10
cm
Quick Check.
1 Orthographic Drawing Make an orthographic drawing of the isometric drawing at right.
Orthographic drawings flatten the depth of a figure. An orthographic
three
drawing shows
views. Because
no edge of the isometric drawing is hidden in the top, front,
and right views, all lines are solid.
2. The drawing shows one possible net for the Graham Crackers box.
ht
t Rig
on
Front
Top
Right
Quick Check.
1. Make an orthographic drawing from this isometric drawing.
Front
Top
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Fr
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
14 cm
20 cm
7 cm
AM
AH
GR KERS
AC
CR
AM
AH
GR KERS
AC
CR
20 cm
7 cm
14 cm
Draw a different net for this box. Show the dimensions in your diagram.
Answers may vary. Example:
14 cm
20 cm
Fr
ht
on
t Rig
Right
7 cm
4
L1
Geometry Lesson 1-2
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 1-2
Geometry
5
1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 1-3
1
A plane is a flat surface that has no thickness.
Points, Lines, and Planes
Lesson Objectives
2
Name_____________________________________ Class____________________________ Date ________________
NAEP 2005 Strand: Geometry
Understand basic terms of geometry
Understand basic postulates of
geometry
B
C
A
Plane ABC
Two points or lines are coplanar if they lie on the same plane.
Topic: Dimension and Shape
A postulate or axiom is an accepted statement of fact.
Local Standards: ____________________________________
Examples.
Vocabulary and Key Concepts.
name three points that are collinear and three points
that are not collinear.
Postulate 1-1
t
Line t is the only line that passes through points A and B .
B
A
Points
All rights reserved.
All rights reserved.
Through any two points there is exactly one line.
*AE )
B
C
E
*
,
Z
, and
W
lie on a line, so they are
m
Z
Y
2 Using Postulate 1-4 Shade the plane that contains X, Y, and Z.
Postulate 1-2
A
Y
collinear.
)
and BD intersect at C .
Y
V
Postulate 1-4
Through any three noncollinear points there is exactly one plane.
A point is a location.
is the set of all points.
Space
A line is a series of points that extends in two opposite directions
without end.
t
b. Name line m in three different ways.
*
)*
)* )
Answers may vary. Sample: ZW , WY , YZ .
2. a. Shade plane VWX.
Z
b. Name a point that is coplanar
with points V, W, and X.
Y
Y
V
X
W
are points that lie on the same line.
Collinear points
6
B
A
no
All rights reserved.
ST
Plane RST and plane STW intersect in ST .
1. Use the figure in Example 1.
a. Are points W, Y, and X collinear?
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
* )
R
T W
S
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
If two planes intersect, then they intersect in exactly one line.
X
W
Quick Check.
Postulate 1-3
W
Z
Points X, Y, and Z are the vertices of one of the four triangular
faces of the pyramid. To shade the plane, shade the interior of
the triangle formed by X , Y , and Z .
If two lines intersect, then they intersect in exactly one point.
D
X
1 Identifying Collinear Points In the figure at right,
Geometry Lesson 1-3
Daily Notetaking Guide
L1
Daily Notetaking Guide
L1
Geometry Lesson 1-3
7
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 1-4
Examples.
1 Naming Segments and Rays Name the segments and rays in the figure.
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
The labeled points in the figure are A, B, and C.
Local Standards: ____________________________________
A segment is a part of a line consisting of two endpoints and all points
between them. A segment is named by its two endpoints. So the
BA (or AB)
segments are
and
.
BC (or CB)
B
A ray is a part of a line consisting of one endpoint and all the points of
the line on one side of that endpoint. A ray is named by its endpoint first, followed
) .
)
by any other point on the ray. So the rays are
and
BA
BC
Vocabulary.
Segment AB
AB
A
B
A
Endpoint
Endpoint
)
is the part of a line consisting of one endpoint and
ray
Ray YX
YX
all the points of the line on one side of the endpoint.
X
Y
Endpoint
All rights reserved.
endpoints and all points between them.
All rights reserved.
A segment is the part of a line consisting of two
)
RQ
and
)
R
RS
Parallel planes
A
J
parallel
←
→
AB and CG are
←
→
to EF.
skew
lines.
F
are planes that do not intersect.
G
D
AB is
←→
G
H
B
Plane ABCD is
parallel
to plane GHIJ.
I
C
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
E
←→
. If the walls of your classroom
)
)
1. Critical Thinking Use the figure in Example 1. CB and BC form a line. Are
they opposite rays? Explain.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
B
walls are parts of parallel planes. If the ceiling and
opposite
Quick Check.
are opposite rays.
Skew lines are noncoplanar; therefore, they are not parallel and do not intersect.
H
do not intersect
are vertical,
S
are coplanar lines that do not intersect.
D
Planes are parallel if they
floor of the classroom are level, they are parts of parallel planes.
Q
endpoint.
A
C
2 Identifying Parallel Planes Identify a pair of parallel planes in your classroom.
Opposite rays are two collinear rays with the same
Parallel lines
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Segments, Rays, Parallel Lines and Planes
Lesson Objectives
1 Identify segments and rays
2 Recognize parallel lines
No; they do not have the same endpoint.
2. Use the diagram to the right.
a. Name three pairs of parallel planes.
PSWT RQVU, PRUT SQVW, PSQR TWVU
S
Q
R
P
W
* )
b. Name a line that is parallel to PQ .
T
V
U
* )
TV
c. Name a line that is parallel to plane QRUV.
* )
Answers may vary. Sample: PS
8
2
Geometry Lesson 1-4
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 1-4
All-In-One Answers Version B
9
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 1-5
Examples.
Measuring Segments
Lesson Objectives
1
Name_____________________________________ Class____________________________ Date ________________
1 Using the Segment Addition Postulate If AB 25,
NAEP 2005 Strand: Measurement
Find the lengths of segments
2x 6
find the value of x. Then find AN and NB.
Local Standards: ____________________________________
NB
AN
(
Vocabulary and Key Concepts.
)(
2x 6
AB
3x that the distance between any two points is the absolute value of the difference of the
corresponding numbers.
Simplify the left side.
24
Subtract 1 from each side.
x 8
All rights reserved.
All rights reserved.
correspondence with the real numbers so
Substitute.
1 25
3x
( )6
NB x 7 ( 8 ) 7 AN 2x 6 2 8
AN and NB 10
15
B
Segment Addition Postulate
) 25
x7
Postulate 1-5: Ruler Postulate
The points of a line can be put into one-to-one
x7
A
N
Use the Segment Addition Postulate (Postulate 1-6) to write an equation.
Topic: Measuring Physical Attributes
Divide each side by 3 .
10
Substitute 8 for x.
15
, which checks because the sum equals 25.
Postulate 1-6: Segment Addition Postulate
If three points A, B, and C are collinear and B
B
Find RM, MT, and RT.
C
8x 36
5x 9
2 Finding Lengths M is the midpoint of RT.
A
is between A and C, then AB ⫹ BC ⫽ AC.
R
M
T
Use the definition of midpoint to write an equation.
RM A coordinate is a point’s distance and direction from zero on a number line.
All rights reserved.
R
AB 5 u
a
Q
A
B
a
b
u
b
a
coordinate of A
coordinate of B
Congruent (艑) segments are segments with the same length.
2 cm
A
B
A
D
C
2 cm
C
B
AB CD
AB
CD
D
midpoint
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the length of AB
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
5x 9
5x segments.
B
|
|
AB MT
(
RM 5x 9 5
(
MT 8x 36 8
15
15
Substitute.
8x
RT RM MT 84
RM and MT are each
36
Add
Subtract
x
)9
) 36 midpoint
Definition of
8x 36
45 3 x
15
to each side.
5x
from each side.
Divide each side by 3 .
84
Substitute
84
168
15
for x.
Segment Addition
, which is half of
168
Postulate
, the length of RT.
Quick Check.
1. EG 100. Find the value of x. Then find EF and FG.
4x – 20
x ⫽ 15, EF ⫽ 40; FG ⫽ 60
A midpoint is a point that divides a segment into two congruent
A
45
E
2x + 30
F
G
C
2. Z is the midpoint of XY, and XY 27. Find XZ.
BC
13.5
10
Geometry Lesson 1-5
Daily Notetaking Guide
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 1-6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Examples.
1 Naming Angles Name the angle at right in four ways.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
m⬔
BOC
A
B
m⬔AOC.
0
180
m⬔AOB m⬔BOC 180
T 1 Q
⬔TBQ
90
x5
x°
x°
x°
angle
90
88
Substitute 42
for m⬔ABC.
m⬔2 46
Subtract
42
Postulate
for m⬔1 and
1
88
2
B
from each side.
C
l2, lDEC
A O C
right
Angle Addition
m⬔2 1. a. Name ⬔CED two other ways.
.
B
are the sides of the angle and the endpoint is the vertex of the angle.
0,x,
42
Quick Check.
If ⬔AOC is a straight angle, then
An angle (⬔) is formed by two rays with the same endpoint. The rays
angle
m⬔ 1 m⬔ 2 m⬔ABC
B
B
acute
A
O
O C
x°
All rights reserved.
14
0
30
15
0
20
16 0
17 0
10
D
obtuse
90
angle
,x,
straight angle
x 5 180
180
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
AOB
13
0
50
y
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
m⬔
12
0
60
10
Postulate 1-8: Angle Addition Postulate
If point B is in the interior of ⬔AOC, then
70
17 0
A
80
x
20
.
90
1
A
Use the Angle Addition Postulate (Postulate 1-8) to solve.
16 0
)
7 0 10 0
6 0 11 0
20
50 0 1
3
10 0 1
10
All rights reserved.
40
C
80
0
15
0
30
14 0
4
)
mlCOD 5 u x 2 y u
3
G
2 Using the Angle Addition Postulate Suppose that m⬔1 42
and m⬔ABC 88. Find m⬔2.
b. If OC is paired with x and OD is paired with y,
then
.
.
Finally, the name can be a point on one side, the vertex, and a point
on the other side of the angle: lAGC or lCGA .
a. OA is paired with 0 and OB is paired with 180 .
)
lG
The name can be the vertex of the angle:
Vocabulary and Key Concepts.
)
C
l3
The name can be the number between the sides of the angle:
Local Standards: ____________________________________
Postulate 1-7: Protractor Postulate
)
)
Let OA and OB be opposite rays in a plane.
) )
OA , OB , and all the rays with endpoint O that can
* )
be drawn on one side of AB can be paired with the
real numbers from 0 to 180 so that
11
Geometry Lesson 1-5
Name_____________________________________ Class____________________________ Date ________________
Measuring Angles
Lesson Objectives
1 Find the measures of angles
2 Identify special angle pairs
Daily Notetaking Guide
L1
b. Critical Thinking Would it be correct to name any of the
angles ⬔E? Explain.
No, 3 angles have E for a vertex, so you
need more information in the name to
distinguish them from one another.
G
2. If m⬔DEG 145, find m⬔GEF.
35
D
E
F
An acute angle has measurement between 0 and 90.
A right angle has a measurement of exactly 90⬚.
An
obtuse angle
has measurement between 90 and 180.
A straight angle has a measurement of exactly 180⬚.
Congruent angles are two angles with the same measure.
12
L1
Geometry Lesson 1-6
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 1-6
Geometry
13
3
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
2 Constructing the Perpendicular Bisector
Lesson 1-7
Basic Constructions
Lesson Objectives
Use a compass and a straightedge to
construct congruent segments and
congruent angles
Use a compass and a straightedge to
bisect segments and angles
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
same
Step 2 With the
compass setting, put the compass
point on point B and draw another long arc. Label the
intersect
points where the two arcs
as X and Y.
is a ruler with no markings on it.
A compass is a geometric tool used to draw circles and parts of circles called arcs.
Perpendicular lines
are two lines that intersect to form right angles.
A perpendicular bisector of a segment is a line, segment, or ray that is
perpendicular to the segment at its midpoint, thereby bisecting the segment
A
D
C
B
All rights reserved.
All rights reserved.
Construction is using a straightedge and a compass to draw a geometric figure.
straightedge
J
* )
T
T
W
KM TW
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
M
Step 2 Open the compass the length of KM.
Step 3 With the same compass setting, put the compass point
on point T. Draw an arc that intersects the ray. Label the
point of intersection W.
* )
Geometry Lesson 1-7
M
B
Y
Quick Check.
1. Use a straightedge to draw XY. Then construct RS so that RS 2XY.
X
R
Y
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
S
2. Draw ST. Construct its perpendicular bisector.
S
14
X
A
XY ⬜ AB at the midpoint of AB, so XY is the
of AB.
perpendicular bisector
L
K
Step 1 Draw a ray with endpoint T.
B
* )
* )
Examples.
to KM.
X
A
K
N
1 Constructing Congruent Segments Construct TW congruent
B
Step 3 Draw XY . The point of intersection of AB and XY is M,
the
of AB.
midpoint
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
coplanar angles.
A
Y
into two congruent segments.
An angle bisector is a ray that divides an angle into two congruent
B
Step 1 Put the compass point on point A and draw a long arc.
greater
Be sure that the opening is
than 12AB.
Vocabulary.
A
A
All rights reserved.
2
Given: AB.
* )
* )
Construct: XY so that XY ⬜ AB at the midpoint M of AB.
NAEP 2005 Strand: Geometry
T
Daily Notetaking Guide
Geometry Lesson 1-7
15
Name_____________________________________ Class____________________________ Date ________________
2 Finding an Endpoint The midpoint of DG is M(1, 5). One endpoint is
Lesson 1-8
The Coordinate Plane
Lesson Objectives
1 Find the distance between two points
in the coordinate plane
2 Find the coordinates of the midpoint
of a segment in the coordinate plane
D(1, 4). Find the coordinates of the other endpoint G.
Use the Midpoint Formula. Let (x 1, y 1 ) be
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
x 1x y 1y
a 1 2 2 , 1 2 2 b be
coordinates of G.
Local Standards: ____________________________________
(1, 5)
. Solve for
(1, 4)
x2
and the midpoint
and
y2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
Name_____________________________________ Class____________________________ Date ________________
, the
Find the x-coordinate of G.
d (x2 x1 )2
(y2 y1 )2
All rights reserved.
Formula: The Distance Formula
The distance d between two points A(x 1, y 1 ) and B(x 2, y 2 ) is
.
All rights reserved.
Key Concepts.
(
x1 x2
)
y1 y2
,
2
2
1 x2
2
d Use the Midpoint Formula. S
5
2
1 x2 d Multiply each side by 2 . S
10
3 x2
Formula: The Midpoint Formula
The coordinates of the midpoint M of AB with endpoints A(x 1, y 1 ) and
B(x 2, y 2 ) are the following:
M
1
d Simplify. S
6
4 y2
2
4 y2
x2
The coordinates of G are (3, 6) .
Quick Check.
.
1. Find the coordinates of the midpoint of XY with endpoints X(2, 5) and Y(6, 13).
Examples.
1 Finding the Midpoint AB has endpoints (8, 9) and (6, 3). Find the coordinates of its
midpoint M.
Use the Midpoint Formula. Let (x 1, y 1 ) be
(8, 9)
and (x 2, y 2 ) be
(6, 3) .
The midpoint has coordinates
(
x1
x2
,
y1
2
The x-coordinate is
The y-coordinate is
y2
)
6
2
) 1
Substitute 8 for x1 and
6
for x2. Simplify.
6
) 3
Substitute 9 for y1 and
3
for y2. Simplify.
(1, 3)
.
2
8 (
2
9 (
2
3
2
The coordinates of the midpoint M are
16
4
Midpoint Formula
.
Geometry Lesson 1-8
Geometry
2
Daily Notetaking Guide
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
(4, 4)
2. The midpoint of XY has coordinates (4, 6). X has coordinates (2, 3).
Find the coordinates of Y.
(6, 9)
L1
Daily Notetaking Guide
Geometry Lesson 1-8
All-In-One Answers Version B
17
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 1-9
2 Finding Area of a Circle Find the area of 䉺B in terms of π.
Perimeter, Circumference, and Area
Lesson Objectives
1
Find perimeters of rectangles and
squares, and circumferences of circles
Find areas of rectangles, squares, and
circles
2
NAEP 2005 Strand: Measurement
In 䉺B, r Topic: Measuring Physical Attributes
Aπ
Local Standards: ____________________________________
Aπ
(
A
r
dO
h
b
Square with side length s.
Perimeter P 4s
s2
Rectangle with base b and
height h.
C
Circle with radius r and
diameter d.
Perimeter P Circumference C Area A All rights reserved.
b
h
s
2b ⫹ 2h
All rights reserved.
Perimeter and Area
s
yd.
B
Formula for the area of a circle
1.5
)2
2.25
π
The area of 䉺B is
Key Concepts.
Area A 1.5
r2
Substitute
2.25p
1.5
1.5 yd
for r.
yd 2.
Quick Check.
1. a. Find the circumference of a circle with a radius of 18 m in terms of π.
36p m
pd
or C 2pr
bh
Area b. Find the circumference of a circle with a diameter of 18 m to the nearest tenth.
pr 2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
equal
If two figures are congruent, then their areas are
.
All rights reserved.
Postulate 1-10
The area of a region is the sum of the areas of its non-overlapping parts.
Examples.
1 Finding Circumference 䉺G has a radius of
6.5
cm. Find the
circumference of 䉺G in terms of π. Then find the circumference to the
nearest tenth.
C2 p
(
6.5
C
13
π
C
13
C 2π
Formula for circumference of a circle
)
Substitute
6.5
2. You are designing a rectangular banner for the front of
a museum. The banner will be 4 ft wide and 7 yd high.
How much material do you need in square yards?
2
91
3 yd
for r.
Exact answer
40.840704
The circumference of 䉺G is
18
6.5 cm
G
r
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
56.5m
Postulate 1-9
13p
Use a calculator.
, or about
40.8
cm.
Geometry Lesson 1-9
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 1-9
19
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 2-1
2 Writing the Converse of a Conditional Write the converse of the following
Conditional Statements
conditional.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objectives
1 Recognize conditional statements
2 Write converses of conditional
statements
NAEP 2005 Strand: Geometry
Topic: Mathematical Reasoning
If x 9, then x 3 12.
The converse of a conditional exchanges the hypothesis and the conclusion.
Local Standards: ____________________________________
Conditional
Vocabulary and Key Concepts.
If an angle is a straight angle,
Converse
If the measure of an angle is
Symbolic Form
If p ,
then q .
then its measure is 180.
If q ,
qSp
180, then it is a straight angle.
Conclusion
Hypothesis
Conclusion
x9
x 3 12
x ⫹ 3 ⫽ 12
x⫽9
So the converse is: If x ⫹ 3 ⫽ 12, then x ⫽ 9.
You read it
pSq
All rights reserved.
Conditional
All rights reserved.
Conditional Statements and Converses
Statement
Example
Converse
Hypothesis
then p .
Quick Check.
1. Identify the hypothesis and the conclusion of this conditional statement:
If y 3 5, then y 8.
Hypothesis:
A conditional is an if-then statement.
y⫺3⫽5
hypothesis
is the part that follows if in an if-then statement.
truth value
The
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The conclusion is the part of an if-then statement (conditional) that follows then.
of a statement is “true” or “false” according to whether the
statement is true or false, respectively.
The converse of the conditional “if p, then q” is the conditional “if q, then p.”
Examples.
1 Identifying the Hypothesis and the Conclusion Identify the hypothesis
and conclusion: If two lines are parallel, then the lines are coplanar.
In a conditional statement, the clause after if is the hypothesis and the
clause after then is the conclusion.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The
Conclusion:
y⫽8
2. Write the converse of the following conditional:
If two lines are not parallel and do not intersect, then they are skew.
If two lines are skew, then they are not parallel and do not intersect.
Hypothesis: Two lines are parallel.
Conclusion: The lines are coplanar.
20
L1
Geometry Lesson 2-1
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 2-1
Geometry
21
5
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
2 Identifying a Good Definition Is the following statement a good
Lesson 2-2
Biconditionals and Definitions
Lesson Objectives
1
definition? Explain.
An apple is a fruit that contains seeds.
NAEP 2005 Strand: Geometry
Write biconditionals
Recognize good definitions
2
Name_____________________________________ Class____________________________ Date ________________
The statement is true as a description of an apple.
Topics: Dimension and Shape; Mathematical Reasoning
Exchange “An apple” and “a fruit that contains seeds.” The converse reads:
Local Standards: ____________________________________
A fruit that contains seeds is an apple.
Vocabulary and Key Concepts.
There are many fruits that contain seeds but are not apples, such as lemons
and peaches. These are
Biconditional Statements
A biconditional combines p S q and q S p as p 4 q.
Symbolic Form
An angle is a straight angle if
You read it
p if and
p4q
only if q .
and only if its measure is 180°.
All rights reserved.
Example
Biconditional
All rights reserved.
Statement
statement is
biconditional
The original statement
statement
is not
counterexamples
, so the converse of the
.
is not
a good definition because the
reversible.
Quick Check.
1. Consider the true conditional statement. Write its converse. If the converse
is also true, combine the statements as a biconditional.
A biconditional statement is the combination of a conditional statement and its converse.
A
false
contains the words “if and only if.”
Conditional: If three points are collinear, then they lie on the same line.
Converse:
Examples.
If three points lie on the same line, then they are collinear.
1 Writing a Biconditional Consider the true conditional statement. Write its
To write the converse, exchange the hypothesis and conclusion.
Converse: If x ⫹ 15 ⫽ 20, then x ⫽ 5.
When you subtract 15 from each side to solve the equation, you get x 5. Because
true
both the conditional and its converse are
a
biconditional
using the phrase
, you can combine them in
if and only if
.
Biconditional: x ⫽ 5 if and only if x ⫹ 15 ⫽ 20.
22
Geometry Lesson 2-2
Daily Notetaking Guide
L1
true
The converse is
.
Biconditional:
All rights reserved.
Conditional: If x 5, then x 15 20.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
converse. If the converse is also true, combine the statements as a biconditional.
Three points are collinear if and only if they lie on the same line.
2. Is the following statement a good definition? Explain.
A square is a figure with four right angles.
It is not a good definition because a rectangle has four right angles
and is not necessarily a square.
L1
Daily Notetaking Guide
Geometry Lesson 2-2
23
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 2-3
3 Using the Law of Syllogism Use the Law of Syllogism to draw a
Deductive Reasoning
conclusion from the following true statements:
If a quadrilateral is a square, then it contains four right angles.
If a quadrilateral contains four right angles, then it is a rectangle.
NAEP 2005 Strand: Geometry
Topic: Mathematical Reasoning
Local Standards: ____________________________________
The conclusion of the first conditional is the hypothesis of the second
conditional. This means that you can apply the
Vocabulary and Key Concepts.
The Law of Syllogism: If
then
is true.
All rights reserved.
conclusion
In symbolic form:
If p S q is a true statement and p is true, then q is true.
Law of Syllogism
All rights reserved.
Law of Detachment
If a conditional is true and its hypothesis is true, then its
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objectives
1 Use the Law of Detachment
2 Use the Law of Syllogism
pSr
pSq
and
qSr
Law of Syllogism
.
are true statements,
is a true statement.
So you can conclude:
If a quadrilateral is a square, then it is a rectangle.
Quick Check.
1. Suppose that a mechanic begins work on a car and finds that the car will not
start. Can the mechanic conclude that the car has a dead battery? Explain.
If p S q and q S r are true statements, then p S r is a true statement.
No, there could be other things wrong with the car, such as a faulty starter.
Deductive reasoning is a process of reasoning logically from given facts to a conclusion.
1 Using the Law of Detachment A gardener knows that if it rains, the
garden will be watered. It is raining. What conclusion can he make?
The first sentence contains a conditional statement. The hypothesis is
it rains.
Because the hypothesis is true, the gardener can conclude that
the garden will be watered.
2 Using the Law of Detachment For the given statements, what can you
conclude?
Given: If ⬔A is acute, then m⬔A 90°. ⬔A is acute.
A conditional and its hypothesis are both given as true.
By the
Law of Detachment
conclusion of the conditional, m⬔A 90°, is
24
6
Geometry Lesson 2-3
Geometry
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Examples.
2. If a baseball player is a pitcher, then that player should not pitch a complete
game two days in a row. Vladimir Nuñez is a pitcher. On Monday, he pitches
a complete game. What can you conclude?
Answers may vary. Sample: Vladimir Nuñez should not pitch a complete
game on Tuesday.
3. If possible, state a conclusion using the Law of Syllogism. If it is not possible
to use this law, explain why.
a. If a number ends in 0, then it is divisible by 10.
If a number is divisible by 10, then it is divisible by 5.
If a number ends in 0, then it is divisible by 5.
b. If a number ends in 6, then it is divisible by 2.
If a number ends in 4, then it is divisible by 2.
, you can conclude that the
true
Not possible; the conclusion of one statement is not the hypothesis of
the other statement.
.
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 2-3
All-In-One Answers Version B
25
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 2-4
Examples.
Reasoning in Algebra
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
1 Justifying Steps in Solving an Equation Justify each step used to solve
NAEP 2005 Strand: Algebra and Geometry
Connect reasoning in algebra and
geometry
5x 12 32 x for x.
Topics: Algebraic Representations; Mathematical
Reasoning
Given: 5x 12 32 x
Local Standards: ____________________________________
5x 44 x
Addition Property of Equality
4x 44
Subtraction Property of Equality
Key Concepts.
x 11
If a b, then a c b c .
Multiplication Property If a b, then a c b c .
If a b and c 0, then
b
c
If a b, then b a .
Transitive Property
If a b and b c, then a c .
Substitution Property
If a b, then b can replace a in any expression.
Distributive Property
a(b c) ac .
AB
⬔A ⬔A
If AB CD, then CD AB .
If ⬔A ⬔B, then ⬔B ⬔A .
If AB CD and CD EF , then AB Transitive Property
⬔P 艑 ⬔S
If ⬔P ⬔R and ⬔R ⬔S, then
Properties of Congruence
Symmetric Property
.
for
⬔P ⬔R and the third part of the hypothesis:
.
EF
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
ab
⬔P 艑 ⬔R
Transitive Property of Congruence
Use the
a a .
for
the first two parts of the hypothesis:
If ⬔P ⬔Q and ⬔Q ⬔R, then
Symmetric Property
AB Transitive Property of Congruence
Use the
.
Reflexive Property
Reflexive Property
If ⬔P ⬔Q, ⬔Q ⬔R, and ⬔R ⬔S, then ⬔P ⬔S.
c
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
a
justifies each statement.
All rights reserved.
If a b, then a c b c .
Subtraction Property
All rights reserved.
Addition Property
Division Property
Division Property of Equality
2 Using Properties of Equality and Congruence Name the property that
Properties of Equality
.
Quick Check.
M
1. Fill in each missing reason.
)
(2x40)
Given: LM bisects ⬔KLN.
)
LM
bisects ⬔KLN
4x
Given
K
m⬔MLN m⬔KLM Definition of
Angle
4x 2x 40
Substitution Property of Equality
2x 40
Subtraction Property of Equality
x 20
N
L
Bisector
Division Property of Equality
2. Name the property of equality or congruence illustrated.
a. XY XY
Reflexive Property of Congruence
If ⬔A ⬔B and ⬔B ⬔C, then ⬔A ⬔C .
b. If m⬔A 45 and 45 m⬔B, then m⬔A m⬔B
Transitive or Substitution Property of Equality
26
Geometry Lesson 2-4
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 2-5
27
Geometry Lesson 2-4
Name_____________________________________ Class____________________________ Date ________________
complementary angles
Proving Angles Congruent
Lesson Objectives
1 Prove and apply theorems about
angles
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Daily Notetaking Guide
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
supplementary angles
2
4
3
1
Local Standards: ____________________________________
⬔1 and ⬔2 are complementary angles.
Vocabulary and Key Concepts.
Theorem 2-1: Vertical Angles Theorem
3
4
2
⬔1 ⬔2 and ⬔3 ⬔4
Theorem 2-2: Congruent Supplements Theorem
If two angles are supplements of the same angle (or of congruent angles),
then the two angles are
congruent
All rights reserved.
1
.
All rights reserved.
congruent
Vertical angles are
.
congruent
Two angles are complementary angles if
Two angles are supplementary angles if
the sum of their measures is 90°.
the sum of their measures is 180°.
A theorem is a conjecture that is proven.
A
paragraph proof
is a convincing argument that uses deductive reasoning in which
statements and reasons are connected in sentences.
Examples.
1 Using the Vertical Angles Theorem Find the value of x.
The angles with labeled measures are vertical angles. Apply the
Vertical Angles Theorem to find x.
Theorem 2-3: Congruent Complements Theorem
If two angles are complements of the same angle (or of congruent angles),
then the two angles are
4x 101 2x 3
.
angle.
right
adjacent angles
1
3
4
2
⬔1 and ⬔2 are vertical angles,
as are ⬔3 and
3
1
⬔4 .
2
4
⬔1 and ⬔2 are adjacent angles,
as are ⬔3 and
⬔4 .
Vertical angles are two angles whose
Adjacent angles are two coplanar angles
sides form two pairs of opposite rays.
that have a common side and a common
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
vertical angles
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
angles are congruent.
Theorem 2-5
If two angles are congruent and supplementary, then each is a
(2x 3)
(4x 101)
Vertical Angles Theorem
4x 2x 104
Theorem 2-4
right
All
⬔3 and ⬔4 are supplementary angles.
Addition Property of Equality
2x 104
Subtraction Property of Equality
x 52
Division Property of Equality
2 Proving Theorem 2-2 Write a paragraph proof of Theorem 2-2
1
using the diagram at the right.
Start with the given: ⬔1 and ⬔2 are supplementary, ⬔3 and ⬔2 are
supplementary. By the definition of
supplementary angles
2
3
,
m⬔1 m⬔2 180 and m⬔3 m⬔2 180. By substitution,
m⬔1 m⬔2 subtract
m⬔2
m⬔2
m⬔3
. Using the
from each side. You get m⬔1 Subtraction Property of Equality
m⬔3
,
, or ⬔ 1 ⬔ 3 .
Quick Check.
1. Refer to the diagram for Example 1.
a. Find the measures of the labeled pair of vertical angles.
107°
b. Find the measures of the other pair of vertical angles.
vertex but no common interior points.
73°
c. Check to see that adjacent angles are supplementary.
107° ⫹ 73° ⫽ 180°
28
L1
Geometry Lesson 2-5
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 2-5
Geometry
29
7
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 3-1
Properties of Parallel Lines
Lesson Objectives
1
Examples.
NAEP 2005 Strand: Measurement
Identify angles formed by two lines
and a transversal
Prove and use properties of parallel
lines
2
Name_____________________________________ Class____________________________ Date ________________
1 Applying Properties of Parallel Lines In the diagram of
Topic: Measuring Physical Attributes
Lafayette Regional Airport, the black segments are runways
and the gray areas are taxiways and terminal buildings.
Local Standards: ____________________________________
Compare ⬔2 and the angle vertical to ⬔1. Classify the angles
as alternate interior angles, same-side interior angles, or
corresponding angles.
Vocabulary and Key Concepts.
2
3
1
/
m
⬔1 ⬔2
a
t
b
3 2
1
are nonadjacent interior angles that
t
56
13
42
78
lie on opposite sides of the transversal.
Same-side interior angles are interior angles that lie on the same side
of the transversal.
⬔1
and ⬔2 are
alternate interior angles.
of the transversal and in corresponding positions relative to the
corresponding angles
Because ᐉm, m⬔1 by the
42
Angle Addition Postulate
for m⬔1, the equation becomes
Subtract
p
q
.
Daily Notetaking Guide
4
.
5
42
/
m
42
42
. If you substitute
42
m⬔2 180 .
from each side to find m⬔2 138 .
1. Use the diagram in Example 1. Classify ⬔2 and ⬔3 as alternate interior angles,
same-side interior angles, or corresponding angles.
same-side interior angles
2. Using the diagram in Example 2 find the measure of each angle. Justify each answer.
a. ⬔3 42; Vertical angles are congruent
b. ⬔4 138; Corresponding angles are congruent
c. ⬔5 138; Same-side interior angles are supplementary
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 3-2
8 6
7
2 1
3
Quick Check.
⬔1 and ⬔7
corresponding angles.
coplanar lines.
Daily Notetaking Guide
Geometry Lesson 3-1
31
Name_____________________________________ Class____________________________ Date ________________
Examples.
Proving Lines Parallel
1 Proving Theorem 3-5 If two lines and a transversal form alternate
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
3
/
interior angles that are congruent, then the two lines are parallel.
Given: ⬔1 ⬔2
Prove: ᐉ m
Local Standards: ____________________________________
1
m
2
Write the flow proof below of the Alternate Interior Angles
Theorem as a paragraph proof.
Vocabulary and Key Concepts.
ᐉ m .
Theorem 3-5: Converse of the Alternate Interior Angles Theorem
If two lines and a transversal form alternate interior angles that are
congruent, then the two lines are parallel.
⬔1 ⬔3
All rights reserved.
then the two lines are parallel.
2
All rights reserved.
m
If two lines and a transversal form corresponding angles that are congruent,
⬔1 ⬔2
Given
1
/
Postulate 3-2: Converse of the Corresponding Angles Postulate
Vertical angles are congruent
⬔3 then ᐉ m .
congruent, then the two lines are parallel.
Theorem 3-8: Converse of the Same-Side Exterior Angles Theorem
If ⬔3 and ⬔6 are
If two lines and a transversal form same-side exterior angles that are
supplementary,
supplementary, then the two lines are parallel.
then ᐉ m .
A flow proof uses arrows to show the logical connections between the statements.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
If two lines and a transversal form alternate exterior angles that are
by the
⬔3
and
Transitive
⬔2
Property of Congruence.
are corresponding angles, ᐉ m by the
Postulate.
Use the diagram at the right. Which lines, if any, must be parallel
if ⬔3 and ⬔2 are supplementary? Justify your answer with
a theorem or postulate.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
supplementary,
If ⬔3 ⬔5,
?
2 Using Postulate 3-2
If two lines and a transversal form same-side interior angles that are
then ᐉ m .
/ || m
Converse of the Corresponding Angles
then ᐉ m .
supplementary, then the two lines are parallel.
⬔2
Because
If ⬔2 and ⬔4 are
Theorem 3-7: Converse of the Alternate Exterior Angles Theorem
⬔3 ⬔2
Transitive
Property of
Congruence
By the Vertical Angles Theorem, ⬔3 ⬔1 . ⬔1 ⬔2, so
5 6
1 4
2
m
3
If ⬔1 ⬔2,
/
Theorem 3-6: Converse of the Same-Side Interior Angles Theorem
Reasons
.
straight angle, m⬔1 m⬔2 180 by the
⬔1 and ⬔4 are
same-side interior angles.
are angles that lie on the same side
Geometry Lesson 3-1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
/
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
180
m
Lesson Objectives
1 Use a transversal in proving lines
parallel
⬔1 and the 42 angle are
Because ⬔1 and ⬔2 are adjacent angles that form a
distinct points.
30
alternate interior angles
Corresponding Angles Postulate
A transversal is a line that intersects two coplanar lines at two
Corresponding angles
⬔2 and the angle vertical to ⬔1 are
In the diagram at right, ᐉm and pq. Find m⬔1 and m⬔2.
Theorem 3-2: Same-Side Interior Angles Theorem
If a transversal intersects two parallel lines, then same-side interior
angles are supplementary .
m⬔1 m⬔2 are not
adjacent and lie between the lines on opposite sides of the transversal,
⬔3
⬔1
alternate interior angles
transversal runway. Because
2 Finding Measures of Angles
Theorem 3-1: Alternate Interior Angles Theorem
If a transversal intersects two parallel lines, then alternate interior
congruent
angles are
.
Alternate interior angles
⬔2 is between the runway segments and on the opposite side of the
All rights reserved.
1
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
t
All rights reserved.
Postulate 3-1: Corresponding Angles Postulate
If a transversal intersects two parallel lines, then corresponding
congruent
angles are
.
All rights reserved.
The angle vertical to ⬔1 is between the runway segments.
E
It is given that ⬔3 and ⬔2 are supplementary. The diagram
shows that ⬔4
and ⬔2
are supplementary. Because
supplements of the same angle are
3
C
4K
2
congruent
(Congruent Supplements Theorem), ⬔3
and
D1
⬔4 . Because ⬔3
⬔4 are congruent corresponding angles, EC
Converse of the Corresponding Angles
)
DK
)
by the
Postulate.
Quick Check.
1. Supply the missing reason in the flow proof from Example 1.
If corresponding angles are congruent, then the lines are parallel.
2. Use the diagram from Example 1. Which lines, if any, must be parallel if
⬔3 ⬔4? Explain.
are written below the statements.
u u
EC 储 DK ; Converse of Corresponding Angles Postulate
32
8
Geometry Lesson 3-2
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 3-2
All-In-One Answers Version B
33
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 3-3
Examples.
Parallel and Perpendicular Lines
Lesson Objectives
1
Name_____________________________________ Class____________________________ Date ________________
1 Proof of Theorem 3-10
NAEP 2005 Strand: Geometry
Relate parallel and perpendicular
lines
Local Standards: ____________________________________
Key Concepts.
By the Vertical Angles Theorem, ⬔1 is
Because ⬔1 they are
parallel
a
b
, then
All rights reserved.
parallel to the same line
If two lines are
c
to each other.
a b .
perpendicular to the same line
In a plane, if two lines are
then they are
t
m
Theorem 3-10
parallel
,
⬔1 by the
Converse of the Alternate Interior Angles
and the
Theorem to prove that
the lines are parallel.
2 Using Algebra
n
m n .
Find the value of x for which ᐉ m.
The labeled angles are
alternate interior angles
.
If ᐉm, the alternate interior angles are
,
equal
and their measures are
m
All rights reserved.
m
5x
66 5x
equation
ᐉ
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
n
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 3-11
In a plane, if a line is perpendicular to one of two
parallel lines, then it is also perpendicular to the other.
⊥
congruent
(14 3x)
. Write and solve the
14
3x
5x
80
3x
2x
80
Daily Notetaking Guide
L1
Subtract 3x from each side.
x
40
Divide each side by 2.
Quick Check.
1. Critical Thinking In a plane, if two lines form congruent angles with a third
line, must the lines be parallel? Draw a diagram to support your answer.
No; answers may vary. Sample:
115°
115°
L1
Name_____________________________________ Class____________________________ Date________________
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objectives
1 Classify triangles and find the
measures of their angles
2 Use exterior angles of triangles
Daily Notetaking Guide
Examples.
48
m⬔Z 67
180
m⬔Z 180
115
Local Standards: ____________________________________
2
1
measures of its two remote interior angles.
Back
B
A
All rights reserved.
.
All rights reserved.
C
Theorem 3-13: Triangle Exterior Angle Theorem
The measure of each exterior angle of a triangle equals the sum of the
65
x
67° Y
X 48°
Subtract 115 from each side.
3
x
(180180
Arm
)
x adjacent angles
straight angle
measure
decreases
three acute angles.
one right angle.
__________________
__________________
__________________ __________________
An equilateral
triangle has
An equiangular
triangle has
one obtuse angle.
__________________
__________________
three congruent angles.
______________________
______________________
An isosceles
triangle has
A scalene
triangle has
at least two congruent sides.
__________________________
__________________________
no congruent sides.
_____________________
_____________________
An exterior angle of a polygon is an angle formed by a side
and an extension of an adjacent side.
Remote
interior
All-In-One Answers Version B
increases
as you make a lounge chair recline more.
32
b. Reasoning Explain why this statement must be true:
If a triangle is a right triangle, its acute angles are complementary.
The sum of the measures of the angles of a triangle is 180. If you
subtract the measure of the right angle from 180, you get 90. The
sum of the measures of the other two angles is 90, so they are
complementary.
2. a. Find m⬔3.
90
45
b. Critical Thinking Is it true that if two acute angles of a triangle are
complementary, then the triangle must be a right triangle? Explain.
45
3
True, because the measures of the two complementary angles add to
90, leaving 90 for the third angle.
angles
Daily Notetaking Guide
, which together form a
. As one measure increases, the other
. The angle formed by the back of the chair and
1. a. 䉭MNP is a right triangle. ⬔M is a right angle and m⬔N is 58. Find m⬔P.
2
3
)
x Quick Check.
angle
Remote interior angles are the two nonadjacent interior
angles corresponding to each exterior angle of a triangle.
Exterior
1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
An obtuse
triangle has
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the armrest
A right
triangle has
(180180
The exterior angle and the angle formed by the back of the chair and
the armrest are
L1
Simplify.
the angle formed by the back of the chair and the armrest as you make a
lounge chair recline more.
180
m⬔A m⬔B m⬔C 180
Geometry Lesson 3-4
Z
Triangle Angle-Sum Theorem
2 Applying the Triangle Exterior Angle Theorem Explain what happens to
Theorem 3-12: Triangle Angle-Sum Theorem
The sum of the measures of the angles of a triangle is
36
35
Geometry Lesson 3-3
m⬔Z three congruent sides.
________________________
________________________
(7x 8)
62
1 Applying the Triangle Angle-Sum Theorem Find m⬔Z.
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Vocabulary and Key Concepts.
An acute
triangle has
a
b
Name_____________________________________ Class____________________________ Date ________________
Parallel Lines and the
Triangle Angle-Sum Theorem
Lesson 3-4
m
Add 66 to each side.
x 18; 7(18) 8 118°, and 62° 118° 180°
Geometry Lesson 3-3
/
(5x 66)
3x.
14
66 2. Find the value of x for which a b. Explain how you can check your answer.
34
t
2
1
Property of Congruence. Because alternate interior angles
are congruent, you can use the vertical angle of ⬔1
to each other.
n
r
to its vertical angle.
congruent
⬔2 , ⬔2 is congruent to the vertical angle of
Transitive
All rights reserved.
Theorem 3-9
s
Use the diagram at the right. Which angle would you use with ⬔1 to
prove the theorem In a plane, if two lines are perpendicular to
the same line, then they are parallel to each other (Theorem 3-10)
using the Converse of the Alternate Interior Angles Theorem
instead of the Converse of the Corresponding Angles Postulate?
Topic: Relationships Among Geometric Figures
L1
L1
Daily Notetaking Guide
Geometry Lesson 3-4
Geometry
37
9
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 3-5
Name_____________________________________ Class____________________________ Date ________________
An equilateral polygon has all sides congruent.
The Polygon Angle-Sum Theorems
An equiangular polygon has all angles congruent .
Lesson Objectives
1
NAEP 2005 Strand: Geometry
Classify polygons
Find the sums of the measures of the
interior and exterior angles of
polygons
2
A
Topic: Relationships Among Geometric Figures
is both equilateral and equiangular.
regular polygon
Local Standards: ____________________________________
Examples.
1 Classifying Polygons Classify the polygon at the right by its sides. Identify it
as convex or concave.
Vocabulary and Key Concepts.
Starting with any side, count the number of sides clockwise around the figure.
Because the polygon has 12 sides, it is a dodecagon. Think of the
.
Theorem 3-15: Polygon Exterior Angle-Sum Theorem
The sum of the measures of the exterior angles of a polygon,
polygon as a star. Draw a diagonal connecting two adjacent points of the star.
That diagonal lies
2
4
360
2 Finding a Polygon Angle Sum Find the sum of the measures of the angles
1
.
5
of a decagon.
A polygon is a closed plane figure with at least three sides that are segments. The sides
B
A
C
E
D
C
D
D
Not a polygon;
A polygon
not a
closed
E
Not a polygon;
figure
two sides
intersect
between endpoints.
A convex polygon does not have diagonal points
A
outside of the polygon.
D
R
Y
A convex
polygon
A concave polygon has at least one diagonal
T
Diagonal
P
M
K
W
Q
G
with points outside of the polygon.
Geometry Lesson 3-5
(
)
2 (180)
10
.
Polygon Angle-Sum
Substitute
8 ? 180
Subtract.
Simplify.
1440
10
Theorem
for n.
Quick Check.
1. Classify each polygon by its sides. Identify each as convex or concave.
a.
b.
hexagon; convex
octagon; concave
2. Find the sum of the measures of the angles of a 13-gon.
1980
A concave
polygon
S
38
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
E
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
B
A
C
10
Sum (n 2)(180)
intersect only at their endpoints, and no two adjacent sides are collinear.
B
sides, so n 10
A decagon has
All rights reserved.
.
For the pentagon, m⬔1 m⬔2 m⬔3 m⬔4 m⬔5 A
the polygon, so the dodecagon is
outside
.
concave
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 3-5
39
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 3-6
2 Using Point-Slope Form Write an equation in point-slope form
Lines in the Coordinate Plane
Lesson Objectives
1 Graph lines given their equations
2 Write equations of lines
of the line with slope 8 that contains P(3, 6).
NAEP 2005 Strand: Algebra
Topics: Patterns, Relations, and Functions;
Algebraic Representations
y
y
Slope-intercept
1
Ax By C.
Standard
The point-slope form for a nonvertical line is
form
y ⫺ y 1 ⫽ m(x ⫺ x 1 ).
8
x1
(x (x 3
3
)
)
)
Use point-slope form.
Substitute 8 for m and (3, 6) for (x1, y1).
Simplify.
1. Graph each equation.
form of a linear equation is
standard
O
4
2x y 1
4
3
x
All rights reserved.
The
y
All rights reserved.
y 2x 3
y ⫽ mx ⫹ b.
(
m x
)
Quick Check.
form
The slope-intercept form of a linear equation is
y1
6
y 6 8
Local Standards: ____________________________________
Vocabulary.
(
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
360
one at each vertex, is
3
All rights reserved.
(n ⫺ 2)180
The sum of the measures of the angles of an n-gon is
All rights reserved.
Theorem 3-14: Polygon Angle-Sum Theorem
a. y 12x 2
1
x
O y
1
2
4
c. 5x y 3
2
x
O
3
y 2 2(x 1)
Point-slope
b. 2x 4y 8
y
O
y
2
x
2
form
Examples.
2. Write an equation of the line that contains the points P(5, 0) and Q(7, 3).
1 Graphing Lines Using Intercepts Use the x-intercept and y-intercept to
5x 6y 30
( ) 6y 30
5 0
5x 0 30
0 6y 30
6y 30
x 6
30
y 5
The x-intercept is 6 .
A point on the line is
To find the y-intercept, substitute
0 for x and solve for y.
5x 6y 30
( ) 30
5x 6 0
5x © Pearson Education, Inc., publishing as Pearson Prentice Hall.
To find the x-intercept, substitute
0 for y and solve for x.
( 6 , 0).
The y-intercept is 5 .
(
A point on the line is 0, 5
Plot (6, 0) and (0, 5). Draw the line containing the two points.
).
y
2
(6, 0)
2 O
2
4
6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
3
y 0 3
2(x 5) or y 3 2(x 7)
graph 5x 6y 30.
8x
2
4
6
40
10
Geometry Lesson 3-6
Geometry
(0, 5)
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 3-6
All-In-One Answers Version B
41
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 3-7
2 Finding Slopes for Perpendicular Lines Find the slope of a line
Slopes of Parallel and Perpendicular Lines
Lesson Objectives
1
Relate slope and parallel lines
Relate slope and perpendicular lines
2
perpendicular to 5x 2y 1.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
To find the slope of the given line, rewrite the equation in
slope-intercept form.
Local Standards: ____________________________________
5x 2y 1
2y 5 x 1
Key Concepts.
5
y 2
Slopes of Parallel Lines
If the slopes of two distinct nonvertical lines are equal, the lines are parallel.
Any two vertical lines are parallel.
Slopes of Perpendicular Lines
If the slopes of two lines have a product of 1, the lines are perpendicular.
5
( )
Examples.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
slope-intercept
All rights reserved.
form. Write the equation x 5y 4 in slope-intercept form.
x 5y 4
x 4 5y
5
4
5
Subtract 4 from each side.
y
y ⫺
Divide each side by 5.
1
5
x
4
5
The line x 5y 4 has slope
⫺1
5
.
The line y 5x 4 has slope
⫺5
.
The lines
42
are not
parallel because their slopes are
not equal
Geometry Lesson 3-7
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
m
x 5y 4 parallel? Explain.
x
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objectives
1 Construct parallel lines
2 Construct perpendicular lines
2
2
Multiply each side by 5
.
5
2
Simplify.
5
Quick Check.
1. Are the lines y 12x 5 and 2x 4y 9 parallel? Explain.
Yes; Each line has slope 1
2 and the y-intercepts are different.
2. Find the slope of a line perpendicular to 5y x 10.
m 5
.
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 3-8
.
The product of the slopes of perpendicular lines is 1.
m 1 ? 1 Determining Whether Lines are Parallel Are the lines y 5x 4 and
1
5
2
m 1
2
Any horizontal line and vertical line are perpendicular.
⫺
Divide each side by 2.
2
Find the slope of a line perpendicular to 5x 2y 1. Let m be the slope
of the perpendicular line.
If two nonvertical lines are perpendicular, the product of their slopes is ⫺1.
The equation y 5x 4 is in
Subtract 5x from each side.
1
The line 5x 2y 1 has slope
All rights reserved.
All rights reserved.
If two nonvertical lines are parallel, their slopes are equal.
x
Daily Notetaking Guide
43
Geometry Lesson 3-7
Name_____________________________________ Class____________________________ Date________________
Example.
Constructing Parallel and Perpendicular Lines
R
the
2 Perpendicular From a Point to a Line * Examine
)
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
construction. At what special point does RG meet line ᐉ?
Point R is the same distance from point E as it is from point F
Local Standards: ____________________________________
/
E
F
because the arc was made with one compass opening.
G
Point G is the same distance from point E as it is from point F
Example.
because both arcs were made with the same compass opening.
1 Constructing ᐉm Examine the diagram at right. Explain
m
/
H
J
All rights reserved.
Step 1 With the compass point on point H, draw an arc that
intersects the sides of ⬔H .
Step 4 Use a straightedge to draw line m through the point
you located and point N.
Quick Check.
1. Use Example 1. Explain why lines ᐉ and m must be parallel.
If corresponding angles are congruent, the lines are parallel by the
Converse of Corresponding Angles Postulate.
L1
)
)
Quick Check.
midpoint
of EF,
perpendicular bisector
* )
* )
of EF.
* ) * )
Daily Notetaking Guide
All-In-One Answers Version B
L1
G
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Step 3 Put the compass point below point N where the arc
* )
intersects HN . Open the compass to the length where
the arc intersects line ᐉ . Keeping the same compass
setting, put the compass point above point N where the
* )
arc intersects HN . Draw an arc to locate a point.
Geometry Lesson 3-8
*
and that RG is the
2. Use a straightedge to draw EF . Construct FG so that FG ⬜ EF at point F.
Step 2 With the same compass setting, put the compass point
on point N . Draw an arc.
44
*
This means that RG intersects line ᐉ at the
N 1
Use the method learned for constructing congruent angles.
All rights reserved.
how to construct ⬔1 congruent to ⬔H. Construct the angle.
F
E
L1
Daily Notetaking Guide
Geometry Lesson 3-8
Geometry
45
11
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Congruent Figures
List corresponding vertices in the same order.
If 䉭ABC 䉭CDE, then ⬔BAC NAEP 2005 Strand: Geometry
Recognize congruent figures and their
corresponding parts
The diagram above shows ⬔BAC Topic: Transformation of Shapes and Preservation of
Properties
The statement 䉭ABC 䉭CDE
Local Standards: ____________________________________
Notice that BC Also, ⬔CBA Vocabulary and Key Concepts.
two angles of another triangle, then the third angles
B
C
E
F
⬔ C ⬔ F
Congruent polygons are polygons that have corresponding
⬔B ⬔ T
䉭
DEF
GHI
m⬔K Q
T
B
⬔ C ⬔J
TJ
AC C
A
QJ
J
Use the Triangle Angle-Sum Theorem and the definition of congruent
polygons to find m⬔X.
Triangle Angle-Sum Theorem
m⬔Z m⬔M
Corresponding angles of congruent triangles are congruent.
m⬔Z 48
Substitute 48 for m⬔M.
115
180
Substitute.
180
Simplify.
m⬔X 46
180
65
⬔ACB
.
䉭EDC
.
YS O KV
lWYS O lMKV
lSWY O lVMK
35
Corresponding angles of congruent triangles are congruent and
have the same measure.
3. Can you conclude that 䉭JKL 䉭MNL? Justify your answer.
N
J
L
K
M
No. The corresponding sides are not necessarily equal.
Subtract 115 from each side.
Geometry Lesson 4-1
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson Objective
1 Prove two triangles congruent using
the SSS and SAS Postulates
47
Geometry Lesson 4-1
Name_____________________________________ Class____________________________ Date ________________
2 Using SAS AD > BC . What other information do you
Triangle Congruence
by SSS and SAS
Lesson 4-2
Daily Notetaking Guide
B
A
need to prove 䉭ADC 䉭BCD by SAS?
It is given that AD > BC. Also, DC > CD by the
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
.
Reflexive Property of Congruence
⬔ADC ⬵ ⬔BCD
Therefore, if you know
䉭ADC 䉭BCD by
SAS
C
D
You now have two pairs of corresponding congruent sides.
Local Standards: ____________________________________
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
m⬔X .
2. It is given that 䉭WYS 䉭MKV. If m⬔Y 35, what is m⬔K? Explain.
2 Using Congruency 䉭XYZ 䉭KLM, m⬔Y 67, and m⬔M 48. Find m⬔X.
48
and ⬔BAC A
.
EC
G
F
List the corresponding sides and angles in the same order.
⬔DEC
WS O MV
Angles: ⬔WSY ⬔MVK
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
䉭
1 Naming Congruent Parts 䉭ABC 䉭QTJ. List the
congruent corresponding parts.
67
true.
D
4
1. 䉭WYS 䉭MKV. List the congruent corresponding parts. Use three letters
for each angle.
Sides: WY MK
Examples.
m⬔X is not
DE , and AC 3
C
3
Quick Check.
I
E
m⬔X m⬔Y m⬔Z , not ⬔DCE.
are congruent. The correct way to state this is 䉭ABC H
D
sides congruent and corresponding angles congruent.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
are congruent.
All rights reserved.
All rights reserved.
A
If two angles of one triangle are congruent to
BC
.
Since all of the corresponding sides and angles are congruent, the triangles
D
Sides: AB > QT
⬔CDE
⬔DEC
4
B
Using Theorem 4-1, you can conclude that ⬔ECD Theorem 4–1
Angles: ⬔A ⬔Q
DC , BA ⬔DCE
All rights reserved.
Lesson Objective
1
E
3 Finding Congruent Triangles Can you conclude that 䉭ABC 䉭CDE?
Lesson 4-1
, you can prove
.
Key Concepts.
Q
P
congruent.
䉭
GHF
䉭
F
PQR
Postulate 4-2: Side-Angle-Side (SAS) Postulate
If two sides and the included angle of one triangle
are congruent to two sides and the included angle
of another triangle, then the two triangles
All rights reserved.
Quick Check.
H
All rights reserved.
Postulate 4-1: Side-Side-Side (SSS) Postulate
If the three sides of one triangle are
congruent to the three sides of another
G
triangle, then the two triangles are
1. Given: HF HJ, FG JK,
H is the midpoint of GK.
G
D
E
C
FDE
A
Examples.
A
1 Proving Triangles Congruent
Given: M is the midpoint of XY , AX > AY
Prove: 䉭AMX 䉭AMY
Write a paragraph proof.
You are given that M is the midpoint of XY , and AX AY .
Midpoint M implies that MX MY . AM > AM by the
Reflexive Property of Congruence
䉭AMX 䉭AMY by the
48
12
Geometry Lesson 4-2
Geometry
X
, so
SSS Postulate
M
Y
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
䉭
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
BCA
H
Statements
are congruent.
䉭
J
K
R
F
B
F
Prove: 䉭FGH 䉭JKH
Reasons
1. HF > HJ, FG > JK
1. Given
2. H is the midpoint of GK.
2. Given
3. GH > HK
3.
Definition of Midpoint
4. 䉭FGH ⬵ 䉭JKH
4.
SSS Postulate
2. What other information do you need to prove 䉭ABC 䉭CDA by SAS?
A
8
⬔DCA 艑 ⬔BAC
7
7
6
D
B
C
.
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 4-2
All-In-One Answers Version B
49
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 4-3
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
Triangle Congruence
by ASA and AAS
Example.
2 Writing a Proof Write a two-column proof that uses AAS.
NAEP 2005 Strand: Geometry
Prove two triangles congruent using
the ASA Postulate and the AAS
Theorem
B
A
Given: ⬔B ⬔D, AB6CD
Prove: 䉭ABC 䉭CDA
Topic: Transformation of Shapes and Preservation
of Properties
Local Standards: ____________________________________
C
D
Key Concepts.
Statements
1. ⬔B ⬔D, AB6CD
P
to two angles and the included side of another triangle, then
䉭
HGB
䉭
B
K
.
G
NKP
N
H
1. Given
2. lBAC O lDCA
All rights reserved.
If two angles and the included side of one triangle are congruent
All rights reserved.
Postulate 4-3: Angle-Side-Angle (ASA) Postulate
the two triangles are congruent
Reasons
alternate interior
2. If lines are parallel, then
2. angles are congruent.
3. AC > AC
3. Reflexive Property of Congruence
4. 䉭ABC 䉭CDA
4. AAS Theorem
Theorem 4-2: Angle-Angle-Side (AAS) Theorem
C
If two angles and a nonincluded side of one triangle are congruent
Quick Check.
to two angles and the corresponding nonincluded side of another
CDM
䉭
G
XGT
All rights reserved.
D
O
C
and ⬔I is not congruent to ⬔C. Name the triangles
that are congruent by the ASA Postulate.
The diagram shows ⬔N ⬔A ⬔D and
FN > CA > GD.
F
G
T
N
A
I
If ⬔F ⬔C, then ⬔F ⬔C ⬔ G .
Therefore, 䉭FNI 䉭CAT 䉭GDO by
2. In Example 2, explain how you could prove 䉭ABC 䉭CDA using ASA.
ASA
It is given that lB O lD. You know that lBAC O lDCA by the
Alternate Interior Angles Theorem. By Theorem 4-1, lBCA O lDAC .
By the Reflexive Property of Congruence, AC O AC . Therefore,
kABC O kCDA by ASA.
X
Example.
1 Using ASA Suppose that ⬔F is congruent to ⬔C
T
D
.
Quick Check.
1. Using only the information in the diagram, can you conclude that 䉭INF is
congruent to either of the other two triangles? Explain.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
䉭
M
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the triangles are congruent
triangle, then
No; Only one angle and one side are shown to be congruent. At least
one more congruent side or angle is necessary to prove congruence
with SAS, ASA, or AAS.
50
Geometry Lesson 4-3
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Geometry Lesson 4-3
51
Name_____________________________________ Class____________________________ Date ________________
2 Using Right Triangles According to legend, one of Napoleon’s followers used
Lesson 4-4
Using Congruent Triangles: CPCTC
Lesson Objective
1 Use triangle congruence and CPCTC
to prove that parts of two triangles are
congruent
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Daily Notetaking Guide
congruent triangles to estimate the width of a river. On the riverbank, the officer
stood up straight and lowered the visor of his cap until the farthest thing he could
see was the edge of the opposite bank. He then turned and noted the spot on his
side of the river that was in line with his eye and the tip of his visor.
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
Local Standards: ____________________________________
G
F
Vocabulary.
D
CPCTC stands for:
of C ongruent
.
Examples.
E
All rights reserved.
P arts
are C ongruent
All rights reserved.
C orresponding
T riangles
Given: ⬔DEG and ⬔DEF are right angles; ⬔EDG ⬔EDF.
The officer then paced off the distance to this spot and declared that
distance to be the width of the river!
1 Congruence Statements The diagram shows the frame of an umbrella.
What congruence statements besides ⬔3 ⬔4 can you
prove from the diagram, in which SL > SR and
⬔1 ⬔2 are given?
by the Reflexive Property of Congruence,
䉭RSC
by SAS . ⬔3 ⬔4 because
corresponding parts of congruent triangles are congruent.
12
When two triangles are congruent, you can form congruence
statements about three pairs of corresponding angles and
three pairs of corresponding sides. List the congruence
statements.
S
So the officer must stand
ground must be
Stretcher
Shaft
Sides:
SL > SR
Given
SC SC
Reflexive Property of Congruence
CL CR
Other congruence statement
Angles:
⬔1 ⬔2
Given
⬔3 ⬔ 4
Corresponding Parts of Congruent Triangles
lCLS O lCRS
Other congruence statement
right angles
.
⬔DEG and ⬔DEF are the angles that the officer makes with the ground.
R
6
L
5
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
SC
and 䉭LSC Rib
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
SC The given states that ⬔DEG and ⬔DEF are
What conditions must hold for that to be true?
C
34
perpendicular
level or flat
to the ground, and the
.
Quick Check.
1. In Example 1, what can you say about ⬔5 and ⬔6? Explain.
They are congruent, because supplements of congruent angles are
congruent.
2. Recall Example 2. About how wide was the river if the officer paced off
20 paces and each pace was about 212 feet long?
50 feet
The congruence statements that remain to be proved are
lCLS O lCRS and CL O CR.
52
L1
Geometry Lesson 4-4
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 4-4
Geometry
53
13
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 4-5
Example.
Isosceles and Equilateral Triangles
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
Using the Isosceles Triangle Theorems Explain why 䉭ABC is
isosceles.
NAEP 2005 Strand: Geometry
Use and apply properties of isosceles
triangles
Topic: Relationships Among Geometric Figures
⬔ABC and ⬔XAB are
Local Standards: ____________________________________
alternate interior
X
A
angles formed
* )
←
→ ←
→
←
→
←
→
by XA , BC , and the transversal AB . Because XA 6 BC ,
⬔ABC ⬔
Vocabulary and Key Concepts.
XAB
.
B
C
The diagram shows that ⬔XAB ⬔ACB. By
Theorem 4-3: Isosceles Triangle Theorem
⬔ A ⬔ B
B
A
C
All rights reserved.
All rights reserved.
the angles opposite those sides are congruent.
.
to
AC .
By the definition of an isosceles triangle, 䉭ABC is isosceles.
a. In the figure, suppose m⬔ACB 55. Find m⬔CBA and m⬔BAC.
the sides opposite those sides are congruent.
conclude that AB >
ACB
Quick Check.
If two angles of a triangle are congruent, then
AC
the Converse of the Isosceles Triangle Theorem
You can use
If two sides of a triangle are congruent, then
Theorem 4-4: Converse of Isosceles Triangle Theorem
, ⬔ABC ⬔
the Transitive Property of Congruence
C
Theorem 4-5
m⬔CBA 55; m⬔BAC 70
B
A
BC
C
and
CD
A
bisects AB.
The legs of an isosceles triangle
D
The
are the congruent sides.
B
vertex angle
of an
isosceles triangle is formed by
Vertex
the two congruent sides (legs).
Angle
The base angles of an isosceles
The
of an
base
and the legs.
Base
non-congruent, side.
54
triangle are formed by the base
Legs
isosceles triangle is the third, or
Base Angles
Geometry Lesson 4-5
Daily Notetaking Guide
L1
b. In the figure, can you deduce that 䉭ABX is isosceles?
No; neither lAXB nor lXBA can be shown to be
congruent to lA.
Daily Notetaking Guide
L1
Name_____________________________________ Class____________________________ Date________________
All rights reserved.
AB
Geometry Lesson 4-5
Name_____________________________________ Class____________________________ Date ________________
2 Two-Column Proof—Using the HL Theorem
Lesson 4-6
Congruence in Right Triangles
Lesson Objective
1 Prove triangles congruent using the
HL Theorem
55
C
Given: ⬔ABC and ⬔DCB are right angles, AC > DB
Prove: 䉭ABC 䉭DCB
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
Statements
Reasons
1. ⬔ABC and ⬔DCB are
Local Standards: ____________________________________
D
E
B
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
CD '
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
of the base.
perpendicular bisector
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The bisector of the vertex angle of an isosceles triangle is the
A
1. Given
right angles
2. 䉭ABC and 䉭
Vocabulary and Key Concepts.
DCB
2. Definition of Right Triangle
are right triangles
All rights reserved.
the triangles are congruent.
All rights reserved.
3. AC >
Theorem 4-6: Hypotenuse-Leg (HL) Theorem
If the hypotenuse and a leg of one right triangle are congruent to the
hypotenuse and leg of another right triangle, then
3. Given
DB
BC O CB
4.
5. 䉭ABC 䉭
4. Reflexive Property of Congruence
DCB
5.
HL
The hypotenuse of a right triangle is its longest side, or the side opposite the right angle.
Quick Check.
The
of a right triangle are the two shortest sides, or the sides that are not
legs
1.
R
L
opposite the right angle.
3
S
O
3
Examples.
1 Proving Triangles Congruent One student wrote “䉭CPA 䉭MPA
A
by the HL Theorem” for the diagram. Is the student correct? Explain.
The diagram shows the following congruent parts.
CA >
MA
⬔CPA lMPA
PA PA
C
hypotenuse
Since AC is the
of right triangle CPA, and MA is the
leg
the
56
14
hypotenuse
M
leg
5
3
M
5
5
Q
P
N
T
Which two triangles are congruent by the HL Theorem? Write a correct
congruence statement.
䉭LMN 艑 䉭OQP
2. You know that two legs of one right triangle are congruent to two legs of
another right triangle. Explain how to prove the triangles are congruent.
Since all right angles are congruent, the triangles are congruent by SAS.
and PA is a
of right triangle MPA, the triangles are congruent by
HL Theorem
The student
and PA is a
P
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Legs
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Hypotenuse
is
.
correct.
Geometry Lesson 4-6
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 4-6
All-In-One Answers Version B
57
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 4-7
Lesson Objectives
1
Quick Check.
1. The diagram shows triangles from the scaffolding that workers used
when they repaired and cleaned the Statue of Liberty.
a. Name the common side in 䉭ACD and 䉭BCD.
NAEP 2005 Strand: Geometry
Identify congruent overlapping
triangles
Prove two triangles congruent by first
proving two other triangles congruent
2
Name_____________________________________ Class____________________________ Date ________________
Using Corresponding Parts
of Congruent Triangles
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
E
E
D
, respectively.
FG
2 Using Two Pairs of Triangles Write a paragraph proof.
X
All rights reserved.
and
HG
All rights reserved.
EG
These parts are
F
H
are shared by 䉭DFG and 䉭EHG.
Parts of sides DG and
Answers may vary. Sample: 䉭ABD and 䉭CBD; BD
G
䉭DFG and 䉭EHG share.
Identify the overlapping triangles.
2. Write a two-column proof.
Given: PS > RS, ⬔PSQ ⬔RSQ
Prove: 䉭QPT 䉭QRT
Plan: 䉭XPW 䉭YPZ by AAS if ⬔WXZ SAS
S
lZYW
. These angles
P
Proof: You are given XW > YZ. Because ⬔XWZ and ⬔YZW are
Reflexive Property of Congruence
䉭XPW 䉭YPZ by
. Therefore, 䉭XWZ 䉭YZW by
by CPCTC, and ⬔XPW ⬔
ZYW
vertical angles
because
W
, ⬔XWZ ⬔YZW. WZ > ZW by the
right angles
SAS. ⬔WXZ ⬔
Z
YPZ
are congruent. Therefore,
AAS
.
Geometry Lesson 4-7
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
congruent by
R
T
Statements
if 䉭XWZ 䉭YZW. These triangles are
CPCTC
are congruent by
Q
P
Y
Given: XW > YZ, ⬔XWZ and ⬔YZW are right angles.
Prove: 䉭XPW 䉭YPZ
All rights reserved.
D
b. Name another pair of triangles that share a common side. Name the
common side.
1 Identifying Common Parts Name the parts of the sides that
L1
L1
Name_____________________________________ Class____________________________ Date________________
Reasons
1. PS O RS, lPSQ O lRSQ
1. Given
2. QS > QS
2. Reflexive Property of Congruence
3. 䉭PSQ 䉭RSQ
3. SAS
4. PQ > RQ
4. CPCTC
5. ⬔PQT lRQT
5. CPCTC
6. QT > QT
6. Reflexive Property of Congruence
7. 䉭QPT 䉭QRT
7. SAS
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Midsegments of Triangles
Lesson Objective
1 Use properties of midsegments to
solve problems
59
Geometry Lesson 4-7
2 Identifying Parallel Segments Find m⬔AMN and m⬔ANM. MN and BC
Lesson 5-1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
F
B
CD
Examples.
58
A
are cut by transversal AB, so ⬔AMN and ⬔B are
NAEP 2005 Strand: Geometry
Topics: Relationships Among Geometric Figures
angles. MN6BC by the
Triangle Midsegment
⬔AMN ⬔B by the
Corresponding Angles
Local Standards: ____________________________________
m⬔AMN 75
Theorem, so
Postulate.
N
because congruent angles have the same measure.
In 䉭AMN, AM AN , so m⬔ANM Vocabulary and Key Concepts.
A
corresponding
Isosceles Triangle
m⬔AMN
M
by the
Theorem. m⬔ANM 75
C
75
B
by substitution.
parallel
is
half
to the third side, and is
All rights reserved.
If a segment joins the midpoints of two sides of a triangle, then the segment
its length.
A midsegment of a triangle is a segment connecting the midpoints of two sides.
A
coordinate proof
All rights reserved.
Theorem 5-1: Triangle Midsegment Theorem
Quick Check.
1. AB 10 and CD 18. Find EB, BC, and AC.
A
E
B
D
is a form of proof in which coordinate geometry and algebra are
used to prove a theorem.
C
EB ⫽ 9; BC ⫽ 10; AC ⫽ 20
Examples.
1 Finding Lengths In 䉭XYZ, M, N, and P are midpoints.
NP MN MP 60
M
22
P
Definition of perimeter
24
NP 24
NP 22
46
60
60
NP Substitute
22
24
for MN and
Y
for MP.
N
Z
Simplify.
14
46
Subtract
from each side.
Use the Triangle Midsegment Theorem to find YZ.
MP 60
L1
1
YZ Triangle Midsegment Theorem
2
1
22
44
YZ
2
YZ
Substitute
22
Multiply each side by
Geometry Lesson 5-1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Because the perimeter of 䉭MNP is 60, you can find NP.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The perimeter of 䉭MNP is 60. Find NP and YZ.
2. Critical Thinking Find m⬔VUZ. Justify your answer.
X
65
Y
U
V
Z
65; UV6XY so ⬔VUZ and ⬔YXZ are corresponding and congruent.
for MP.
2
.
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 5-1
Geometry
61
15
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 5-2
Example.
Bisectors in Triangles
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
Using the Angle Bisector Theorem
Find x, FB, and FD in the diagram at right.
NAEP 2005 Strand: Geometry
Use properties of perpendicular
bisectors and angle bisectors
Topics: Relationships Among Geometric Figures
FD
Local Standards: ____________________________________
FB
7x 37 2x 5
Vocabulary and Key Concepts.
7x 2x equidistant from the endpoints of the segment.
Theorem 5-3: Converse of the Perpendicular Bisector Theorem
If a point is equidistant from the endpoints of a segment, then it is
x
All rights reserved.
All rights reserved.
If a point is on the perpendicular bisector of a segment, then it is
on the perpendicular bisector of the segment.
Substitute.
42
D
2x 5
A
Add 37 to each side.
42
5x
Theorem 5-2: Perpendicular Bisector Theorem
C
B
Angle Bisector Theorem
8.4
Substitute.
FD 7( 8.4 ) 37 21.8
Substitute.
Quick Check.
)
)
D
10; 10
2x
If a point is on the bisector of an angle, then it is
K
angle, then it is on the angle bisector.
The distance from a point to a line is the length of the perpendicular
segment from the point to the line.
* )
D is 3 in. from AB and
* )
AC .
C
0
B
)
(x 20)
H
b. What can you conclude about EK ?
)
EK is the angle bisector of lDEH.
All rights reserved.
If a point in the interior of an angle is equidistant from the sides of the
E
C
10
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
equidistant from the sides of the angle.
Theorem 5-5: Converse of the Angle Bisector Theorem
1
2
3
4
5
6
D
E
Divide each side by 5.
a. According to the diagram, how far is K from EH ? From ED ?
A
F
Subtract 2x from each side.
FB 2( 8.4 ) 5 21.8
Theorem 5-4: Angle Bisector Theorem
7x 37
c. Find the value of x.
20
d. Find m⬔DEH.
80
Geometry Lesson 5-2
Daily Notetaking Guide
L1
Name_____________________________________ Class____________________________ Date________________
Lesson Objective
1 Identify properties of
perpendicular bisectors and
angle bisectors
2 Identify properties of medians
and altitudes of a triangle
2 Finding Lengths of Medians M is the centroid of 䉭WOR, and WM 16.
W
Find WX.
NAEP 2005 Strand: Geometry
Topics: Relationships Among Geometric Figures
The
is the point of concurrency of the medians of
centroid
Z
Y
a triangle.
M
Local Standards: ____________________________________
The medians of a triangle are concurrent at a point that is
O
R
X
the distance from each vertex to the midpoint
two-thirds
of the opposite side. (Theorem 5-8)
Vocabulary.
of 䉭WOR, WM centroid
Because M is the
is the point at which concurrent lines intersect.
point of concurrency
A circle is circumscribed about a polygon when the vertices
Circumscribed
of the polygon are on the circle.
circumcenter
S
of a triangle is the point of
concurrency of the perpendicular bisectors of a triangle.
Median
A median of a triangle is a segment that
circle
QC SC RC
All rights reserved.
The
All rights reserved.
Concurrent lines are three or more lines that meet in one point.
The
63
Geometry Lesson 5-2
Name_____________________________________ Class____________________________ Date ________________
Concurrent Lines,
Medians, and Altitudes
Lesson 5-3
Daily Notetaking Guide
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
62
circumcenter
Examples.
1 Finding the Circumcenter Find the center of the circle that
(1, 1)
and Y has
, XY lies on the vertical line x 1 .
The perpendicular bisector of XY is the horizontal line that
7
passes through (1, 1 2 ) or
(1, 4)
, so the
equation of the perpendicular bisector of XY is y 4 .
(1, 1)
Because X has coordinates
coordinates
(5, 1)
and Z has
y
8Y
6
(3, 4)
4
2
X
O
Z
2
4
6
x
, XZ lies on the horizontal line
y 1 . The perpendicular bisector of XZ is the vertical line that
passes through
(1 2 5, 1)
or
(3, 1)
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
side.
(1, 7)
Theorem
WX
Substitute
2WX
3
.
5-8
2
16
24
WX
16
for WM.
3
Multiply each side by
3
2
.
y
Quick Check.
triangle and the midpoint of the opposite
coordinates
WX
R
Q
circumscribes 䉭XYZ.
2
3
C
has as its endpoints a vertex of the
Because X has coordinates
WM 1. a. Find the center of the circle that you can
circumscribe about the triangle with
vertices (0, 0), (8, 0), and (0, 6).
5
(⫺4, 3)
5
5x
5
b. Critical Thinking In Example 1, explain why it is not necessary to find
the third perpendicular bisector.
Theorem 5-6: All the perpendicular bisectors of the sides of a
triangle are concurrent.
2. Using the diagram in Example 2, find MX. Check that WM MX WX.
, so the equation
of the perpendicular bisector of XZ is x 3 .
8
Draw the lines y 4 and x 3 . They intersect at the point
(3, 4)
. This point is the center of the circle that
circumscribes 䉭XYZ.
64
16
Geometry Lesson 5-3
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 5-3
All-In-One Answers Version B
65
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date ________________
Lesson 5-4
Lesson Objectives
1
2 Writing the Inverse and Contrapositive Write the inverse and contrapositive
of the conditional statement “If 䉭ABC is equilateral, then it is isosceles.”
To write the inverse of a conditional, negate both the hypothesis and the
conclusion.
NAEP 2005 Strand: Geometry
Topics: Mathematical Reasoning
Write the negation of a statement
and the inverse and contrapositive
of a conditional statement
2
Name_____________________________________ Class____________________________ Date ________________
Inverses, Contrapositives,
and Indirect Reasoning
Hypothesis
Inverse:
If an angle is a straight angle,
then its measure is 180.
Symbolic Form
You Read It
pSq
p,
If
q.
then
Negation (of p)
An angle is not a straight angle.
p
Not
Inverse
If an angle is not a straight angle,
then its measure is not 180.
p S q
If not
p, then not q.
Contrapositive
If an angle’s measure is not 180,
then it is not a straight angle.
q S p
If not
q, then not p.
The contrapositive of a conditional switches the hypothesis and the conclusion and negates
both.
contrapositive
always have the same truth value.
Equivalent statements have the same truth value.
Examples.
1 Writing the Negation of a Statement Write the negation of “ABCD is
not a convex polygon.”
opposite
The negation of a statement has the
.
Hypothesis
Conclusion
T
T
Switch and negate both.
, then it is
not isosceles
.
not equilateral
Quick Check.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
conclusion.
and its
T
not isosceles
then it is
If 䉭ABC is equilateral, then it is isosceles.
Conditional:
Contrapositive: If 䉭ABC is
of a conditional statement negates both the hypothesis and the
conditional
,
p.
The negation of a statement has the opposite truth value from that of the original statement.
inverse
not equilateral
To write the contrapositive of a conditional, switch the hypothesis and
conclusion, then negate both.
All rights reserved.
Example
Conditional
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Statement
All rights reserved.
Negation, Inverse, and Contrapositive Statements
If 䉭ABC is
then it is isosceles.
Negate both.
T
Vocabulary and Key Concepts.
A
T
If 䉭ABC is equilateral,
Conditional:
Use indirect reasoning
The
Conclusion
T
Local Standards: ____________________________________
1. Write the negation of each statement.
a. m⬔XYZ 70.
The measure of ⬔XYZ is not more than 70.
b. Today is not Tuesday.
Today is Tuesday.
2. Write (a) the inverse and (b) the contrapositive of Maya Angelou’s
statement “If you don’t stand for something, you’ll fall for anything.”
a.
b.
If you stand for something, you won’t fall for anything.
If you won’t fall for anything, then you stand for something.
truth value. The
negation of is not in the original statement removes the word
not
.
The negation of “ABCD is not a convex polygon” is
“ABCD is a convex polygon.”
66
Geometry Lesson 5-4
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Geometry Lesson 5-4
67
Name_____________________________________ Class____________________________ Date ________________
2 Using the Triangle Inequality Theorem Can a triangle have sides with the given lengths?
Lesson 5-5
Inequalities in Triangles
Lesson Objectives
1 Use inequalities involving angles of
triangles
2 Use inequalities involving sides of
triangles
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Daily Notetaking Guide
Explain.
a. 2 cm, 2 cm, 4 cm
NAEP 2005 Strand: Geometry
Topics: Relationships Among Geometric Figures
Triangle Inequality
According to the
Theorem, the
sum of the lengths of any two sides of a triangle is
Local Standards: ____________________________________
greater
than the length of the third side.
22 ⬐ 4
Key Concepts.
The sum of 2 and 2
No , a triangle
cannot
is not
greater than 4.
have sides with these lengths.
b. 8 in., 15 in., 12 in.
m⬔1
m⬔2 and
3
angle of a triangle is greater
exterior
than the measure of each of its remote
angles.
interior
2
m⬔3
m⬔1
1
All rights reserved.
The measure of an
All rights reserved.
Corollary to the Triangle Exterior Angle Theorem
8 15 ⬎ 12
Yes , a triangle
8 12 ⬎ 15
can
15 12 ⬎ 8
have sides with these lengths.
Quick Check.
1. List the angles of 䉭ABC in order from smallest to largest.
Theorem 5-10
If two sides of a triangle are not congruent, then
A
⬔A, ⬔C, ⬔B
Y
27 ft
C
the larger angle lies opposite the longer side.
X
If XZ XY, then m⬔Y ⬎ m⬔Z.
21 ft
Z
18 ft
greater than the length of the third side.
LM MN LN
MN LN LM
LN LM MN
M
N
Examples.
1 Applying Theorem 5-10 In 䉭RGY, RG 14, GY 12, and RY 20.
G
List the angles from largest to smallest.
No two sides of 䉭RGY are congruent, so the largest angle lies
opposite the longest side.
Find the angle opposite each side.
The longest side is
20
12
Y
20
. The opposite angle, ⬔G ,
is the largest. The shortest side is
12
14
R
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
L
The sum of the lengths of any two sides of a triangle is
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 5-12: Triangle Inequality Theorem
B
2. Can a triangle have sides with the given lengths? Explain.
a. 2 m, 7 m, and 9 m
no; 2 7 is not greater than 9
b. 4 yd, 6 yd, and 9 yd
yes; 4 6 ⬎ 9, 6 9 ⬎ 4, and 4 9 ⬎ 6
. The opposite angle, ⬔R , is the
smallest. From largest to smallest, the angles are ⬔G , ⬔Y , ⬔R .
68
L1
Geometry Lesson 5-5
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 5-5
Geometry
69
17
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 6-1
Example.
Classifying Quadrilaterals
Classifying by Coordinate Methods Determine the most precise name for
the quadrilateral with vertices Q(4, 4), B(2, 9), H(8, 9), and A(10, 4).
Graph quadrilateral QBHA.
NAEP 2005 Strand: Geometry
Define and classify special types of
quadrilaterals
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
Find the slope of each side.
94
2 (4)
99
Slope of BH 8 2
y
B
Slope of QB H
Key Concepts.
8
Special Quadrilaterals
Parallelogram
Rhombus
Isosceles
Trapezoid
All rights reserved.
All rights reserved.
Rectangle
Q
Trapezoid
Kite
(
6
Quadrilateral
Quadrilateral
of
irs es
2 pairs of
pa
1 pair of
id
No llel s
parallel
sides parallel sides
ra
pa
Square
A
Slope of HA 2
4
6
8
equal
BH is parallel to QA because their slopes are
square
A
is a parallelogram with
congruent sides ___
four__________________
______________ and four
right
angles
___
___________________
.
rectangle
A
is a parallelogram with
right angles
four______________________
__________________________.
isosceles trapezoid
An _____________________
________________________
is a trapezoid whose
nonparallel sides are
congruent.
Geometry Lesson 6-1
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
kite
A
is a
quadrilateral with two pairs
adjacent sides
of ______________________
opposite
congruent and no ________
sides
________________________
congruent.
trapezoid
A
is a quadrilateral with
exactly one pair of
________________________
parallel sides
________________________
________________________ .
L1
not equal
"29
"29
(2 (4))2 (9 4 )2 4 25
(10 8)2 ( 4 9 )2 4 25
BH (8 (2))2 ( 9 9 )2 100 0
2
2
QA (4 10 ) ( 4 4 ) 196 0
Because QB 0
HA
10
14
an isosceles trapezoid
, QBHA is
.
Quick Check.
y
a. Graph quadrilateral ABCD with vertices A(3, 3), B(2, 4),
C(3, 1), and D(2, 2).
b. Classify ABCD in as many ways as possible.
B
A
x
ABCD is a quadrilateral because it has four sides. It is a
parallelogram because both pairs of opposite sides are
parallel. It is a rhombus because all four sides are
congruent. It is a rectangle because it has four right angles
and its opposite sides are congruent. It is a square because
it has four right angles and its sides are all congruent.
C
D
Daily Notetaking Guide
2x 5
L
x
A
7y 16
5y
Local Standards: ____________________________________
M
N
x 7y 16
A
Opposite angles of a parallelogram are
congruent
.
C
D
R
14y
14y
32
5 5y
27
bisect
U
each other.
2x 120
Add x to each side.
60
Divide each side by 2.
15 75
Substitute
60
75
180
105
D
C
for x.
Consecutive angles of a parallelogram are
supplementary
m⬔A (135 x)
Subtract 15 from each side.
x
m⬔A m⬔B 180
(x 15)
.
Substitute
Subtract
75
75
.
for m⬔B.
from each side.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
135
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
congruent
B
A
Subtract
14y
Divide each side by
from each side.
9 .
x7 3
Substitute 3 for y in the first equation
to solve for x.
x 5
Simplify.
( ) 16
T
Opposite angles of a
parallelogram are
Simplify.
9y
3 y
Examples.
1 Using Algebra Find the value of x in ⵥABCD. Then find m⬔A.
15 each other.
Distribute.
5y
27 S
Theorem 6-3
x 15 135 x
bisect
Substitute 7y 16 for x in
the second equation to solve for y.
2(7y 16) 5 5y
All rights reserved.
Theorem 6-2
All rights reserved.
B
sides of a parallelogram are congruent.
The diagonals of a parallelogram
The diagonals of a parallelogram
2x 5 5y
Theorem 6-1
Opposite
So x 5 and y 3 .
Quick Check.
1. Find the value of y in ⵥEFGH. Then find m⬔E, m⬔F, m⬔G, and
m⬔H.
F
y 5 11; mlE 5 70, mlF 5 110, mlG 5 70, and mlH 5 110
(6y 4)
E
2. Find the values of a and b.
(3y 37)
X
a
a ⫽ 16, b ⫽ 14
b
0
1
18
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
G
H
Y
8
2a b
2
W
Geometry Lesson 6-2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Vocabulary and Key Concepts.
72
71
Geometry Lesson 6-1
Name_____________________________________ Class____________________________ Date ________________
K
m⬔A 25
2
.
HA Properties of Parallelograms
Lesson Objectives
1 Use relationships among sides and
among angles of parallelograms
2 Use relationships involving diagonals
of parallelograms and transversals
60
2 Using Algebra Find the values of x and y in ⵥKLMN.
Lesson 6-2
m⬔B 10
0
.
QB L1
Name_____________________________________ Class____________________________ Date________________
2x
4 5
2
trapezoid
One pair of opposite sides is parallel, so QBHA is a
.
Next, use the distance formula to see whether any pairs of sides are congruent.
parallelogram
A
is a quadrilateral with
both pairs of opposite
parallel
sides ___________________
________________________ .
70
4 4
Slope of QA x
HA is not parallel to QB because their slopes are
rhombus
A
is a parallelogram with
congruent sides
four_____________________
______________________.
10 8
2
4 2 O
)
4 9
All rights reserved.
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
Z
Geometry Lesson 6-2
All-In-One Answers Version B
73
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 6-3
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
Proving That a Quadrilateral
Is a Parallelogram
2 Is the Quadrilateral a Parallelogram? Can you prove the quadrilateral is
a parallelogram from what is given? Explain.
NAEP 2005 Strand: Geometry
Determine whether a quadrilateral is
a parallelogram
Given: m⬔E m⬔G 75°, m⬔F 105°
Prove: EFGH is a parallelogram.
Topic: Geometry
Local Standards: ____________________________________
(
the quadrilateral is a parallelogram.
of a quadrilateral are congruent,
then the quadrilateral is a parallelogram.
bisect
congruent
and
parallel
All rights reserved.
Examples.
A
B
8x 12
for which ABCD must be a parallelogram.
y9
2y 80
G
10x 24
C
D
Diagonals of parallelograms
bisect each other.
10x 24 8x 12
2x
If x 74
24 12
Collect the variables on one side.
2x 36
Solve.
x
18
and y 18
2y 80
y9
1. Find the values of a and c for which PQRS must be a
parallelogram.
Given: PQ > SR, PQ 6SR
Prove: PQRS is a parallelogram.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Q
Yes; a pair of opposite sides are parallel
and congruent (Theorem 6-8).
R
x
P
L1
S
Daily Notetaking Guide
L1
Geometry Lesson 6-3
Examples.
Special Parallelograms
1 Finding Angle Measures Find the measures of the
B
numbered angles in the rhombus.
Theorem 6-9 states that each diagonal of a rhombus
bisects two angles of the rhombus, so m⬔1 78.
Local Standards: ____________________________________
78
3
the diagonals of the rhombus are perpendicular
so m⬔2 90
. Because the four angles formed by the diagonals
A
Theorem 6-10
B
perpendicular
BD
A
Rectangles
Theorem 6-11
The diagonals of a rectangle are
congruent
the
complementary
.
. Finally, because BC DC,
78 .
2 Finding Diagonal Length One diagonal of a rectangle has length 8x 2.
The other diagonal has length 5x 11. Find the length of each diagonal.
C
By Theorem 6-11, the diagonals of a rectangle are
5x 11
D
A
D
B
C
.
BD
12
allows you to conclude that ⬔ 1 ⬔4.
Isosceles Triangle Theorem
So m⬔4 .
Parallelograms
Theorem 6-12
If one diagonal of a parallelogram bisects two angles of the parallelogram,
the parallelogram is a rhombus.
Theorem 6-13
If the diagonals of a parallelogram are perpendicular, then
the parallelogram is a rhombus.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The diagonals of a rhombus are
D
Because m⬔ABD 78, m⬔3 90 78 All rights reserved.
, so ⬔3 ⬔4
4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
BCD
1
2
All rights reserved.
Each diagonal of a rhombus bisects two angles of the rhombus.
AC bisects ⬔BAD, so ⬔ 1 ⬔ 2
C
4
D
,
all must have measure 90, ⬔3 and ⬔ABD must be
3
C
1
2
A
Theorem 6-10 states that
B
75
Name_____________________________________ Class____________________________ Date ________________
NAEP 2005 Strand: Geometry
Topic: Geometry
Rhombuses
Theorem 6-9
c1
S
2. Can you prove the quadrilateral is a parallelogram? Explain.
Key Concepts.
AC
3c 3
89
Daily Notetaking Guide
Lesson Objectives
1 Use properties of diagonals of
rhombuses and rectangles
2 Determine whether a parallelogram is
a rhombus or a rectangle
'
a
x
Lesson 6-4
AC
R
(a 40)
P
Name_____________________________________ Class____________________________ Date________________
bisects ⬔
Q
a 70, c 2
89 , then ABCD is a parallelogram.
Geometry Lesson 6-3
AC
.
Theorem 6-6
y 80 9
y
H
Quick Check.
, then the quadrilateral is a parallelogram.
If the diagonals of quadrilateral ABCD bisect each other,
Theorem 6-7
then ABCD is a parallelogram by
.
Write and solve two equations to find values of x and y for
which the diagonals bisect each other.
75
E
Because both pairs of opposite angles are congruent, the quadrilateral
is a parallelogram by
1 Finding Values for Parallelograms Find values for x and y
G
congruent, then the quadrilateral is a parallelogram.
each other,
Theorem 6-8
If one pair of opposite sides of a quadrilateral is both
75
Theorem 6-6 states If both pairs of opposite angles of a quadrilateral are
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 6-7
If the diagonals of a quadrilateral
then the quadrilateral is a parallelogram.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
opposite angles
105
)
If x represents the measure of the unmarked angle,
360
105
x 75 105 75 , so x =
.
All rights reserved.
All rights reserved.
Theorem 6-5
If both pairs of opposite sides of a quadrilateral are congruent, then
Theorem 6-6
If both pairs of
F
The sum of the measures of the angles of a polygon is
(n 2)180 where n represents the number of sides,
so the sum of the measures of the angles of a
360
quadrilateral is 4 2 180 .
Key Concepts.
8x 2
11 3x
9 3x
2
Subtract
.
5x
from each side.
Subtract 2 from each side.
3 x
( )2
5x 11 5( 3 ) 11 8x 2 8 3
congruent
Diagonals of a rectangle are congruent.
Divide each side by 3 .
26
Substitute.
26
The length of each diagonal is
Substitute.
26
.
Quick Check.
1. Find the measures of the numbered angles in the rhombus.
50
1
m⬔1 90, m⬔2 50, m⬔3 50, m⬔4 40
4
2. Find the length of the diagonals of rectangle GFED if FD 5y 9
and GE = y 5.
2
3
F
E
G
D
81
2
Theorem 6-14
If the diagonals of a parallelogram are congruent, then
the parallelogram is a rectangle.
76
L1
Geometry Lesson 6-4
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 6-4
Geometry
77
19
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
2 Finding Angle Measures in Kites Find m⬔1, m⬔2, and m⬔3 in the kite.
Lesson 6-5
Trapezoids and Kites
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
S
1
NAEP 2005 Strand: Geometry
Verify and use properties of
trapezoids and kites
Topic: Relationships Among Geometric Figures
D
R
3
Local Standards: ____________________________________
U
Vocabulary and Key Concepts.
m⬔2 90
RU .
Theorem 6-16
diagonals
of an isosceles trapezoid are congruent.
m⬔1 All rights reserved.
congruent
The base angles of an isosceles trapezoid are
All rights reserved.
Trapezoids
Theorem 6-15
The
T
2
72
Diagonals of a kite are
Definition of a kite
72
Isosceles Triangle Theorem
m⬔3 m⬔RDU 72 180
m⬔RDU m⬔3 m⬔3 90
72 180
90
162
180
m⬔3 The base angles of a trapezoid are two angles that share a base of the trapezoid.
perpendicular
RS
18
.
Triangle Angle-Sum Theorem
Diagonals of a kite are perpendicular.
Substitute.
Simplify.
Subtract
162
from each side.
Quick Check.
angles
angles
1. In the isosceles trapezoid, m⬔S 70. Find m⬔P, m⬔Q, and m⬔R.
Leg
perpendicular
.
Examples.
1 Finding Angle Measures in Trapezoids WXYZ is
X
156
an isosceles trapezoid, and m⬔X 156. Find m⬔Y,
m⬔Z, and m⬔W.
Y
Z
W
Two angles that share
a leg of a trapezoid
180
are
156
m⬔W 180
m⬔W .
supplementary
S
70
R
2. Find m⬔1, m⬔2, and m⬔3 in the kite.
m⬔1 90, m⬔2 46, and m⬔3 44
46
2
1
3
Substitute.
24
156
Subtract
from each side.
Because the base angles of an isosceles trapezoid are
m⬔Y m⬔X 156 and m⬔Z 78
Q
m⬔W
24
,
congruent
.
Geometry Lesson 6-5
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson Objective
1 Name coordinates of special figures
by using their properties
79
Geometry Lesson 6-5
Name_____________________________________ Class____________________________ Date________________
Placing Figures in
the Coordinate Plane
Lesson 6-6
Daily Notetaking Guide
Example.
2 Naming Coordinates Use the properties of parallelogram
NAEP 2005 Strand: Geometry
Topic: Position and Direction
y
OCBA to find the missing coordinates. Do not use any new
variables.
(0, 0)
The vertex O is the origin with coordinates O
.
Local Standards: ____________________________________
(p, q)
A
B
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
m⬔X m⬔W © Pearson Education, Inc., publishing as Pearson Prentice Hall.
Kites
Theorem 6-17
The diagonals of a kite are
P
110, 110, 70
All rights reserved.
Base
Base
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Leg
Because point A is p units to the left of point O, point B
Example.
is also p units to the left of point C , because OCBA is a
If both pairs of opposite sides of a quadrilateral are
congruent
T(a, b)
x
.
Theorem 6-6
You can prove that TWVU is a parallelogram by showing that
TW VU and WV U(e, 0)
O
E
TU . Use the distance formula.
(
) (
)
) (
) 2
2
ce
e) ( d 0 ) c2 d2
2
2
ce ) ( bd d ) ac
2
2
a e ) ( b 0 ) (a e) 2 b 2
2
bd
Because TW VU and WV TU, TWVU is a
b
2
c2
parallelogram
Geometry
O
x
(0, 0)
and B
(p x, q)
.
y R(b, c)
Q(?, ?)
Q(s b, c)
O
(a e) 2 b 2
.
Midpoint of TV 5 aa 1 c 1 e , b 1 d b midpoint of UW . Thus, the
2
2
diagonals bisect each other, and TWVU is a parallelogram.
20
The missing coordinates are O
2. Use the properties of parallelogram OPQR to find the missing
coordinates. Do not use any new variables.
1. Use the diagram above. Use a different method: Show that TWVU is a
parallelogram by finding the midpoints of the diagonals.
Geometry Lesson 6-6
C (x, 0)
. So point B has the same second
coordinate, q , as point A.
d2
Quick Check.
80
horizontal
Daily Notetaking Guide
L1
P(s, 0)
x
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
a
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
(
VU (
WV (
TU (
ac
.
Quick Check.
( )
Use the coordinates T (a, b), W a c , b d , V c e , d , and U e, 0 .
TW p x
Because AB 6 CO and CO is horizontal, AB is also
V(c e, d)
, then the quadrilateral is a
parallelogram by
parallelogram. So the first coordinate of point B is
W(a c, b d)
C
All rights reserved.
A
y
proving pairs of opposite sides congruent.
All rights reserved.
1 Proving Congruency Show that TWVU is a parallelogram by
L1
Daily Notetaking Guide
Geometry Lesson 6-6
All-In-One Answers Version B
81
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 6-7
If the diagonals of a parallelogram are
Proofs Using Coordinate Geometry
Lesson Objective
1
parallelogram is a
NAEP 2005 Strand: Geometry
Prove theorems using figures in the
coordinate plane
Topics: Position and Direction; Mathematical Reasoning
To show that XYZW is a rectangle, find the lengths of its diagonals, and then
Local Standards: ____________________________________
compare them to show that they are
XZ Vocabulary and Key Concepts.
Theorem 6-18: Trapezoid Midsegment Theorem
All rights reserved.
to the bases.
(2) The length of the midsegment of a trapezoid is half the sum of the
.
All rights reserved.
parallel
(1) The midsegment of a trapezoid is
lengths of the bases
The midsegment of a trapezoid is the segment that joins the midpoints of the nonparallel
R
RA )
All rights reserved.
y
D(0, 2b)
W(a, b)
A(2a, 0)
x
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
TP
Example.
C(2a, 0)
.
))
2
⫺2a 2 4a2 4b2
(a a )2 (b (
( ⫺2a )2 ( ⫺2b )2
4a2 4b2
YW b
))2
rectangle
.
Quick Check.
P
1
RA , and MN 2 (
Using Coordinate Geometry Use coordinate geometry to prove that the
quadrilateral formed by connecting the midpoints of rhombus ABCD is
a rectangle.
Draw quadrilateral XYZW by connecting the midpoints of ABCD.
Z(a, b)
)
b ⫺b
N
T
TP , MN 6
equal
) ( (
( 2b )2
2
Because the diagonals are congruent, parallelogram XYZW is a
A
M
(
(
a a
XZ ⫽ YW
opposite sides of the trapezoid.
MN 6
, then the
congruent
from Theorem 6–14.
rectangle
Use the diagram from the Example on page 82. Explain why the proof using
A(2a, 0), B(0, 2b), C(2a, 0), and D(0, 2b) is easier than the proof using
A(a, 0), B(0, b), C(a, 0), and D(0, b).
Using multiples of 2 in the coordinates for A, B, C, and D eliminates
the use of fractions when finding midpoints, since finding midpoints
requires division by 2.
X(a, b)
Y(a, b)
B(0, 2b)
From Lesson 6-6, you know that XYZW is a parallelogram.
82
Geometry Lesson 6-7
Daily Notetaking Guide
L1
Name_____________________________________ Class____________________________ Date________________
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
83
Name_____________________________________ Class____________________________ Date ________________
Ratios and Proportions
Lesson Objective
1 Write ratios and solve proportions
a.
NAEP 2005 Strand: Geometry
Topic: Position and Direction
Properties of Proportions
b
d
b
(4) a 1
b c1d
d
All rights reserved.
(3) ac All rights reserved.
d
c
A proportion is a statement that two ratios are equal.
)
Cross-Product Property
Simplify.
n5
14
Divide each side by 5 .
(
2x
x 1 1
1 2
x5 2
)
Cross-Product Property
Distributive Property
Subtract
2x
from each side.
Quick Check.
1. A photo that is 8 in. wide and 531 in. high is enlarged to a poster that is 2 ft
c
a
b d and a : b c : d are examples of proportions.
extended proportion
35
70
3x 5
c
a
b d is equivalent to
(2) ba (
5n 5
1
x
b. x 1
3 52
3x 5 2
Vocabulary and Key Concepts.
bc
25 n
5
35
5n 5 2
Local Standards: ____________________________________
An
Geometry Lesson 6-7
2 Solving for a Variable Solve each proportion.
Lesson 7-1
(1) ad Daily Notetaking Guide
L1
wide and 113 ft high. What is the ratio of the height of the photo to the height
is a statement that three or more ratios are equal.
of the poster?
6
1
4
24 16 4 is an example of an extended proportion.
The Cross-Product Property states that the product of the extremes of a proportion is equal
1:3
a
c
b d
a d b c
extremes
A
scale drawing
is a drawing in which all lengths are proportional to
corresponding actual lengths.
A scale is the ratio of any length in a scale drawing to the corresponding actual length.
The lengths may be in different units.
Examples.
1 Finding Ratios A scale model of a car is 4 in. long. The actual car is 15 ft
long. What is the ratio of the length of the model to the length of the car?
Write both measurements in the same units.
180
15 ft 15 12 in. in.
length of model
4 in.
4 in.
4
length of car 5 15 ft 5 180 in. 5 180 5
L1
Geometry Lesson 7-1
2. Solve each proportion.
a. 5z 5 20
3
z 0.75
6
b. n 18
165n
n3
1
45
The ratio of the length of the scale model to the length of the car is
84
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
means
a:b c:d
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
to the product of the means.
1
:
45 .
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 7-1
Geometry
85
21
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date ________________
Lesson 7-2
2 Using Similar Figures If 䉭ABC 䉭YXZ, find the value of x.
NAEP 2005 Strand: Geometry and Measurement
Identify similar polygons
Apply similar polygons
Because 䉭ABC 䉭YXZ, you can write and solve
a proportion.
Topics: Transformation of Shapes and Preservation of
Properties; Measuring Physical Attributes
Local Standards: ____________________________________
AC 5
YZ
BC
Corresponding sides are
XZ
proportional
40
Vocabulary.
x
Similar figures have the same shape but not necessarily the same size. Two polygons are
All rights reserved.
is ⬃.
similarity
The mathematical symbol for
The similarity ratio is the ratio of the lengths of corresponding sides of similar figures.
A
All rights reserved.
50
similar if corresponding angles are congruent and corresponding sides are proportional.
is a rectangle that can be divided into a square and a
golden rectangle
x The golden ratio is the ratio of the length to the width of any golden rectangle, about
Z
40
12
Substitute.
50
Solve for x.
15
Simplify.
B
Complete each statement.
a. m⬔B Y
A
⬔B ⬔ Y and m⬔Y 78, so m⬔B 78
because congruent angles have the same measure.
C
X
BC 5
b. YZ
XZ
78
42
Z
BC AC .
corresponds to XZ, so YZ
XZ
Quick Check.
1. Refer to the diagram for Example 1. Complete:
42
BC and YZ
All rights reserved.
Example.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
9
1 Understanding Similarity 䉭ABC 䉭XYZ.
86
12
2. Refer to the diagram for Example 2. Find AB.
1.618 : 1.
m⬔A .
50
X
C
Quick Check.
rectangle that is similar to the original rectangle.
AC
B
30
x
12
40
x5
Y
A
AB
XY
Geometry Lesson 7-2
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date ________________
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
W
2 Using Similarity Theorems Explain why the triangles
Lesson 7-3
Proving Triangles Similar
Lesson Objectives
1 Use AA, SAS, and SSS similarity
statements
2 Apply AA, SAS, and SSS similarity
statements
87
Geometry Lesson 7-2
must be similar. Write a similarity statement.
24
⬔YVZ > ⬔WVX because they are vertical angles.
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
VY
12
VW 5 24 1
2
VZ 5
and VX
18
36
Z
18
1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
Example.
Similar Polygons
Lesson Objectives
2
Name_____________________________________ Class____________________________ Date________________
V
,
2
12
Local Standards: ____________________________________
36
Y
so corresponding sides are proportional.
X
Vocabulary and Key Concepts.
䉭TRS 䉭
R
PLM
P
T
S
L
M
All rights reserved.
Postulate 7-1: Angle-Angle Similarity (AA⬃) Postulate
If two angles of one triangle are congruent to
two angles of another triangle, then the
triangles are similar.
All rights reserved.
Therefore, 䉭YVZ 䉭WVX
Side-Angle-Side Similarity (SAS)
by the
Theorem.
Quick Check.
1. In Example 1, you have enough information to write a similarity statement.
Do you have enough information to find the similarity ratio? Explain.
No; none of the side lengths are given.
Theorem 7-1: Side-Angle-Side Similarity (SAS⬃) Theorem
If an angle of one triangle is congruent to an angle of a second triangle,
and the sides including the two angles are proportional, then
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the triangles are similar.
Indirect measurement is a way of measuring things that are difficult to measure directly.
Examples.
1 Using the AA⬃ Postulate MX ' AB. Explain why the triangles
M
are similar. Write a similarity statement.
Because MX ' AB, ⬔AXM and ⬔BXK are
right angles
, so ⬔AXM > ⬔
BXK
K
.
⬔A > ⬔ B because their measures are equal.
䉭AMX 䉭BKX
Angle-Angle Similarity (AA)
by the
88
22
Geometry Lesson 7-3
Geometry
Postulate.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the triangles are similar.
Theorem 7-2: Side-Side-Side Similarity (SSS⬃) Theorem
If the corresponding sides of two triangles are proportional, then
2. Explain why the triangles at the right must be similar.
Write a similarity statement.
G
8
AC 5 CB 5 AB 5 3 , so the triangles are similar by the SSS~ Theorem;
EF
4
EG
GF
nABC , nEFG
8
12
E
F
9
A
6
B
6
C
58
A
58
X
Daily Notetaking Guide
B
L1
L1
Daily Notetaking Guide
Geometry Lesson 7-3
All-In-One Answers Version B
89
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date ________________
Name_____________________________________ Class____________________________ Date________________
Lesson 7-4
2 Finding Distance At a golf course, Maria drove her ball 192 yd straight
Similarity in Right Triangles
Lesson Objective
1
Find and use relationships in similar
right triangles
Topic: Transformation of Shapes and Preservation of
Properties
x 192
240
Local Standards: ____________________________________
192x The altitude to the hypotenuse of a right triangle divides the triangle into
Write a proportion.
similar
to the original triangle and
to each other.
240
x
M
192
Cross-Product Property
57,600
36,864
G
192
192 x All rights reserved.
All rights reserved.
Theorem 7-3
similar
240
192 (x 1 192) 5 2402
Vocabulary and Key Concepts.
two triangles that are
Cup
y
toward the cup. Her brother Gabriel drove his ball straight 240 yd, but not
toward the cup. The diagram shows the results. Find x and y, their remaining
distances from the cup.
Use Corollary 2 of Theorem 7-3 to solve for x.
NAEP 2005 Strand: Geometry
★
Distributive Property
20,736
Solve for x.
x 5 108
Use Corollary 2 of Theorem 7-3 to solve for y.
x 192
Corollary 1 to Theorem 7-3
y
x
y
Write a proportion.
The length of the altitude to the hypotenuse of a right triangle is the
geometric mean
108 192
of the lengths of the segments
adjacent hypotenuse segment
All rights reserved.
and the length of the
.
hypotenuse
The geometric mean of two positive numbers a and b is the positive number x such
a 5 x.
that x
b
Examples.
1 Finding the Geometric Mean Find the geometric mean of 3 and 12.
x
3
x5
Write a proportion.
12
x2 5
36
x5
x5 6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
so that the length of each leg of the triangle is the geometric mean of the
length of the
y
y
The altitude to the hypotenuse of a right triangle separates the hypotenuse
Substitute 108
for x.
108
300
Corollary 2 to Theorem 7-3
y
y
of the hypotenuse.
Simplify.
108
y2 32,400
Cross-Product Property
!32,400
y
Find the positive square root.
y 180
Maria’s ball is
yd from the cup, and Gabriel’s ball is
108
yd
180
from the cup.
Quick Check.
1. Find the geometric mean of 5 and 20.
2. Recall Example 2. Find the distance
between Maria’s ball and Gabriel’s ball.
144 yd
10
Cross-Product Property.
!36
Find the positive square root.
The geometric mean of 3 and 12 is 6 .
90
Geometry Lesson 7-4
Daily Notetaking Guide
L1
Name_____________________________________ Class____________________________ Date ________________
Geometry Lesson 7-4
91
Name_____________________________________ Class____________________________ Date________________
2 Using the Triangle-Angle-Bisector Theorem Find the value of x.
Lesson 7-5
Proportions in Triangles
Lesson Objectives
1 Use the Side-Splitter Theorem
2 Use the Triangle-Angle-Bisector
Theorem
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Daily Notetaking Guide
L1
G
40
24
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
I
x
Local Standards: ____________________________________
IG GH
Key Concepts.
24
then it divides those sides proportionally.
All rights reserved.
All rights reserved.
If a line is parallel to one side of a triangle and intersects the other two sides,
Theorem 7-5: Triangle-Angle-Bisector Theorem
24
(
40
IK
Triangle-Angle-Bisector Theorem.
HK
x
Substitute.
30
40
Theorem 7-4: Side-Splitter Theorem
H
30
K
30 ) x
x
Solve for x.
18
Quick Check.
If a ray bisects an angle of a triangle, then it divides the opposite side into two segments that
1. Use the Side-Splitter Theorem to find the value of x.
are proportional to the other two sides of the triangle.
1.5
5
x 1.5
2.5
x
transversals are proportional.
a
c
b
d
a
b5
c
d
Examples.
1 Using the Side-Splitter Theorem Find y.
B
CM MB
12
y
10
92
L1
y
12
M
CN
10
Side-Spitter Theorem
N
NA
A
10
C
6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
If three parallel lines intersect two transversals, then the segments intercepted on the
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Corollary to Theorem 7-4
2. Find the value of y.
3.6
5.76
y
5
8
Substitute.
6
y
72
Cross-Product Property
y
7.2
Solve for y.
Geometry Lesson 7-5
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 7-5
Geometry
93
23
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
2 Is It a Right Triangle? Is this triangle a right triangle?
Lesson 8-1
1
4m
The Pythagorean Theorem and Its Converse
a2 1 b2 0 c2
Lesson Objectives
NAEP 2005 Strand: Geometry
Use the Pythagorean Theorem
Use the Converse of the Pythagorean
Theorem
2
Name_____________________________________ Class____________________________ Date ________________
Topic: Relationships Among Geometric Figures
42
62
0
72
Substitute 4 for a, 6 for b, and 7 for c.
16
36
0
49
Simplify.
52
2
49
Local Standards: ____________________________________
Because a2 b2 ⫽ c2, the triangle
Vocabulary and Key Concepts.
Theorem 8-5: Pythagorean Theorem
7m
6m
a right triangle.
is not
c
a
.
b
a2 b2 c 2
Theorem 8-6: Converse of the Pythagorean Theorem
All rights reserved.
hypotenuse
is equal to the square of the length of the
All rights reserved.
Quick Check.
legs
In a right triangle, the sum of the squares of the lengths of the
1. A right triangle has a hypotenuse of length 25 and a leg of length 10. Find the
length of the other leg. Do the lengths of the sides form a Pythagorean triple?
5!21; no
If the square of the length of one side of a triangle is equal to the sum of
the squares of the lengths of the other two sides, then the triangle is a
right
triangle.
2. A triangle has sides of lengths 16, 48, and 50. Is the triangle a right triangle?
A Pythagorean triple is a set of nonzero whole numbers a, b, and c that satisfy
B
Do the lengths of the sides of #ABC form a Pythagorean triple?
1
a2
b2
5
c2
Use the
52 122 c2
Î 169
Pythagorean
Theorem.
12
Substitute 5 for a, and 12 for b.
169 c2
Simplify.
c
Take the square root.
13 c
A
C
5
Simplify.
The length of the hypotenuse is 13 . The lengths of the sides, 5, 12, and 13
form a Pythagorean triple because they are
whole
All rights reserved.
Examples.
1 Pythagorean Triples Find the length of the hypotenuse of #ABC.
no
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
the equation a 2 ⫹ b 2 ⫽ c 2.
numbers that
satisfy a2 1 b2 5 c2.
Geometry Lesson 8-1
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 8-2
Geometry Lesson 8-1
2 Using the Length of One Side Find the value of each variable.
Use the 30-60-90 Triangle Theorem to find the lengths of the legs.
4 !3 2 ? x
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
x
Key Concepts.
hypotenuse ⴝ 2 ? shorter leg
4"3
Divide each side by 2 .
"3
y "3 ? 2 !3
s2
hypotenuse "2 ?
45 s
45
s
leg
y 2 ? "3 ? "3
All rights reserved.
and the
All rights reserved.
congruent
length of the hypotenuse is "2 times the length of a leg.
y
y
Simplify.
60
y "3 ? shorter leg
Theorem 8-5: 45-45-90 Triangle Theorem
30
4 3
2
x 2
In a 45-45-90 triangle, both legs are
95
Name_____________________________________ Class____________________________ Date ________________
Special Right Triangles
Lesson Objectives
1 Use the properties of 45-45-90
triangles
2 Use the properties of 30-60-90
triangles
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
94
30°-60°-90°
Substitute
2"3
Triangle Theorem
x
for shorter leg.
Simplify.
6
The length of the shorter leg is
2"3
, and the length of the longer leg is
6
.
Quick Check.
1. Find the length of the hypotenuse of a 45-45-90 triangle with legs of length 5 !3.
Theorem 8-6: 30-60-90 Triangle Theorem
twice
the length of the
hypotenuse 2
?
longer leg "3 ?
5!6
. The length of the longer leg
is "3 times the length of the
shorter leg
2s
.
shorter leg
30 s3
60
s
shorter leg
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
shorter leg
Examples.
1 Finding the Length of the Hypotenuse Find the value of the
variable. Use the 45-45-90 Triangle Theorem to find the
hypotenuse.
h "2 ? 5!6
"12
h 5
h5
hypotenuse ⴝ "2
45
h
? leg
5 6
Simplify.
45
4(3)
( ) 3
5 6
h5 2
h
10
"3
The length of the hypotenuse is
96
24
Geometry Lesson 8-2
Geometry
10"3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
In a 30-60-90 triangle, the length of the hypotenuse is
2. Find the lengths of the legs of a 30-60-90 triangle with hypotenuse of length 12.
6; 6!3
.
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 8-2
All-In-One Answers Version B
97
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
2 Using a Tangent Ratio To measure the height of a tree, Alma walked
Lesson 8-3
The Tangent Ratio
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
125 ft from the tree and measured a 32 angle from the ground to the top
of the tree. Estimate the height of the tree.
NAEP 2005 Strand: Measurement
Use tangent ratios to determine side
lengths in triangles
Topic: Measuring Physical Attributes
Local Standards: ____________________________________
32
125 ft
Vocabulary.
A
Leg
⬔A
C
adjacent to
All rights reserved.
All rights reserved.
B
opposite
125
⬔A
.
All rights reserved.
B
29
98
A
opposite
adjacent
opposite
adjacent
BC
AC
AC
BC
20
21
21
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Examples.
tan B Solve for height.
78.108669
32
78
Use a calculator.
feet tall.
J
3, 7
7 3
3
L
K
7
2. Find the value of w to the nearest tenth.
a.
b.
10
54
w
1.0
28
w
13.8
1.9
20
Geometry Lesson 8-3
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Daily Notetaking Guide
99
Geometry Lesson 8-3
Name_____________________________________ Class____________________________ Date ________________
2 Using the Cosine Ratio A 20-ft wire supporting a flagpole forms a
Lesson 8-4
Sine and Cosine Ratios
Lesson Objective
1 Use sine and cosine to determine side
lengths in triangles
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
)
1. Write the tangent ratios for ⬔K and ⬔J.
1 Writing Tangent Ratios Write the tangent ratios for ⬔A and ⬔B.
tan A tan 32°
Quick Check.
opposite
adjacent
You can abbreviate the equation as tan A 21
Use the tangent ratio.
125
(
height 125
length of leg opposite lA
length of leg adjacent to lA
tangent of ⬔A C
height
tan 32 The tree is about
20
angle with the ground, so you can use the
tangent ratio to estimate the height of the tree.
lA to the length of the leg adjacent to lA.
Leg
right
The tree forms a
The tangent of acute ⬔A in a right triangle is the ratio of the length of the leg opposite
35 angle with the flagpole. To the nearest foot, how high is the flagpole?
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
The flagpole, wire, and ground form a
Local Standards: ____________________________________
measure of its hypotenuse, you can use the
wire as the
hypotenuse
right triangle
35
20 ft
with the
. Because you know an angle and the
cosine
ratio
to find the height of the flagpole.
Vocabulary.
This can be abbreviated sin A cosine of lA
B
leg opposite lA
hypotenuse
hypotenuse
opposite
hypotenuse
opposite
.
leg adjacent to /A
hypotenuse
This can be abbreviated
cos A
Leg
cos T sin G hypotenuse
adjacent
hypotenuse
opposite
hypotenuse
adjacent
cos G hypotenuse 100
L1
Geometry Lesson 8-4
20
16
T
12
20
16
20
4
R
12
G
4
5
20
16
Solve for height.
Use a calculator.
feet tall.
X
48
80
Z
5
20
16
16.383041
⬔A
3
5
35
cos 35°
1. Write the sine and cosine ratios for ⬔X and ⬔Y.
sin G, and cos G. Write your answers in simplest terms.
sin T ?
48
48
64
sin X ⫽ 64
80, cos X ⫽ 80, sin Y ⫽ 80, cos Y ⫽ 80
1 Writing Sine and Cosine Ratios Use the triangle to find sin T, cos T,
12
20
The flagpole is about
adjacent to
adjacent
hypotenuse.
Examples.
opposite
20
Use the cosine ratio.
20
height Quick Check.
⬔A
C
A
Leg
3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
sine of ⬔A All rights reserved.
of ⬔A is the ratio of the length of the leg adjacent to ⬔A to the
hypotenuse.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
cosine
All rights reserved.
hypotenuse.
The
height
cos 35 The sine of ⬔A is the ratio of the length of the leg opposite ⬔A to the length of the
64
Y
2. In Example 2, suppose that the angle the wire makes with the ground is 50°.
What is the height of the flagpole to the nearest foot?
15 feet
20 ft
50
5
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 8-4
Geometry
101
25
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 8-5
2 Aviation An airplane flying 3500 ft above the ground begins a 2 descent
to land at the airport. How many miles from the airport is the airplane when
it starts its descent? (Note: The angle is not drawn to scale.)
NAEP 2005 Strand: Measurement
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
Angles of Elevation
and Depression
Topic: Measuring Physical Attributes
Use angles of elevation and
depression to solve problems
2
x
Local Standards: ____________________________________
3,500 ft
Ground
` Vocabulary.
An angle of elevation is the angle formed by a horizontal line and the line of sight to an
sin 2⬚
3,500
x
is the angle formed by a horizontal line and the line of
angle of depression
sight to an object below the horizontal line.
Horizontal line
Angle of
B
All rights reserved.
An
All rights reserved.
object above the horizontal line.
5280
100287.9792
Angle of
depression
2
3500
Use the
ratio.
sine
3500
x
Solve for x.
sin 2⬚
100287 .9792
18.993935
Use a calculator.
5280
Divide by
to convert
feet to miles.
elevation
A
The airplane is about
Horizontal line
19
miles from the airport when it starts its descent.
Quick Check.
1
2
3
4
One side of the angle of depression is a horizontal line. ⬔1 is the angle of
depression
airplane
from the
building
to the
.
1. Describe each angle as it relates to the situation in Example 1.
a. ⬔3
The angle of depression from the building to the person on the ground
All rights reserved.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
as they relate to the situation shown.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Examples.
1 Identifying Angles of Elevation and Depression Describe ⬔1 and ⬔2
b. ⬔4
The angle of elevation from the person on the ground to the building
2. An airplane pilot sees a life raft at a 26 angle of depression. The airplane’s
altitude is 3 km. What is the airplane’s surface distance d from the raft?
about 6.8 km
One side of the angle of elevation is a horizontal line. ⬔2 is the angle of
102
from the
to the
building
.
airplane
Geometry Lesson 8-5
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
2 Adding Vectors Vectors v 4, 3 and w 4, 3 are shown at the right.
Lesson 8-6
Vectors
Lesson Objectives
1 Describe vectors
2 Solve problems that involve vector
addition
NAEP 2005 Strand: Geometry
Topic: Position and Direction
To find the first coordinate of s, add the
first
y
4
Write the sum of the two vectors as an ordered pair. Then draw s,
the sum of v and w.
2
(4, 3)
coordinates of
2
v and w.
Local Standards: ____________________________________
Vocabulary and Key Concepts.
To find the second coordinate of s, add the second coordinates of
2
v and w.
4
8x
4
6
(4, 3)
s k4, 3l k4, 3l
Adding Vectors
k
All rights reserved.
.
A vector is any quantity with magnitude (size) and direction.
4ⴙ4
,
3 ⴙ (ⴚ3)
k 8 , 0 l
All rights reserved.
kx1 1 x2, y1 1 y2 l
For a x1, y1 and c x2, y2, a c 4
2
Draw vector v using the origin as the initial point. Draw vector w
using the terminal point of v, (4, 3), as the initial point. Draw the
can be represented with an arrow.
vector
l Add the coordinates.
Simplify.
resultant vector s from the
A
103
Geometry Lesson 8-5
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
elevation
the
terminal point
initial point
of w.
of v to
y
v
2
w
4
s8 x
6
2
4
The magnitude of a vector is its size, or length.
Quick Check.
initial point
The
of a vector is the point at which it starts.
is the sum of other vectors.
y
Examples.
1 Describing a Vector Describe OM as an ordered pair.
x
Give coordinates to the nearest tenth.
Use the sine and cosine ratios to find the values of x and y.
cos 40 x
Use sine and cosine.
80
x 80 ( cos 40ⴗ
x
)
61.28355545
Because point M is in the
negative
104
26
sin 40 Solve for the variable.
Use a calculator.
third
Geometry
y
y
80
M
80
y 80( sin 40ⴗ
y
x
)
51.42300878
51
k221.6, 46.2l
65
O
x
2. Write the sum of the two vectors k2, 3l and k24, 22l as an ordered pair.
k22, 1l
quadrant, both coordinates are
. To the nearest tenth, OM Geometry Lesson 8-6
40 O
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
resultant vector
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
y
1. Describe the vector at the right as an ordered pair. Give the coordinates to
the nearest tenth.
The terminal point of a vector is the point at which it ends.
k261.3, 251.4l
.
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 8-6
All-In-One Answers Version B
105
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
2 Finding a Translation Image Find the image of 䉭ABC
Lesson 9-1
1
A(1, 2) → A( ⴚ1 ⴙ 2 , 2 3)
NAEP 2005 Strand: Geometry
Identify isometries
Find translation images of figures
y
under the translation (x, y) → (x 2, y 3).
Translations
Lesson Objectives
2
Name_____________________________________ Class____________________________ Date ________________
A
Use the rule (x, y ) → (x ⴙ 2, y ⴚ 3)
Topic: Transformation of Shapes and Preservation of
Properties; Position and Direction
B(1, 0) → B( 1 2, 0 ⴚ 3 )
Local Standards: ____________________________________
The image of 䉭ABC is 䉭ABC with A( 1 , ⴚ1 ), B( 3 , ⴚ3 ),
x
B
C
C(0, 1) → C( 0 ⴙ 2 , ⴚ1 ⴚ 3 )
and C( 2 , ⴚ4 ).
Vocabulary.
A transformation of a geometric figure is a change in its position, shape, or size.
Quick Check.
is the original image before changes
are made.
In a transformation, the image is the resulting figure after changes are made.
An
isometry
All rights reserved.
preimage
All rights reserved.
In a transformation, the
1. Does the transformation appear to be an isometry? Explain.
a.
b.
Preimage
Preimage
Image
Yes; the figures appear to be
congruent by a flip and a
slide.
Yes; the figures appear to be
congruent by a flip.
is a transformation in which the preimage and the image
Image
are congruent.
A translation (slide) is a transformation that maps all points the same distance and in the
2. Find the image of 䉭ABC for the translation (x 2, y 1).
same direction.
Aⴕ(ⴚ3, 3), Bⴕ(ⴚ1, 1), Cⴕ(ⴚ2, 0)
is a combination of two or more transformations.
All rights reserved.
Examples.
1 Identifying Isometries Does the transformation appear to be an
isometry?
Image
Preimage
The image appears to be the same as the preimage, but
Because the figures appear to be
congruent
turned
.
, the transformation
appears to be an isometry.
106
Geometry Lesson 9-1
L1
L1
Daily Notetaking Guide
107
Geometry Lesson 9-1
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 9-2
Examples.
Reflections
Lesson Objectives
1 Find reflection images of figures
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
composition of transformations
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
2 Drawing Reflection Images 䉭XYZ has vertices X(0, 3), Y(2, 0),
NAEP 2005 Strand: Geometry
Topics: Transformation of Shapes and Preservation of
Properties
y
and Z(4, 2). Draw 䉭XYZ and its reflection image in the line x 4.
First locate vertices X, Y, and Z and draw 䉭XYZ in a coordinate
plane.
Local Standards: ____________________________________
Locate points X, Y, and Z such that the line of reflection x 4
perpendicular bisector
is the
Vocabulary.
x4
‚
4
X
2
O
2
of XXr, YYr, and ZZr.
X
Z Z
Y
6
Y
‚
‚
8x
2
Draw the reflection image XYZ.
itself
B is the point such that r is the
, and if a point B is not on line r, then its image
perpendicular bisector
B
Line of
All rights reserved.
the image of A is
of BBr.
r
A Aⴕ
All rights reserved.
A reflection in line r is a transformation such that if a point A is on line r, then
3 Congruent Angles D is the reflection of D across ᐉ.
Show that PD and PW form congruent angles with line ᐉ.
isometry
Because a reflection is an
D
W
, PD and
PDr form congruent angles with line ᐉ.
P
PDr and PW also form congruent angles with line ᐉ
reflection
because
Bⴕ
艎
angles are congruent.
vertical
Therefore, PD and PW form congruent angles with line ᐉ
by the
Property.
Transitive
D
Example.
1 Finding Reflection Images If point Q(1, 2) is reflected across
y x=1
Quick Check.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Q
Q is 2 units to the left of the reflection line, so its image Q is
x
ⴚ2
2 units to the right of the reflection line. The reflection line is the
2
perpendicular bisector of QQr if Q is at (3, 2) .
Quick Check.
1. What are the coordinates of the image of Q if the reflection
line is y 1?
(ⴚ1, ⴚ4)
108
L1
Geometry Lesson 9-2
Daily Notetaking Guide
All-In-One Answers Version B
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
line x 1, what are the coordinates of its reflection image?
2. 䉭ABC has vertices A(3, 4), B(0, 1), and C(2, 3). Draw 䉭ABC and its
reflection image in the line x 3.
y
8
A
Aⴕ
4 C Cⴕ
Bⴕ
B
ⴚ8 ⴚ4
4
8
12 x
ⴚ4
ⴚ8
3. In Example 3, what kind of triangle is 䉭DPD? Imagine the image of point W
reflected across line ᐉ. What can you say about 䉭WPW and 䉭DPD?
isosceles; they are similar by SAS
L1
Daily Notetaking Guide
Geometry Lesson 9-2
Geometry
109
27
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
A
2 Identifying a Rotation Image Regular hexagon ABCDEF is divided
Lesson 9-3
Rotations
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
into six equilateral triangles.
a. Name the image of B for a 240 rotation about M.
NAEP 2005 Strand: Geometry
Draw and identify rotation images of
figures
Because 360 6 60 , each central angle of ABCDEF measures
60° . A 240 counterclockwise rotation about center M moves point B
across
triangles. The image of point B is point D .
four
Topic: Transformation of Shapes and Preservation of
Properties
Local Standards: ____________________________________
B
F
C
M
D
E
b. Name the image of M for a 60° rotation about F.
Vocabulary.
䉭AMF is equilateral, so ⬔AFM has measure 180 3 A rotation of x about a point R is a transformation for which the following
are true:
RV
R
R
).
R
and m⬔VRV x
x
.
V
All rights reserved.
• For any point V, RV All rights reserved.
(that is, R itself
• The image of R is
60 . A 60
60
rotation of 䉭AMF about point F would superimpose FM on FA, so the
V
image of M under a 60 rotation about point F is point A .
Quick Check.
1. Draw the image of LOB for a 50 rotation about point B. Label the vertices
of the image.
Examples.
1 Drawing a Rotation Image Draw the image of 䉭LOB under a 60 rotation
about C.
Step 1 Use a protractor to draw a 60 angle at vertex C with one side CO.
O
Step 2 Use a compass to construct COr CO.
L
Oⴕ
L
L
L
B
B
C
O
O
L
B
Lⴕ
C
O
Bⴕ
O
L
B
110
B
C
O
B
O
L
B
L
Geometry Lesson 9-3
L
L1
2. Regular pentagon PENTA is divided into 5 congruent triangles. Name the
image of T for a 144 rotation about point X.
E
L1
Name_____________________________________ Class____________________________ Date________________
P
A
Daily Notetaking Guide
E
X
Geometry Lesson 9-3
N
111
Name_____________________________________ Class____________________________ Date ________________
Symmetry
Lesson Objective
1 Identify the type of symmetry in
a figure
L
T
Daily Notetaking Guide
Lesson 9-4
B B
O
All rights reserved.
C
O
60°
Examples.
1 Identifying Lines of Symmetry Draw all lines of symmetry for the
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation of
Properties
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
O
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
O
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Step 3 Locate L⬘ and B⬘ in a similar manner. Then draw 䉭L⬘O⬘B⬘.
isosceles trapezoid.
Local Standards: ____________________________________
Draw any lines that divide the isosceles trapezoid so that
half of the figure is a mirror image of the other half.
Vocabulary.
reflectional symmetry
if there is symmetry that maps the figure onto
itself.
Line symmetry is the same as reflectional symmetry.
All rights reserved.
A figure has
All rights reserved.
A figure has symmetry if there is an isometry that maps the figure onto itself.
There is
one
line of symmetry.
2 Identifying Rotational Symmetry Judging from appearance, do the
letters V and H have rotational symmetry? If so, give an angle of rotation.
The letter V does not have rotational symmetry because it must be rotated
360
Line of
Symmetry
before it is its own image.
The letter H is its own image after one half-turn, so it has rotational
A figure has
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
(
reflectional
)
rotational symmetry
if it is its own image for some rotation of 180°
or less.
A figure has point symmetry if it has 180⬚ rotational symmetry.
The figure is its own image after one half-turn, so it has
rotational symmetry with a
The figure also has
point
180
angle of rotation.
symmetry.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
symmetry with a
The heart-shaped figure has
symmetry.
line
180
angle of rotation.
Quick Check.
1. Draw a rectangle and all of its lines of symmetry.
2. a. Judging from appearance, tell whether the figure at the right has
rotational symmetry. If so, give the angle of rotation.
yes; 180°
b. Does the figure have point symmetry?
yes
112
28
Geometry Lesson 9-4
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 9-4
All-In-One Answers Version B
113
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
2 Scale Drawings The scale factor on a museum’s floor plan is 1 : 200. The
Lesson 9-5
Dilations
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
length and width of one wing on the drawing are 8 in. and 6 in. Find the
actual dimensions of the wing in feet and inches.
NAEP 2005 Strand: Geometry
Locate dilation images of figures
1
Topic: Geometry
The floor plan is a reduction of the actual dimensions by a scale factor of 200 .
Local Standards: ____________________________________
Multiply each dimension on the drawing by 200 to find the actual
dimensions. Then write the dimensions in feet and inches.
Vocabulary.
A dilation is a transformation with center C and scale factor n for which
the following are true:
R
1
C
CR .
R
3
Q
All rights reserved.
)
(that is, C C ).
• For any point R, R is on CR and CR n ft
y J
J
dilation of concentric circle B with 8-cm diameter. Describe the dilation.
The circles are concentric, so the dilation has center C .
Because the diameter of the dilation image is smaller, the dilation is a
.
diameter of dilation image
diameter of preimage 5
3
8
3
8
The dilation is a reduction with center C and scale factor
.
Geometry Lesson 9-5
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Examples.
1 Finding a Scale Factor Circle A with 3-cm diameter and center C is a
reduction
Lesson 9-6
by
in.
100 ft
.
2
K
x
K
4
L
L
The dilation is a reduction with center (0, 0) and scale factor 1
2.
2. The height of a tractor-trailer truck is 4.2 m. The scale factor for a model
1 . Find the height of the model to the nearest centimeter.
truck is 54
8 cm
L1
L1
Name_____________________________________ Class____________________________ Date________________
Daily Notetaking Guide
Geometry Lesson 9-5
115
Name_____________________________________ Class____________________________ Date ________________
Example.
Compositions of Reflections
Lesson Objectives
1 Use a composition of reflections
2 Identify glide reflections
O
M
M
3
is a dilation with a scale factor less than 1.
reduction
133 ft, 4 in.
4
1. Quadrilateral JKLM is a dilation image of quadrilateral JKLM.
Describe the dilation.
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
ft,
100
with center C and scale factor 3 .
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
133
in. dilation
RrQr is the image of RQ under a
An enlargement is a dilation with a scale factor greater than 1.
114
in. 1200
Quick Check.
Q
Scale factor:
1600
6 in. 200 The museum wing measures
All rights reserved.
itself
• The image of C is
8 in. 200 Composition of Reflections in Intersecting Lines The letter D is
reflected in line x and then in line y. Describe the resulting rotation.
Find the image of D through a reflection across line x.
Find the image of the reflection through another reflection across line y.
NAEP 2005 Strand: Geometry
Topic: Transformation of Shapes and Preservation
of Properties
Local Standards: ____________________________________
D
D
Vocabulary and Key Concepts.
D
43
A
y
reflections
.
Theorem 9-2
A composition of reflections across two parallel lines is a
.
translation
All rights reserved.
x
The composition of two reflections across intersecting lines is a
The center of rotation is
is
twice the angle formed by the intersecting lines
86 clockwise, or
Theorem 9-3
is rotated
A composition of reflections across two intersecting lines is a
rotation at point A .
rotation
.
R
R
R
R
R
A glide reflection is the composition of a glide (translation) and a reflection across a line
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
R
R
R R
Glide reflection
Rotation
L1
Geometry Lesson 9-6
c
b. Use parallel lines ᐉ and m. Draw R between ᐉ and m. Find the image of R
for a reflection across line ᐉ and then across line M. Describe the resulting
translation.
All-In-One Answers Version B
L1
R
Answers may vary. The result is a translation twice the distance
between ᐉ and m.
R
艎
Daily Notetaking Guide
b
R
parallel to the direction of translation.
116
a
Answers may vary. The result is a clockwise rotation about point c
through an angle of 2mlacb.
R
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Translation
. So, the letter D
counterclockwise, with the center of
R
.
Theorem 9-5: Isometry Classification Theorem
There are only four isometries. They are the following:
Reflection
274
a. Reflect the letter R across a and then b. Describe the resulting rotation.
In a plane, one of two congruent figures can be mapped onto the other by a
reflections
.
Quick Check.
Theorem 9-4: Fundamental Theorem of Isometries
composition of at most three
rotation
the point where the lines intersect , and the angle
R
A translation or rotation is a composition of two
All rights reserved.
Theorem 9-1
L1
Daily Notetaking Guide
m
Geometry Lesson 9-6
Geometry
117
29
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
2 Identifying Symmetries in Tessellations List the
Lesson 9-7
Tessellations
Lesson Objectives
1
2
Name_____________________________________ Class____________________________ Date ________________
Identify transformation in tessellations,
and figures that will tessellate
Identify symmetries in tessellations
symmetries in the tessellation.
NAEP 2005 Strand: Geometry
Starting at any vertex, the tessellation can be mapped
Topic: Geometry
onto itself using a
Local Standards: ____________________________________
has
point
180⬚
rotation, so the tessellation
symmetry centered at any vertex.
translational
The tessellation also has
Vocabulary and Key Concepts.
symmetry, as can be seen by sliding any triangle
Theorem 9-6
Every triangle tessellates.
All rights reserved.
All rights reserved.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
onto a copy of itself along any of the lines.
Theorem 9-7
Every quadrilateral tessellates.
A tessellation, or tiling, is a repeating pattern of figures that completely covers a plane,
Quick Check.
1. Identify a transformation and outline the smallest repeating figure in the
tessellation below.
without gaps or overlaps.
is the type of symmetry for which there is a translation that
Translational symmetry
maps a figure onto itself.
maps a figure onto itself.
Examples.
1 Identifying the Transformation in a Tessellation Identify the
repeating figures and a transformation in the tessellation.
octagon
and one adjoining
will completely cover the plane without gaps or
square
overlaps. Use arrows to show a translation.
118
Geometry Lesson 9-7
Daily Notetaking Guide
L1
The height corresponding to the 9-in. base is 15 in. Find the area of the parallelogram.
A bh
Local Standards: ____________________________________
h
.
b
bh
Theorem 10-2: Area of a Parallelogram
height
and the corresponding
A
h
.
b
of a parallelogram corresponding
to a given base is the segment perpendicular to the line
Altitude
Base
containing that base drawn from the side opposite the base.
The height of a parallelogram is the length of its altitude.
A
base
of a triangle is any of its sides.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
height
1bh
2
altitude
for h.
135
in.2
A 5 12bh
A 5 12
(
Area of a
30
)(
13
)
195
Substitute
X
triangle
30
for b and
13
31 cm
for h.
13 cm
Simplify.
195
cm2.
Y
Z
30 cm
1. A parallelogram has sides 15 cm and 18 cm. The height corresponding to a
15-cm base is 9 cm. Find the area of the parallelogram.
the product of
A base of a parallelogram is any of its sides.
The
15
Quick Check.
half
and the corresponding
A
Substitute 9 for b and
Simplify.
䉭XYZ has area
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
base
135
)
b
Theorem 10-3: Area of a Triangle
a
A
15
2 Finding the Area of a Triangle Find the area of 䉭XYZ.
h
.
bh
The area of a triangle is
(
A5
The area of a parallelogram is the product of a
base
All rights reserved.
height
All rights reserved.
and
A
Area of a parallelogram
A 9
The area of the parallelogram is
The area of a rectangle is the product of
119
Examples.
Theorem 10-1: Area of a Rectangle
base
Geometry Lesson 9-7
1 Finding the Area of a Parallelogram A parallelogram has 9-in. and 18-in. sides.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Vocabulary and Key Concepts.
its
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Areas of Parallelograms and Triangles
Lesson Objectives
1 Find the area of a parallelogram
2 Find the area of a triangle
line symmetry, rotational symmetry,
glide reflectional symmetry,
translational symmetry
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 10-1
translation
2. List the symmetries in the tessellation.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A repeated combination of an
All rights reserved.
Glide reflectional symmetry is the type of symmetry for which there is a glide reflection that
135 cm2
2. Find the area of the triangle.
5 cm
30 cm2
13 cm
12 cm
The height of a triangle is the length of the altitude to the line
h
containing that base.
b
120
30
Geometry Lesson 10-1
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 10-1
All-In-One Answers Version B
121
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 10-2
1
Find the area of a trapezoid
Find the area of a rhombus or a kite
R
2 Finding the Area of a Rhombus Find the area of rhombus RSTU.
Areas of Trapezoids, Rhombuses, and Kites
Lesson Objectives
2
Name_____________________________________ Class____________________________ Date ________________
NAEP 2005 Strand: Measurement
To find the area, you need to know the lengths of both diagonals.
Draw diagonal SU, and label the intersection of the diagonals point X.
Topic: Measuring Physical Attributes
䉭SXT is a
Local Standards: ____________________________________
are perpendicular.
right
S
24
triangle because the diagonals of a rhombus
The diagonals of a rhombus bisect each other, so TX 12
ft
13 ft
X
U
T
ft.
You can use the Pythagorean triple 5, 12, 13 or the Pythagorean Theorem
Vocabulary and Key Concepts.
to conclude that SX SU The area of a trapezoid is half the product of the height and the sum of
A 5 12
A 5 12
All rights reserved.
b1
the bases.
h
1h(b 1 b )
1
2
2
A
b2
Theorem 10-5: Area of a Rhombus or a Kite
The area of a rhombus or a kite is half the product of the lengths of its
d2
d1
diagonals.
All rights reserved.
Theorem 10-4: Area of a Trapezoid
d1
(
5
ft.
ft because the diagonals of a rhombus bisect each other.
24
d2
)(
Area of a rhombus
)
10
120
24
Substitute
for d1 and
10
for d2 .
Simplify.
ft 2.
120
The area of rhombus RSTU is
Quick Check.
1d d
2 1 2
A
A5
10
1. Find the area of a trapezoid with height 7 cm and bases 12 cm and 15 cm.
94.5 cm 2
The height of a trapezoid is the perpendicular
Leg
h
All rights reserved.
b2
Base
Examples.
1 Applying the Area of a Trapezoid A car window is
20 in.
shaped like the trapezoid shown. Find the area of the
window.
18 in.
36 in.
A 5 12 h
A 5 12
(
A5
(
b1
)(
18
)
b2
20
36
)
18
Substitute
504
5 2 ⫹ 12 2 ⫽ 13 2
for h,
20
for b1 , and
36
for b2 .
Simplify.
in.2.
504
Geometry Lesson 10-2
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Name_____________________________________ Class____________________________ Date ________________
Lesson 10-3
2 .Finding the Area of a Regular Polygon A library is in the shape
Areas of Regular Polygons
a2 a
p
9.0
1 ap
2
81
)2 (
a2 All rights reserved.
perimeter.
All rights reserved.
The area of a regular polygon is half the product of the apothem and the
A
(
a2 Theorem 10-6: Area of a Regular Polygon
a<
of a regular polygon is the distance from the center to a vertex.
18.0
A 12 ap
21.7
A 12
1562.4
A
Examples.
1 Finding Angle Measures This regular hexagon has an apothem and
radii drawn. Find the measure of each numbered angle.
m⬔2 1
2
1
2
3
Divide 360 by the number of sides.
The apothem bisects the vertex angle of
m⬔1
isosceles
the
triangle formed by
the radii.
m⬔2 1
2
(
60
)
90
m⬔3 180 (
m⬔1 , m⬔2 124
60
Geometry Lesson 10-3
30
30
Substitute
)
60
30 , and m⬔3 60
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Apothem
6
471.25
21.7
)
144
)(
Substitute 8 for n and
)
144
To the nearest 10 ft 2, the area is
Radius
60
9.0 ft
Solve for a.
18.0
for s, and simplify.
Step 3 Find the area A.
(
9.0 ft
Pythagorean Theorem
Find the perimeter, where n ⴝ the number of sides
of a regular polygon.
(
Center
360
)2
p ns
p (8)
The apothem of a regular polygon is the perpendicular distance from the center to a side.
m⬔1 23.5
552.25
Step 2 Find the perimeter p.
The center of a regular polygon is the center of the circumscribed circle.
radius
23.5 ft
Step 1 Find the apothem a.
Vocabulary and Key Concepts.
The
18.0 ft
of a regular octagon. Each side is 18.0 ft. The radius of the octagon is
23.5 ft. Find the area of the library to the nearest 10 ft 2.
Consecutive radii form an isosceles triangle, so an apothem bisects
the side of the octagon.
To apply the area formula A 12ap, you need to find a and p.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Local Standards: ____________________________________
L1
123
Geometry Lesson 10-2
Name_____________________________________ Class____________________________ Date________________
Lesson Objective
1 Find the area of a regular polygon
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2. Critical Thinking In Example 2, explain how you can use a Pythagorean
triple to conclude that XU 5 ft.
Area of a trapezoid
The area of the car window is
122
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
b1
Leg
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Base
distance h between the bases.
Area of a
regular
Substitute
21.7
polygon
for a and
144
for p.
Simplify.
1560
ft 2.
Quick Check.
1. At the right, a portion of a regular octagon has radii and an apothem drawn.
Find the measure of each numbered angle.
ml1 5 45; ml2 5 22.5; ml3 5 67.5
1
2
3
2. Find the area of a regular pentagon with 11.6-cm sides and an 8-cm
apothem.
232 cm 2
for m⬔1.
The sum of the measures of the angles of a triangle is 180.
60
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 10-3
Geometry
125
31
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Name_____________________________________ Class____________________________ Date ________________
Lesson 10-4
3 Using Similarity Ratios Benita plants the same crop in two rectangular
Perimeters and Areas of Similar Figures
Lesson Objective
1
fields. Each dimension of the larger field is 3 12 times the dimension of the
smaller field. Seeding the smaller field costs $8. How much money does
seeding the larger field cost?
NAEP 2005 Strand: Measurement and Number Properties
and Operations
Find the perimeters and areas of
similar figures
Topics: Systems of Measurement; Ratios and Proportional
Reasoning
The similarity ratio of the fields is 3.5 : 1, so the ratio of the areas of the
.
fields is (3.5) 2 : (1) 2, or
12.25 to 1
Local Standards: ____________________________________
Because seeding the smaller field costs $8, seeding 12.25 times as much
land costs
$8 .
12.25
Key Concepts.
All rights reserved.
If the similarity ratio of two similar figures is ba , then
a
b
(1) the ratio of their perimeters is
a2
b2
(2) the ratio of their areas is
and
All rights reserved.
Seeding the larger field costs
Theorem 10-7: Perimeters and Areas of Similar Figures
.
$98
.
Quick Check.
1. Two similar polygons have corresponding sides in the ratio 5 : 7.
a. Find the ratio of their perimeters.
b. Find the ratio of their areas.
5:7
25 : 49
Examples.
5
the shortest side of the right-hand triangle has length
5
4
From larger to smaller, the similarity ratio is
6
5
6.25
5
, and
7.5
.
.
By the Perimeters and Areas of Similar Figures Theorem, the ratio of
the perimeters is
5
4
52
42
, and the ratio of the areas is
, or
25
16
.
2 Finding Areas Using Similar Figures The ratio of the length of the
corresponding sides of two regular octagons is 83. The area of the larger
octagon is 320 ft 2. Find the area of the smaller octagon.
All regular octagons are similar.
Because the ratio of the lengths of the corresponding sides of the regular
82
32
octagons is 83, the ratio of their areas is
320
A
9
64 A A
64
9
.
3. The similarity ratio of the dimensions of two similar pieces of window glass
is 3 : 5. The smaller piece costs $2.50. What should be the cost of the larger
piece?
$6.94
Write a proportion.
2880
45
Use the Cross-Product Property.
64 .
Divide each side by
The area of the smaller octagon is
126
, or
96 in.2
ft 2.
45
Geometry Lesson 10-4
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 10-5
Geometry Lesson 10-4
127
Name_____________________________________ Class____________________________ Date ________________
A 12 Trigonometry and Area
A
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Lesson Objectives
1 Find the area of a regular polygon
using trigonometry
2 Find the area of a triangle using
trigonometry
Daily Notetaking Guide
180(sin 36ⴗ)
Substitute for a and p.
(cos 36) (sin 36)
Simplify.
770.355778
A
The area is about
Local Standards: ____________________________________
18(cos 36ⴗ)
1620
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
64
2. The corresponding sides of two similar parallelograms are in the ratio 34.
The area of the smaller parallelogram is 54 in.2. Find the area of the larger
parallelogram.
All rights reserved.
4
The shortest side of the left-hand triangle has length
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
4
similar. Find the ratio (larger to smaller) of their perimeters and of
their areas.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1 Finding Ratios in Similar Figures The triangles at the right are
Use a calculator.
770
ft2.
2 Surveying A triangular park has two sides that measure 200 ft and 300 ft
Key Concepts.
and form a 65 angle. Find the area of the park to the nearest hundred
square feet.
Theorem 10-8: Area of a Triangle Given SAS
c
.
A
1 bc(sin A)
2
Area of 䉭ABC a
C
b
All rights reserved.
the sine of the included angle
B
and
the lengths of two sides
All rights reserved.
Use Theorem 10-8: The area of a triangle is
The area of a triangle is one half the product of
product of the lengths of two sides and the
one half
sine
the
300 ft
of the included
angle.
Area 12 side length side length sine of included angle
Theorem 10-8
Area 12 Substitute.
Area Examples.
200 ft
200
30,000
300
sin 65ⴗ
sin 65ⴗ
Substitute.
27189.23361
The area of the park is approximately
65
Use a calculator.
27,200
ft2.
1 Finding Area The radius of a garden in the shape of a regular pentagon
formula A 12 ap.
C
18 ft
360
Because a pentagon has five sides, m⬔ACB 5
72 .
A
M
B
CA and CB are radii, so CA CB . Therefore, 䉭ACM 䉭BCM by the
HL Theorem, so m⬔ACM 12m⬔ACB Use the cosine ratio to find a.
cos 36 a
18
a 18(cos 36ⴗ)
36 .
Use the sine ratio to find AM.
sin
Use the ratio.
36 AM
18
AM Solve.
18(sin 36ⴗ)
Use AM to find p. Because 䉭ACM 䉭BCM, AB 2 ? AM . Because
the pentagon is regular, p 5 ? AB .
So p 5(2 ? AM ) 10
? AM 10
?
18(sin 36ⴗ)
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Find the perimeter p and apothem a, and then find the area using the
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
is 18 feet. Find the area of the garden.
Quick Check.
1. Find the area of a regular octagon with a perimeter of 80 in. Give the area
to the nearest tenth.
482.8 in.2
2. Two sides of a triangular building plot are 120 ft and 85 ft long. They include
an angle of 85. Find the area of the building plot to the nearest square foot.
5081 ft 2
180(sin 36ⴗ) .
Finally, substitute into the area formula A 12ap.
128
32
Geometry Lesson 10-5
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 10-5
All-In-One Answers Version B
129
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 10-6
Topic: Measuring Physical Attributes
0
0
0
mXY mXD mDY
0
0
mXY m/ XCD mDY
0
mXY 56 40
0
mXY 96
Local Standards: ____________________________________
C
B
measures of the two arcs.
1
0
0
mABC mAB 1 mBC
A
Theorem 10-9: Circumference of a Circle
p times the diameter
d r
O
2pr
C
Theorem 10-10: Arc Length
The length of an arc of a circle is the product of the ratio
measure of the arc
and the
A
circumference of the circle
.
O
0
mAB ? 2pr
360
All rights reserved.
A circle is the set of all points equidistant from a given point called the center.
A
of a circle is the point from which all points are equidistant.
center
A radius is a segment that has one endpoint at the center and the other endpoint on the
circle.
A
is an angle whose vertex is the center of the circle.
central angle
Geometry Lesson 10-6
Daily Notetaking Guide
D
Arc Addition Postulate
Substitute.
56
C
W
X
Simplify.
2π
360
(
)
18
B
150
18 cm
M
D
A
Substitute.
Quick Check.
1
0
1
1. Use the diagram in Example 1. Find m⬔YCD, mYMW , mMW, and mXMY.
40; 180; 96; 264
2. Find the length of a semicircle with radius 1.3 m in terms of π.
1.3p m
L1
Daily Notetaking Guide
Geometry Lesson 10-6
131
Name_____________________________________ Class____________________________ Date ________________
Lesson 10-7
2 Finding the Area of a Sector of a Circle Find the area of sector ACB.
Areas of Circles and Sectors
Leave your answer in terms of π.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
area of sector ACB mAB
Local Standards: ____________________________________
Vocabulary and Key Concepts.
Theorem 10-11: Area of a Circle
measure of the arc
The area of a sector of a circle is the product of the ratio
360
.
area of the circle.
A
r
O
B
5
18
The area of sector ACB is
)2
π( 6
360
10
6 m 100
36
π
π
10π
B
A
Substitute.
Simplify.
Simplify.
m 2.
about 41 in.2
Segment
intercepted arc.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
of a circle is the part bounded by an arc and
Examples.
Sector
1 Applying the Area of a Circle A circular archery target has a 2-ft diameter.
of a circle
It is yellow except for a red bull’s-eye at the center with a 6-in. diameter.
Find the area of the yellow region to the nearest whole number.
First find the areas of the archery target and the red bull’s-eye.
The radius of the archery target is 12 2 ft 1 ft 12 in.
(
The area of the archery target is
πr 2
)
π
(
)
2
12
(6
π(
)2 The area of the red region is
area of
archery target πr 2
in. 3
9p
9p
135
424.11501
The area of the yellow region is about
Geometry Lesson 10-7
3
in.
in.2
area of
red region
144p
)
The radius of the red bull’s-eye region is 12
in.2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
of a circle
segment
the segment joining its endpoints.
132
100
C
πr 2
1. How much more pizza is in a 14-in.-diameter pizza than in a 12-in. pizza?
A sector of a circle is a region bounded by two radii and their
144p
5
360
Quick Check.
0
mAB ? pr2
360
Area of sector AOB A
r
O
Theorem 10-12: Area of a Sector of a Circle
and the
.
pr 2
All rights reserved.
A
All rights reserved.
the product of p and the square of the radius
The area of a circle is
L1
Y
40
Simplify.
Name_____________________________________ Class____________________________ Date________________
Lesson Objective
1 Find the areas of circles, sectors, and
segments of circles
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
L1
Substitute.
21 π
1
The length of ADB is 21π cm.
Circumference of a circle is the distance around the circle.
Pi ()
is the ratio of the circumference of a circle to its diameter.
130
210
M
The measure of a minor arc is the measure
of its corresponding central angle.
1
2 Finding Arc Length Find the length of ADB in circle M in terms of π.
0
Because mAB 150,
1
. Arc Addition Postulate
mADB 360
150
210
1
1
length of ADB mADB
Arc Length Postulate
360 ? 2 pr
B
360
0
length of AB 1
Arc Addition Postulate
0
1
1
mDXM mDX mXWM
1
mDXM 56 180
1
236
mDXM r
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
or C pd
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
All rights reserved.
Postulate 10-1: Arc Addition Postulate
The measure of the arc formed by two adjacent arcs is the sum of the
All rights reserved.
Vocabulary and Key Concepts.
The circumference of a circle is
0
1 Finding the Measures of Arcs Find mXY and mDXM in circle C.
NAEP 2005 Strand: Measurement
Find the measures of central angles
and arcs
Find circumference and arc length
2
Examples.
Circles and Arcs
Lesson Objectives
1
Name_____________________________________ Class____________________________ Date ________________
2. Critical Thinking A circle has a diameter of 20 cm. What is the area of a sector
bounded by a 208 major arc? Round your answer to the nearest tenth.
181.5 cm 2
area of
yellow region
424
π
Simplify.
in.2
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 10-7
Geometry
133
33
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 10-8
Example.
Geometric Probability
2 Finding Probability Using Area A circle is inscribed in a square target
NAEP 2005 Strand: Data Analysis and Probability
Use segment and area models to find
the probabilities of events
Local Standards: ____________________________________
Find the area of the square.
A s2 Vocabulary.
20 2
400
cm 2
Find the area of the circle. Because the square has sides of length 20 cm,
the circle’s diameter is 20 cm, so its radius is 10 cm.
Example.
1 Finding Probability Using Segments A gnat lands at random on the
A πr 2 π
All rights reserved.
All rights reserved.
Geometric probability is a model in which you let points represent outcomes.
(
10
)2 cm 2
100π
Find the area of the region between the square and the circle.
(
A 400 100π
) cm 2
Use areas to calculate the probability that a dart landing randomly in the
square does not land within the circle. Use a calculator. Round to the
nearest thousandth.
area between square and circle
P(between square and circle) area of square
edge of the ruler below. Find the probability that the gnat lands on a point
between 2 and 10.
400 ⴚ 100π
3
4
5
6
7
8
9
The length of the segment between 2 and 10 is
12
The length of the ruler is
10
10
11
400
2
8
.
.
P(landing between 2 and 10)
length of
favorable
segment
length of
entire
segment
8
12
23
Quick Check.
1. A point on AB is selected at random. What is the probability that it is
a point on CD?
A
1
2
3
4
5
6
7
D
B
8
9 10
1
4
0.215
The probability that a dart landing randomly in the square does not land
21.5
within the circle is about
%.
Quick Check.
2. Use the diagram in Example 2. If you change the radius of the circle as
indicated, what then is the probability of hitting outside the circle?
a. Divide the radius by 2.
about 80.4%
b. Divide the radius by 5.
2
5
134
π
about 96.9%
Geometry Lesson 10-8
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date ________________
Lesson 11-1
Geometry Lesson 10-8
135
Name_____________________________________ Class____________________________ Date ________________
2 Using Euler’s Formula Use Euler’s Formula to find the number of
Space Figures and Cross Sections
Lesson Objective
1 Recognize polyhedra and their parts
2 Visualize cross sections of space
figures
Daily Notetaking Guide
edges on a solid with 6 faces and 8 vertices.
FVE2
NAEP 2005 Strand: Geometry
Topic: Dimension and Shape
6 8 E 2
12
Local Standards: ____________________________________
E
Euler’s Formula
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
0
C
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
20 cm
with 20-cm sides. Find the probability that a dart landing randomly within
the square does not land within the circle.
Topic: Probability
All rights reserved.
Lesson Objective
1
Name_____________________________________ Class____________________________ Date ________________
Substitute the number of faces and vertices.
Simplify.
A solid with 6 faces and 8 vertices has
12
edges.
Vocabulary and Key Concepts.
Quick Check.
Euler’s Formula
F⫹V⫽E⫹2
.
A polyhedron is a three-dimensional figure whose surfaces are polygons.
A
face
1. List the vertices, edges, and faces of the polyhedron.
All rights reserved.
are related by the formula
All rights reserved.
The numbers of faces (F ), vertices (V), and edges (E) of a polyhedron
Rectangular
prism
R
S
T
U
2. Use Euler’s Formula to find the number of edges on a polyhedron
with eight triangular faces.
is a flat surface of a polyhedron in the shape
of a polygon.
R, S, T, U, V; RS, RU, RT, VS, VU, VT, SU, UT, TS;
kRSU, kRUT, kRTS, kVSU, kVUT, kVTS
V
12 edges
An edge is a segment that is formed by the intersection of two
Faces
faces.
is a point where three or more edges intersect.
Vertex
Examples.
1 Identifying Vertices, Edges, and Faces How many vertices, edges,
and faces are there in the polyhedron shown? Give a list of each.
There are
four
There are
six
There are
four
vertices:
B
A
.
A, B, C, and D
C
136
34
edges:
AB, BC, CD, DA, AC, and BD
.
D
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
vertex
A cross section is the intersection of a solid and a plane.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A
Edge
faces: kABC, kABD, kACD, and kBCD .
Geometry Lesson 11-1
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 11-1
All-In-One Answers Version B
137
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date ________________
Lesson 11-2
Name_____________________________________ Class____________________________ Date________________
A cylinder
Surface Areas of Prisms and Cylinders
is a three-dimensional figure with two
congruent circular bases that lie in parallel planes.
Lesson Objectives
1
Find the surface area of a prism
Find the surface area of a cylinder
2
In a cylinder, the bases are parallel circles.
Topic: Measuring Physical Attributes
h
Local Standards: ____________________________________
The
.
height of the cylinder
L.A. 2πrh, or L.A. πdh
The surface area of a right cylinder is the sum of the
areas of the two circular bases
.
Area
5 2prh 1 2
r
r
Area of a base
B π r2
prism
Pentagonal
faces.
of a polyhedron are the parallel
bases
faces.
Bases
Lateral
edges
Lateral faces are the faces on a prism that are not bases.
The
of a prism is the sum of the areas
lateral area
of the lateral faces.
Lateral
faces
The surface area of a prism or a cylinder is the sum of the
p
Substitute the formulas for lateral
area of a cylinder and area of a circle.
r2
Bases
lateral area and the areas of the two bases.
5
24
p 1 8
5
32
p
S.A. 5 L.A. 1 2
Use the formula for surface area.
( )( 6 ) 2π ( 2 2)
A prism is a polyhedron with exactly two congruent, parallel
The
Barley Container
B
2π 2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Surface
Area
2πr
All rights reserved.
h
Lateral
Find the surface area of each container. Remember that r d2.
S.A. 5 L.A. 1 2
r
h
Finding Surface Area of Cylinders A company sells cornmeal and barley in
cylindrical containers. The diameter of the base of the 6-in.-high cornmeal
container is 4 in. The diameter of the base of the 4-in.-high barley container is 6
in. Which container has the greater surface area?
Cornmeal Container
2πr 2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
S.A. L.A. 2B, or S.A. 2πrh lateral area
All rights reserved.
All rights reserved.
and the
cylinders
right
Example.
Theorem 11-1: Lateral and Surface Area of a Cylinder
The lateral area of a right cylinder is the product of the
and the
of a cylinder is the area of the curved
lateral area
surface.
Vocabulary and Key Concepts.
circumference of the base
h
Bases
NAEP 2005 Strand: Measurement
5 2 p
2π
Substitute for r and h.
B
h 2pr2
( 3 )( 4 ) 2π ( 3 2)
Simplify.
5
24
p 1
Combine like terms.
5
42
p
p
Because 42π in.2 ⬎ 32π in.2, the
surface area.
r
barley
18
p
container has the greater
Quick Check.
a. Find the surface area of a cylinder with height 10 cm and radius 10 cm in
terms of π.
400p cm2
b. The company in the Example wants to make a label to cover the cornmeal
container. The label will cover the container all the way around, but will not
cover any part of the top or bottom. What is the area of the label to the
nearest tenth of a square inch?
75.4 in.2
prism
Triangular
138
Geometry Lesson 11-2
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 11-3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The perimeter p of the square base is 4 You are given ᐍ 12
find the lateral area.
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
Local Standards: ____________________________________
L.A. 5 12
slant height
ᐉ
.
1 pᐉ
2
L.A. B
The surface area of a regular pyramid is the sum of the
lateral area
S.A. L.A. and the
area of the base
All rights reserved.
and the
(
30
)(
7.5
ft, or
ft and you found that p ᐍ
p
2
L.A. 5
All rights reserved.
perimeter of the base
1
L.A. 5
Vocabulary and Key Concepts.
Theorem 11-3: Lateral and Surface Area of a Regular Pyramid
The lateral area of a regular pyramid is half the product of the
Geometry Lesson 11-2
139
Name_____________________________________ Class____________________________ Date ________________
Surface Areas of Pyramids and Cones
Lesson Objectives
1 Find the surface area of a pyramid
2 Find the surface area of a cone
Daily Notetaking Guide
30
30
ft.
ft, so you can
Use the formula for lateral area of a pyramid.
)
12
180
Substitute.
Simplify.
Find the area of the square base.
Because the base is a square with side length 7.5 ft,
B s2 7.5 2 56.25
S.A. 5 L.A. 1 B
.
S.A. 5
B
S.A. 5
A regular pyramid is a pyramid whose base is a regular polygon and whose lateral faces are
180
1
.
Use the formula for surface area of a pyramid.
56.25
236.25
Substitute.
Simplify.
The surface area of the square pyramid is
236.25
ft2.
congruent isosceles triangles.
vertex to the plane of the base.
The height of a pyramid or a cone is the length of
Lateral
Vertex
edge
Lateral face
the altitude.
The
Altitude
of a regular pyramid
slant height
Base
is the length of the altitude of a lateral face.
The lateral area of a pyramid is is the sum of the
Base
edge
Regular Pyramid
areas of the congruent lateral faces.
The
of a pyramid is the sum of the lateral area and the area of the base.
surface area
Example.
Finding Surface Area of a Pyramid Find the surface area of a square
pyramid with base edges 7.5 ft and slant height 12 ft.
12 ft
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
of a pyramid or cone is the perpendicular segment from the
altitude
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The
Quick Check.
Find the surface area of a square pyramid with base edges 5 m and slant
height 3 m.
55 m 2
7.5 ft
140
L1
Geometry Lesson 11-3
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 11-3
Geometry
141
35
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date ________________
2 Finding Volume of a Triangular Prism Find the volume of the prism.
Lesson 11-4
Volumes of Prisms and Cylinders
Lesson Objectives
Find the volume of a prism
Find the volume of a cylinder
Topic: Measuring Physical Attributes
is the
Local Standards: ____________________________________
Pythagorean Theorem to calculate the length of the other leg.
height of the prism
V
h
20
2
841
)(
h
400
40 m
. Use the
)
20 m
210
441
21
. Use the
Use the formula for the volume of a prism.
V
210
?
V
8400
40
Substitute.
Simplify.
8400
The volume of the triangular prism is
B
m3.
Bh
Quick Check.
h
1. The cylinder shown is oblique.
a. Find its volume in terms of π.
r
B
pr 2h
Volume is the space that a figure occupies.
A
space figure is a three-dimensional figure that is the combination
composite
of two or more simpler figures.
Examples.
1 Finding Volume of a Cylinder Find the volume of the cylinder.
Leave your answer in terms of π.
The formula for the volume of a cylinder is
The diagram shows h and d, but you must find r.
V ⴝ pr 2h
.
9 ft
r 5 12d 5 8
V5 p
r2
h
5p?
82
? 9
5
576
16 ft
Use the formula for the volume of a cylinder.
8m
b. Find its volume to the nearest tenth of a cubic meter.
804.2 m 3
2. Find the volume of the triangular prism.
150 m 3
10 m
5m
Substitute.
6m
p
Simplify.
The volume of the cylinder is
ft3.
576π
Geometry Lesson 11-4
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 11-5
Daily Notetaking Guide
of the oblique cone in terms of π.
1
r5
d 5 3 in.
2
11 in.
Local Standards: ____________________________________
Key Concepts.
height of the pyramid
and the
.
1Bh
3
h
B
Theorem 11-9: Volume of a Cone
All rights reserved.
5
All rights reserved.
area of the base
and the
h
33
Use the formula for the volume of a cone.
11
)
p
Substitute 3 for r and
11
6 in.
for h.
Simplify.
The volume of the cone is
33p
in.3.
Quick Check.
1. Find the volume of a square pyramid with base edges 24 m and slant height 13 m.
960 m 3
The volume of a cone is one third the product of the
area of the base
r2
5 13 p( 3 2)(
Theorem 11-8: Volume of a Pyramid
The volume of a pyramid is one third the product of the
143
2 Finding Volume of an Oblique Cone Find the volume
NAEP 2005 Strand: Measurement
Topic: Measuring Physical Attributes
V 5 13 p
height of the cone
.
h
1pr 2h
3
V 13Bh, or V Geometry Lesson 11-4
Name_____________________________________ Class____________________________ Date ________________
Volumes of Pyramids and Cones
Lesson Objectives
1 Find the volume of a pyramid
2 Find the volume of the cone
V
16 m
256p m3
All rights reserved.
V Bh, or V area of a base
.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
height of the cylinder
and the
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 11-7: Volume of a Cylinder
The volume of a cylinder is the product of the
142
V B
All rights reserved.
All rights reserved.
area of a base
.
(
.
volume
Theorem 11-6: Volume of a Prism
The volume of a prism is the product of the
2
29
where one leg
altitude
and the other leg is the
21
The area B of the base is 21bh 12 20
area of the base to find the volume of the prism.
Theorem 11-5: Cavalieri’s Principle
If two space figures have the same height and the same cross sectional area
at every level, then they have the same
base
29 m
with triangular bases. The
right triangle
base of the triangular prism is a
Vocabulary and Key Concepts.
and the
right triangular prism
The prism is a
NAEP 2005 Strand: Measurement
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
2
Name_____________________________________ Class____________________________ Date________________
r
Examples.
1 Finding Volume of a Pyramid Find the volume of a square pyramid with
base edges 15 cm and height 22 cm.
Because the base is a square, B V 5 13 B
13
V5
(
h
225
)(
22
1650
)
36
?
15
.
225
Substitute
225
for B and
22
for h.
Simplify.
The volume of the square pyramid is
144
15
Use the formula for volume of a pyramid.
Geometry Lesson 11-5
Geometry
1650
cm3.
Daily Notetaking Guide
L1
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
B
2. A small child’s teepee is in the shape of a cone 6 ft tall and 7 ft in diameter.
Find the volume of the teepee to the nearest cubic foot.
77 ft 3
L1
Daily Notetaking Guide
Geometry Lesson 11-5
All-In-One Answers Version B
145
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Date ________________
Lesson 11-6
Examples.
Surface Areas and Volumes of Spheres
Lesson Objective
1
Name_____________________________________ Class____________________________ Date________________
1 Finding Surface Area The circumference of a rubber ball is 13 cm.
NAEP 2005 Strand: Measurement
Find the surface area and volume
of a sphere
Calculate its surface area to the nearest whole number.
Topic: Measuring Physical Attributes
Step 1 First, find the radius.
Local Standards: ____________________________________
C 5 2pr
13
Vocabulary and Key Concepts.
13
Theorem 11-10: Surface Area of a Sphere
2p
square
S.A. of the
radius
r
All rights reserved.
the
of the sphere.
4pr 2
Theorem 11-11: Volume of a Sphere
Substitute
5r
Solve for r.
S.A. 5 4pr2
5 4p ?
The volume of a sphere is four thirds the product of p and
the
of the
cube
radius
of the sphere.
for C.
Use the formula for surface area of a sphere.
(
13
2p
)
2
13
Substitute
for r.
2p
169
5
r
4pr 3
3
S.A. 13
Step 2 Use the radius to find the surface area.
All rights reserved.
The surface area of a sphere is four times the product of p and
Use the formula for circumference.
5 2pr
Simplify.
p
53.794371
<
Use a calculator.
To the nearest whole number, the surface area of the rubber ball is
All rights reserved.
diameter
of a sphere is the given point
from which all points on the sphere are equidistant.
r
The radius of a sphere is a segment that has one endpoint at
the center and the other endpoint on the sphere.
diameter
The
d
of a sphere is a segment passing
circumference
through the center with endpoints on the sphere.
sphere
A great circle is the intersection of a sphere and a plane
containing the center of the sphere.
great circle
The
divides a sphere into two
Leave your answer in terms of π.
V 5 34 p
5
15
4500
Use the formula for volume of a sphere.
Substitute r ⴝ 30 ⴝ
2
3
p
15
.
Simplify.
4500p
The volume of the sphere is
cm3.
Quick Check.
1. Find the surface area of a sphere with d 14 in. Give your answer in two
ways, in terms of π and rounded to the nearest square inch.
196p in.2, 616 in.2
Great
2. Find the volume to the nearest cubic inch of a sphere with diameter 60 in.
circle
The circumference of a sphere is the circumference of its
113,097 in.3
great circle.
Geometry Lesson 11-6
Daily Notetaking Guide
L1
Daily Notetaking Guide
L1
Name_____________________________________ Class____________________________ Date________________
Geometry Lesson 11-6
147
Name_____________________________________ Class____________________________ Date ________________
2 Finding the Similarity Ratio Find the similarity ratio of two similar
Lesson 11-7
cylinders with surface areas of 98π ft 2 and 2π ft 2.
Use the ratio of the surface areas to find the similarity ratios.
Areas and Volumes of Similar Solids
Lesson Objective
1 Find relationships between the ratios
of the areas and volumes of similar
solids
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
r3
5 43p ?
Hemispheres
congruent hemispheres.
146
cm2.
m
30 c
center
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
center
given point.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
A sphere is the set of all points in space equidistant from a
The
54
2 Finding Volume Find the volume of the sphere.
NAEP 2005 Strand: Measurement
Topic: Systems of Measurement
a2
2
Vocabulary and Key Concepts.
b
Theorem 11-12: Areas and Volumes of Similar Solids
If the similarity ratio of two similar solids is a : b, then
5
(2) the ratio of their volumes is a3 : b3 .
All rights reserved.
All rights reserved.
a
b5
(1) the ratio of their corresponding areas is a2 : b2 , and
98p
a2 5
b2
Local Standards: ____________________________________
The ratio of the surface areas is
a2
:
b2
.
2p
49
Simplify.
1
7
Take the
square root
of each side.
1
The similarity ratio is 7 : 1 .
Quick Check.
1. Are the two cylinders similar? If so, give the similarity ratio.
Similar solids have the same shape and all of their corresponding parts are proportional.
yes; 6 : 5
The
10
12
of two similar objects is the ratio of their corresponding linear
similarity ratio
6
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Examples.
1 Identifying Similar Solids Are the two solids similar? If so, give the
similarity ratio.
26 in.
8 in.
9 in.
3 in.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
dimensions.
5
2. Find the similarity ratio of two similar prisms with surface areas 144 m 2 and 324 m 2.
2:3
Both figures have the same shape. Check that the ratios of the
corresponding dimensions are equal.
The ratio of the radii is
The cones are
148
L1
3
9
not similar
Geometry Lesson 11-7
, and the ratio of the heights is
because
8
3
9 ⬆ 26
8
26
.
.
Daily Notetaking Guide
All-In-One Answers Version B
L1
L1
Daily Notetaking Guide
Geometry Lesson 11-7
Geometry
149
37
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 12-1
Applying Tangent Lines A belt fits tightly around
two circular pulleys, as shown. Find the distance
between the centers of the pulleys. Round your
answer to the nearest tenth.
NAEP 2005 Strand: Geometry
Use the relationship between a radius
and a tangent
Use the relationship between two
tangents from one point
2
Example.
Tangent Lines
Lesson Objectives
1
Name_____________________________________ Class____________________________ Date ________________
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
Y
O
Vocabulary and Key Concepts.
P
D
Z
15 cm
Because opposite sides of a rectangle have the same measure,
DW 3 cm and OD A
If a line is tangent to a circle, then the line is perpendicular to the radius drawn
P
to the point of tangency.
* )
B
O
AB ' OP
All rights reserved.
All rights reserved.
Theorem 12-1
Theorem 12-2
the line is tangent to the circle
.
AB is tangent to 䉺 O.
right
of a
supplement
right
Because the radius of 䉺 P is 7 cm, PD 7⫺3⫽4
OD2 PD2 OP2
2
angle,
triangle.
cm.
Pythagorean Theorem
4 2 OP2
Substitute.
OP2
241
W
cm.
⬔ODP is also a right angle, and 䉭OPD is a
B
O
15
Because ⬔ODP is the
15
A
P
If a line in the plane of a circle is perpendicular to a radius at its endpoint
* )
7 cm
3 cm
Draw OP. Then draw OD parallel to ZW to form rectangle
ODWZ. Because OZ is a radius of 䉺 O, OZ 3 cm.
on the circle, then
15 cm
X
Simplify.
15.524175
OP Use a calculator to find the square root.
The distance between the centers of the pulleys is about
15.5
cm.
congruent
B
.
O
AB > CB
C
A tangent to a circle is a line, segment, or ray in the plane of the circle that intersects the
circle in exactly one point.
The point of tangency is the point where a circle and a tangent intersect.
A
tangent
B
point of tangency
Quick Check.
A belt fits tightly around two circular pulleys, as shown. Find the distance
between the centers of the pulleys.
All rights reserved.
are
A
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
The two segments tangent to a circle from a point outside the circle
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Theorem 12-3
35 in.
14 in.
8 in.
about 35.5 in.
A triangle is inscribed in a circle if all vertices of the triangle lie on the circle.
A triangle is circumscribed about a circle if each side of the triangle is tangent
to the circle.
Geometry Lesson 12-1
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 12-2
⬔AOX AX BX
lBOX
congruent
All rights reserved.
congruent
congruent
chords.
arcs.
central angles.
All rights reserved.
congruent
Theorem 12-4
Within a circle or in congruent circles
(3) Congruent arcs have
(1) Chords equidistant from the center are
equidistant
.
congruent
congruent
central angles have
X
O
chords.
congruent
chords have
arcs.
B
B
2 Using Theorem 12-5 Find AB.
Theorem 12-5
Within a circle or in congruent circles
(2) Congruent chords are
by the definition of an angle bisector.
because
0
0
AX BX because
(2) Congruent chords have
A
What can you conclude?
Local Standards: ____________________________________
congruent
151
Examples.
1 Using Theorem 12-4 In the diagram, radius OX bisects ⬔AOB.
NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
Vocabulary and Key Concepts.
(1) Congruent central angles have
Geometry Lesson 12-1
Name_____________________________________ Class____________________________ Date ________________
Chords and Arcs
Lesson Objectives
1 Use congruent chords, arcs, and
central angles
2 Recognize properties of lines through
the center of a circle
Daily Notetaking Guide
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
150
QS QR RS
Segment Addition Postulate
QS 7 7
Substitute.
QS 14
Simplify.
AB QS
Chords that are equidistant from the center of a circle
are congruent.
AB 14
Substitute
4
C
4
A
Q
7
14
R
7
S
for QS.
from the center.
Theorem 12-7
In a circle, a diameter that bisects a chord (that is not a diameter) is
perpendicular
to the chord.
Theorem 12-8
In a circle, the perpendicular bisector of a chord contains the
of the circle.
A chord is a segment whose endpoints are on a circle.
center
P
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Quick Check.
Theorem 12-6
In a circle, a diameter that is perpendicular to a chord bisects the
chord
arcs
and its
.
1. If you are given that BC DF in the circles, what can you conclude?
B
0 0
lO O lP; BC O DF
O
D
P
C
F
䉺O 䉺P
2. Find the value of x in the circle at the right.
16
18
18
16
x
36
Q
O
152
38
Geometry Lesson 12-2
Geometry
Daily Notetaking Guide
L1
L1
Daily Notetaking Guide
Geometry Lesson 12-2
All-In-One Answers Version B
153
L1
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 12-3
1
x 12 mDEF
Topic: Relationships Among Geometric Figures
Local Standards: ____________________________________
(
x 12
Vocabulary and Key Concepts.
half the measure of its intercepted arc.
B
1 mAC
2
C
)
The measure of an angle formed by a tangent and a chord is
B
half the measure of the intercepted arc.
1 1
2mBDC
m⬔C B
D
Substitute.
x°
y 12
C
C
(
right
congruent
.
angle.
3. The opposite angles of a quadrilateral inscribed in a circle are
An inscribed angle has a vertex that is on a
supplementary
A
Intercepted arc
circle and sides that are chords of the circle.
.
B
C
Inscribed angle
intercepted arc
is an arc with endpoints on
the sides of an inscribed angle and its other
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Corollaries to the Inscribed Angle Theorem
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
y
1. Two inscribed angles that intercept the same arc are
0
m FG
70
G
Inscribed Angle Theorem
0
(
y 12 m EF
D
2. An angle inscribed in a semicircle is a
Arc Addition Postulate
D y°
)
70
1
y 12 mEFG
Theorem 12-10
All rights reserved.
80
75
Simplify.
90°
0
1
Because EFG is the intercepted arc of ⬔D, you need to find mFG in order
1
to find mEFG . The arc measure of a circle is 360, so
0
mFG 360 70 80 90 120 .
All rights reserved.
All rights reserved.
A
The measure of an inscribed angle is
m⬔B (
C
x
Theorem 12-9: Inscribed Angle Theorem
An
0
m EF
F
80°
Inscribed Angle Theorem
0
x 12 m DE
70°
E
Using the Inscribed Angle Theorem Find the values of x and y.
NAEP 2005 Strand: Geometry
Find the measure of an inscribed
angle
Find the measure of an angle formed
by a tangent and a chord
2
Example.
Inscribed Angles
Lesson Objectives
1
Name_____________________________________ Class____________________________ Date ________________
120
)
Arc Addition Postulate
)
Substitute.
95
Simplify.
Quick Check.
In the Example, find m⬔DEF.
105
points in the interior of the angle.
154
Geometry Lesson 12-3
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Lesson Objectives
1 Find the measures of angles formed
by chords, secants, and tangents
2 Find the lengths of segments
associated with circles
0
Let mXY x.
1
Then mXAY NAEP 2005 Strand: Geometry
Topic: Relationships Among Geometric Figures
[(
12 (
12
72
72
Key Concepts.
intercepted arcs.
m⬔1 All rights reserved.
x°
y°
1
1 (x 1 y)
2
(2) intersect outside a circle is half the
difference of the measures of the
x°
y° 1
y° 1
x°
All rights reserved.
x
Theorem 12-11
The measure of an angle formed by two lines that
A
)
2x)
360 ⫺ x
360
Theorem 12-11(2)
x
]
Simplify.
72
180
72
180
Add x to both sides.
x
108
Solve for x.
108⬚
x
Substitute.
Distributive Property.
arc will be in the advertising agency’s photo.
2 Finding Segment Lengths A tram travels from
50 ft
point A to point B along the arc of a circle with a
radius of 125 ft. Find the shortest distance from
point A to point B.
y° 1
x°
155
x.
360
1
0
m⬔T 12 (mXAY mXY )
Local Standards: ____________________________________
(1) intersect inside a circle is half the sum of the measures of the
Geometry Lesson 12-3
Name_____________________________________ Class____________________________ Date ________________
Angle Measures
and Segment Lengths
Lesson 12-4
Daily Notetaking Guide
A
x
B
x
intercepted arcs.
m⬔1 The perpendicular bisector of the chord AB contains
1 (x 2 y)
2
Theorem 12-12
For a given point and circle, the product of the lengths of the two segments from
the point to the circle is constant along any line through the point and circle.
a c
b
P
d
abcd
III.
II.
x
w
z
y
t
P
z
(w x)w (y z)y
(y z)y t 2
Examples.
1 Finding Arc Measures An advertising agency wants a frontal photo of a
“flying saucer” ride at an amusement park. The photographer stands at the
vertex of the angle formed by tangents to the “flying saucer.” What is the
measure of the arc that will be in the photograph? In the diagram,0
the
)
)
photographer
1stands at point T. TX and TY intercept minor arc XY and
major arc XAY.
156
L1
Geometry Lesson 12-4
y
P
A
125 250 ft. The length of
2
the other segment along the diameter is
250 ft 50
xx
50
x2 x
ft, or 200 ft.
200
10,000
100
Theorem 12-12(1)
Multiply.
Solve for x.
The shortest distance from point A to point B is 2x or
200
ft.
Quick Check.
1. Find the value of w.
w°
250
2. Find the value of x to the nearest tenth.
Y
X
110°
70°
20
13.8
x
72°
Daily Notetaking Guide
All-In-One Answers Version B
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
diameter is
I.
200 ft
the center of the circle. Because the radius is 125 ft, the
14
16
T
L1
L1
Daily Notetaking Guide
Geometry Lesson 12-4
Geometry
157
39
Geometry: All-In-One Answers Version B (continued)
Name_____________________________________ Class____________________________ Date________________
Lesson 12-5
Lesson Objectives
3 Graphing a Circle Given Its Equation Find the center and radius of the
circle with equation (x 4) 2 (y 1) 2 25. Then graph the circle.
(x 4)2 (y 1)2 25
(x 4)2 (y 1)2 (5)2
NAEP 2005 Strand: Geometry
Write an equation of a circle
Find the center and radius of a circle
Topic: Position and Direction
Local Standards: ____________________________________
c
c
c
h
k
r
The center is
Key Concepts.
y
Theorem 12-13
r
(h, k)
O
x
2
8 6 4 2 O
6
8
x
(x 8
0 )2 "5
(
)2 y 2 Quick Check.
Standard form
)2
5
1. Write the standard equation of the circle with center (2, 1) and radius "2 .
Substitute (ⴚ8, 0) for (h, k) and
(x ⫹ 2) 2 ⫹ (y ⫹ 1) 2 ⫽ 2
Á 5 for r.
Simplify.
2 Using the Center and a Point on a Circle Write the standard equation of
a circle with center (5, 8) that passes through the point (15, 13).
First find the radius.
2(
h )
k x y )2
r #(
Use the Distance Formula to find r.
ⴚ15
5 8
#(
)2 ( ⴚ13
)2
2
)
ⴚ20
ⴚ21
#(
( )2
Substitute (5, 8) for (h, k) and (ⴚ15, ⴚ13) for (x, y).
Simplify.
441
#400
#841
29
Then find the standard equation of the circle with center (5, 8) and radius
(x h )2 (y k )2 r 2
(x 5 )2 (y 8 )2 (x 5 ) 2 (y 8 ) 2 (
29
841
29
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
)] 2 (y © Pearson Education, Inc., publishing as Pearson Prentice Hall.
ⴚ8
(x ⫺ 2) 2 ⫹ (y ⫺ 3) 2 ⫽ 13
y
3. Find the center and radius of the circle with equation
(x 2) 2 (y 3) 2 100. Then graph the circle.
16
center: (2, 3); radius: 10
8
O
.
)2
(2, 3)
4
x
ⴚ8
Standard form
Substitute (5, 8) for (h, k) and
29
for r.
Simplify.
Geometry Lesson 12-5
Daily Notetaking Guide
L1
L1
Name_____________________________________ Class____________________________ Date________________
Lesson 12-6
Daily Notetaking Guide
Geometry Lesson 12-5
159
Name_____________________________________ Class____________________________ Date ________________
2 Describing a Locus in Space Describe the locus of points in space that are
Locus: A Set of Points
Lesson Objective
1 Draw and describe a locus
2. Write the standard equation of the circle with center (2, 3) that passes
through the point (1, 1).
All rights reserved.
(x h )2 (y k )2 r 2
158
4
4
1 Writing the Equation of a Circle Write the standard equation of a circle
with center (8, 0) and radius "5.
(
2
2
Examples.
[x y
4
All rights reserved.
(h, k)
and radius r is (x h) 2 (y k) 2 r 2.
and the radius is 5 .
6
All rights reserved.
The standard form of an equation of a circle with center
(⫺4, 1)
(x, y)
Relate the equation to the standard form
(x ⴚ h)2 ⴙ (y ⴚ k)2 ⴝ r 2.
4 cm from a plane M.
NAEP 2005 Strand: Geometry
Topic: Dimension and Shape
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1
2
Name_____________________________________ Class____________________________ Date ________________
Circles in the
Coordinate Plane
Local Standards: ____________________________________
Vocabulary.
44 cm
A locus is a set of points, all of which meet a stated condition.
M
Examples.
1 Describing a Locus in a Plane Draw and describe the locus of points in a
All rights reserved.
All rights reserved.
44 cm
艎
Imagine plane M as a horizontal plane. Because distance is measured along
a perpendicular segment from a point to a plane, the locus of points in
plane that are 1.5 cm from 䉺 C with radius 1.5 cm.
Use a compass to draw 䉺 C with radius 1.5 cm.
space that are 4 cm from a plane M are a plane 4 cm above and parallel
to plane M and another plane 4 cm below and parallel to plane M.
Quick Check.
*
)
1. Draw and describe the locus: In a plane, the points 2 cm from line XY.
cm
5
1.
cm
The locus of points in the interior of 䉺 C that are 1.5
the center C.
cm from 䉺 C is
The locus of points exterior to 䉺 C that are 1.5 cm from 䉺 C is a circle with
center C and radius 3 cm.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
5
1.
C
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
* )
2 cm
X
40
Geometry Lesson 12-6
Geometry
Daily Notetaking Guide
Y
2 cm
2. Draw and describe the locus of points in space that are equidistant from
two parallel planes.
A plane midway between the two parallel planes,
parallel to and equidistant from each.
The locus of points in a plane that are 1.5 cm from a point on 䉺 C with
radius 1.5 cm is point C and a circle with radius 3 cm and center C.
160
* )
Two lines parallel to XY , each 2 cm from XY .
A
L1
L1
Daily Notetaking Guide
B
Geometry Lesson 12-6
All-In-One Answers Version B
161
L1
Data Analysis and Probability Workbook Answers
Section 1 Graphs
8. Hours Spent Doing Homework
1.
Number of
1
2
3
4
5
6
Tally
|
||||
|||
|||||
|
|
Frequency
1
4
3
6
1
1
Frequency
page 1 Frequency Tables, Line Plots, and Histograms
Rose Bushes
2.
1–
2–1.75
3–2.75
4–3.75
5–4.75
6–5.75
7–6.75
8–7.75
8.
75
Rose Bushes
Wanted
8
7
6
5
4
3
2
1
✗
✗
✗
Number of Hours
✗
page 3 Reteaching 1: Frequency Tables and Line Plots
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
1
2
3
4
5
6
1. Inches
3 4 5 6 7
5 3 1 1 2
Frequency
Charleston Rainfall
✗
✗
✗ ✗
✗ ✗
✗
✗ ✗ ✗ ✗ ✗
3. Answers may vary. Sample:
© Pearson Education, Inc. All rights reserved.
Frequency
Rose Bushes
Wanted
3
9
8
7
6
5
4
3
2
1
0
2. Inches
5
6
7
0 1 2 3 4
5 2 0 4 1
Frequency
San Francisco Rainfall
✗
✗
✗
✗ ✗
✗ ✗
1–2 3–4 5–6
Number of Rose Bushes
0
page 2 Practice: Frequency Tables, Line Plots, and
Histograms
1.
4
1
3. Inches
Frequency
✗
✗
✗
✗ ✗
2
3
4
3 4
4 8
Boxes of Juice Sold
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
26
27
✗
✗
✗
✗
21
22
23
24
25
2. a student 3. 3 students 4. 13 students 5. 80 and 85, 75
and 90 6. No; the interval includes 2–2.75 h. 7. 10 students
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
✗
3
4
page 4 Reteaching 2: Frequency Distributions
1.
Interval
95–100
90–94
85–89
80–84
75–79
70–74
under 70
Data Analysis and Probability Teacher’s Guide
Wilmington Rainfall
Tally
Frequency
2
4
2
7
4
4
2
Answers
11
Data Analysis and Probability Workbook Answers
Interval
Tally
Frequency
60–79
80–99
100–119
120–139
140–159
160–179
180 and over
4
7
6
2
4
1
2
Tally
4
8
4
4
Thorpe
4. Daily Use of Petroleum in the U.S.
2.5
Frequency
||||
|||| |||
||||
||||
25 – 29
30 – 34
35 – 39
40 – 44
Ali
Sports Figure
page 5 Reteaching 3: Histograms
1. Answers may vary. Sample:
Number of
Raisins
30
25
20
15
10
5
0
King
3
4
3
5
4
2
3
Robinson
2.0
1.5
1
0.5
0
Year
5. Answers may vary. Sample: Daily petroleum use has been
declining since 1970.
8
7
6
5
4
3
2
1
0
page 8 Stacked Bar and Multiple Line Graphs
1. grade 7 2. grade 7; grade 8 3. Sample: Find the total,
subtract the bottom part. 4. 1993 5. Find the number of
each item sold and add the two.
page 9 Practice: Stacked Bar and Multiple Line Graphs
44
40
–
34
39
35
–
25
–
29
1.
30
–
Frequency
2.
Average Viewing Time
8:00 P.M.–11:00 P.M. (Mon. – Sun.)
Males
Females
55
Number of Raisins
page 6 Bar and Line Graphs
1. tens 2. fives 3. thousands 4. hundreds 5. fives
6. tens 7. Bar graph; data shows amounts, but not changes
over time. 8. Line graph; data shows change over time.
page 7 Practice: Bar and Line Graphs
1. names of the athletes 2. Answers may vary. Sample: 5
12
Answers
25 – 54
18 – 24
14 12 10 8 6 4 2 0 2 4 6 8 10 12 14
Time (Hours)
Data Analysis and Probability Teacher’s Guide
© Pearson Education, Inc. All rights reserved.
3.
All-Time Favorite Sports Figures
Zaharias
700–749
750–799
800–849
850–899
900–949
950–999
1000 and over
3.
Ruth
Frequency
196
0
196
5
197
0
197
1985
1980
5
199
1990
2005
0
Tally
Number of Votes
Interval
Millions of Barrels
2.
;;
;
;
;;
;;
;
;;;;
;;
;
;;
Data Analysis and Probability Workbook Answers
Average Viewing Time
8:00 P.M. – 11:00 P.M. (Mon. – Sun.)
Time (Hours)
2.
15
12
9
6
3
0
18 – 24
25 – 54
Ages
Male
Misc.
Gas
Food
10.
Children per
Family
Population (100,000s)
Population
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
6.
Population
Population (100,000s)
Clothes
Rent
Female
1960 1970 1980 1990 2000
Year
Hialeah, FL
Birmingham, AL
© Pearson Education, Inc. All rights reserved.
Budget
Phone
55
3. double bar graph 4. sliding bar graph
5.
9.
5.0
4.5
4.0
3.5
3.0
2.5
2.0
1.5
1.0
0.5
1960 1970 1980 1990 2000
Year
Hialeah, FL
Birmingham, AL
1
2
0
5
4
3
page 11 Practice: Circle Graphs
1.–4. Sample graph shown.
Category
Amount
Budgeted
Percent of Degrees in
Total
Central Angle
1.
Gas
$200
10%
36°
2.
Meals
$400
20%
72°
3.
Motels
$500
30%
108°
4.
Other
$800
40%
144°
Rahman Family Budget
Gas
10%
Meals
20%
Other
40%
Motels
30%
7. multiple line graph
page 10 Circle Graphs
1. 72° 2. 144° 3. 180° 4. 18° 5. 126° 6. 36° 7. 72°
8. 30°
Data Analysis and Probability Teacher’s Guide
Answers
13
Data Analysis and Probability Workbook Answers
5–8. Sample graph shown.
Category
Amount
Budgeted
Percent of
Total
Degrees in
Central Angle
5.
Clothing
$50
40%
36°
6.
Entertainment
$40
32%
72°
7.
Savings
$25
20%
108°
8.
Transportation
$10
8%
144°
Transportation
8%
Savings
20%
Milk Drunk
not
sure less
13.1% 10.1%
more
56.3%
the
same
20.5%
page 12 Reteaching: Circle Graphs
1. 97° 2. 86° 3. 47° 4. 65° 5. 65°
page 13 Stem-and-Leaf Plots
1. Times for the 200-M Dash
38
37
36
35
34
33
32
31
30
29
68
569
03447
568
3 8 means 38
56
3
0479
23
4777
4
1225
479
2
35
7 01459
8 1226
9 0169
9 9 means 9.9
5. 8.1; 8.2 and 6.3; 3.6
6. 4 17 36 70
5
6
7
8
21 26 86
34 75 92
19
17
4 17 means 417
6. 586; no mode; 400
7. 17 4 6 9
18 5 5 6
19 4 5
20 4
17 4 means 17.4
7. 18.5; 18.5; 3
8. 15° 9. 64° 10. 69° 11. 76; 75; 79; 80
page 15 Reteaching: Stem-and-Leaf Plots
1. 1 5 6
2. 64; 15; 49
2
3
4
5
6
47
669
25
149
134
1 5 means 15
3. 14 7 8 9
8 234456679
9 012
7 8 means 78
38 8 means 38.8
4. 45 4
5
6
7
8
013
5678
235666
114
1235
8 1 means 81
2. 34.7 s 3. 9.3 s 4. 34.7 s 5. 16 students
14
Answers
Data Analysis and Probability Teacher’s Guide
© Pearson Education, Inc. All rights reserved.
Entertainment
32%
9.
4. 0
1
2
3
4. 23; 24; 32
5. 6 3 3
Lucy’s Budget
Clothing
40%
page 14 Practice: Stem-and-Leaf Plots
1. 10, 11, 12, 13 2. 4, 5, 9, 9 3. 115; 104 and 115; 38
Data Analysis and Probability Workbook Answers
page 16 Activity: Relating Stem-and-Leaf Plots to
Histograms
16 a. Stem
3
4
5
6
7
b.
27
4467
78
1348
69
5
10
15
7.
60
65
70
75
80
85
90
8.
0
Grouping Intervals
Frequency
30–39
40–49
50–59
60–69
70–79
2
4
2
4
2
1st
5 10 15 20 25 30 35 40 45 50
Set
2nd Set
page 20 Reteaching: Box-and-Whisker Plots
1. 1 2 2 2 2 2 3 3 3 3 3 3 3 3 4 4 5 5 6 7 7 7 8 8 8 9 9 10 11 13
2. median 4 3. lower median 3; upper median 8
4.
4
1 2
3 4
1
3 4
5 6
7 8
9 10 11 12 13
8
13
3
2
1
0
30–39 40–49 50–59 60–69 70–79
Score
page 17 Puzzle: Keeping Score
© Pearson Education, Inc. All rights reserved.
0
Leaf
Frequency Table
Number of Teams
c.
6.
Stem
5
6
7
8
7
Leaf
.75, .90
.35, .45, .75
.20, .50, .50, .40, .75
.75, .75, .75
.20 means 7.20
90
110
130
150
170
2. 13, 4, 21
0
10
20
30
40
page 19 Practice: Box-and-Whisker Plots
1. 55 miles, 15 miles 2. 35 miles 3. 75% 4. 6 runners
5.
10
15
20
page 22 Practice: Reading and Understanding Graphs
1. oxygen 2. oxygen, carbon, hydrogen 3. There are other
elements in quantities too small to be labeled individually.
4. Western Europe 5. South and Central America, Middle
East, Africa 6. North America 7. A greater percentage of
men are reaching ages 25–29 without having married.
8. 1960 to 1970 9. C
page 23 Reading Graphs Critically
1. the first graph 2. about 3 times 3. The second graph;
since the scale is smaller, the bars can be read more accurately.
4. By using the break, most of the bar for the United States
has been left out.
page 18 Box-and-Whisker Plots
1. 98, 80.5, 118
70
page 21 Reading and Understanding Graphs
1. 20 books 2. January 3. 10 books 4. October and
November 5. English and Spanish 6. 12 people
7. 12 people 8. Polish
25
30
Data Analysis and Probability Teacher’s Guide
page 24 Practice: Reading Graphs Critically
1. 45–64 year olds 2. 45–64 year olds 3. Answers may vary.
Check students’ graphs. 4. Answers may vary. 5. Answers
may vary. Check students’ graphs. 6. Answers may vary.
Check students’ graphs. 7. Answers may vary. 8. Answers
may vary. Sample: to convince people that one program is
more popular than it is
page 25 Activity: Organizing and Analyzing Data
1. no; too wide a range of data 2. D.C., N.J., and R.I.
3. possible answer: 100 people per square mile 4. People
might assume that the two places have similar densities.
5. You could list the three places individually. 6. nothing
Answers
15
Data Analysis and Probability Workbook Answers
page 26 Scatter Plots and Trends
page 29 Activity 2: Making a Scatter Plot
1.
1.
Test Scores and TV
Latitude (°N)
100
90
80
70
50
30
20
2. Negative; as one value goes up, the other goes down.
3. The more TV students watch, the lower their test scores.
page 27 Practice: Scatter Plots and Trends
1. positive trend 2. negative trend 3. no trend
4. Students’ graphs should show a negative trend. Sample:
Residents of Maintown
1,200
1,150
1,100
1,050
1,000
950
900
70
50
40
30
20
7a.
ERA
5.00
4.00
3.00
60
70
80 90 100
Wins
b. negatively correlated
page 31 Reteaching: Scatter Plots and Trends
1.
6. Answers may vary. Sample: time spent skiing down a hill vs.
speed of skier
page 28 Activity 1: Making a Scatter Plot
1–2. Check student’s work. 3–4. Check students’ work.
There should be a positive correlation.
Answers
Height in Inches
5 10 15 20 25 30 35 40 45 50
Time Spent Studying (minutes)
Weight and Height Survey
70
60
50
40
30
30 40 50 60 70 80 90 100 110
Weight in Pounds
2. about 61 in. 3. about 63 lb 4. yes 5. Sample: As height
increases, weight increases.
Data Analysis and Probability Teacher’s Guide
© Pearson Education, Inc. All rights reserved.
Studying Time vs.
Number of Correct Words
22
20
18
16
14
12
10
8
6
4
2
0
2.0 3.0 4.0 5.0 6.0
Precipitation (in.)
page 30 Analyzing Scatter Plots
1. negatively correlated 2. unrelated 3. positively
correlated 4. positively correlated 5. negatively correlated
6. unrelated
5. Students’ graphs should show a positive trend. Sample:
Number of Words
Spelled Correctly
40 50 60
Temp (°F.)
4. no
1,800 1,900 2,000 2,100
Homeowners
16
30
3.
0 1 2 3 4 5
Hours of TV
Vacation Homeowners
40
60
Latitude (°N)
Test Scores
2. yes, no
50
Data Analysis and Probability Workbook Answers
page 32 Assessment 1
1. 34 2. biking; 12 students 3. sailing and hiking;
4 students 4. 24% 5. Because students picked activities
besides the ones listed, but too few students picked those
activities for each activity to get a separate bar.
6. 100, 50, no mode
7.
0
20
0
20
40
60
50
80
100
80
100
Section 2 Measures of Central Tendency
page 33 Measures of Central Tendency
1. mean: 3; median: 3; mode: 4 2. mean: 8.1; median: 8;
mode 5 3. mean: 5.25; median: 5; mode 0, 1 4. mean: $3.25;
median: $3.25; mode: $4 5. Median; there is no mode and the
median is greater than the mean. 6. Mode; both the median
and the mean are less than the mode.
page 34 Practice 1: Measures of Central Tendency
1. 1.5; 2; 2 2. 3.3; 1; 0 3. 29.1; 29.5; 25, 30, and 35
4. 8.1; 9; 9 5. 73.6; 72; no mode 6. 153.2; 146; no mode
7. Mode; data are nonnumerical. 8. Mode; data are
nonnumerical. 9. Mean; there should be no outliers.
10. Mean or median; use median if there are outliers.
11. 92 12. at least an 87 13. 85 14. 86
© Pearson Education, Inc. All rights reserved.
page 35 Practice 2: Mean, Median, Mode, and Outliers
1. 23 students 2. at least 12 students; up to 11 students
3. mean: 287.5, median: 300, mode: 200 4. 50 5. lower
6. The median, because there is an outlier. 7. 8 points
8. 23 points 9. Answers will vary.
page 36 Activity 1: Choosing an Appropriate Measure
1. Check students’ work. 2. a, b, c, d, g 3. all of them,
where the mode exists 4–6. Check students’ work.
page 37 Activity 2: Average Temperature
1. Stem
Leaf
5
8
6
0233
7
0444555667778888899
8
000011223455666777888899999
9
016
10
page 38 Activity 3: Wink Count
1–6. Answers depend upon class data. EXTRA 7. They
must score at least 12 runs during the 3 games. One way to find
the different ways in which this can be done is to make a chart.
Game 1 Game 2 Game 3
12
0
0
0
12
0
0
0
12
11
1
0
11
0
1
1
11
0
etc.
There are 91 ways in all in which the additional 12 runs may
be scored.
page 39 Puzzle: Mean, Median, and Mode
1. {2, 4, 5, 7, 7} 2. All should have found the same set.
3. One 4. a. 0 b. 1 c. 0 d. 0 e. 0 5. Several solutions
are possible, including {5, 5, 5, 5, 5}, {4, 5, 5, 5, 6}, {3, 5, 5, 5, 7},
etc. 6. Conditions: 0 a b c d e 20, and
2a d e 4c. Several solutions are possible, including
{0, 0, 0, 20, 20}. 7. Conditions: a b c d e 5c, and
a b c d e mode. One possible solution is
{0, 0, 5, 5, 5}. 8. no solution 9. Answers may vary. Samples:
{0, 0, 0, 20, 20} and {0, 0, 20, 20, 20}
page 40 Reteaching: Mean, Median, and Mode
1. 12, 11.5, none 2. 87¢, 86.5¢, none 3. 231, 231, none
4. 50, 49, 43 5. $130, $129, none
page 41 Assessment 2
1. 3:03 2. 2:04; 1:55 3. 1:58 4. 2:03 5. Answers may vary.
Sample: The median, because it eliminates the effect of the
outlier and gives a representative time. 6. Answers may vary.
Sample: The mode, because it is arbitrary that two skiers
happened to have the exact time. 7. 1:18; the difference
between the fastest and the slowest time. 8. below;
2.3 degrees 9. 41 degrees 10. 63 degrees
Section 3 Use and Misuse of Data
Displays
page 42 Choosing an Appropriate Graph
1. multiple line graph: shows changes in two sets of data over
time 2. circle graph: shows how the club’s budget is divided
into parts 3. double bar graph: compares two sets of data
4. circle graph: shows how 100% is divided into parts
5. line graph: shows change over time
27
2. Answers will vary. 3. mean 81.05; median 80.5;
mode 78 and 89 (bimodal) No city has the mean, although
Columbus and Seattle are close. No city has the median,
although Albany, Pittsburgh, St. Louis, and Salt Lake City lie
just below, and Columbus and Seattle lie just above. The data
is bimodal: Denver, Detroit, Spokane, Topeka, and Tulsa all
have 78; Albuquerque, Portland, Raleigh, Richmond, and San
Antonio all have 89. 4–6. Answers depend upon class data.
Data Analysis and Probability Teacher’s Guide
Answers
17
Data Analysis and Probability Workbook Answers
7
6
5
4
3
2
1
1–5
als irds tiles bians Fish nails eans
m
B ep hi
S tac
m
us
Cr
p
Am
Type
page 44 Activity: Appropriate Graphs
1. line graph 2. bar graph 3. no 4. Bar graph; it compares
amounts.
page 45 Misleading Graphs
1. graph A 2. graph B 3. The years on the horizontal axis
are spaced differently; graph A seems to rise more steeply
than graph B. 4. graph D 5. graph C 6. vertical scale is
unbroken on one, and broken on the other; difference in the
bars seems greater on graph D
7
6
5
4
3
2
1
1–4
10
15–24
25–30
page 48 Using Graphs to Persuade 1
1. no 2. 2.0 to 120 3. Answers may vary. Sample:
•
•
Number of
Clear Days
14
13
11–14
Years
Average Number of
Clear Days per Year
2.
•
5–10
2. Lois’ graph has intervals of 5 years and Harold’s has
intervals of varying numbers of years. Lois’ is more useful.
3. You may wish to have students compare their graphs.
page 46 Practice: Misleading Graphs
1. 15
11–15 16–20 21–25
Years
Harold’s Graph
Teachers’ Years of Service
150
100
50
0
J F M
J F M
3. 2 cars 4. Probably not; car sales vary greatly and
Cities
4. Answers may vary. Sample:
Average Number of
Clear Days per Year
Number of
Clear Days
information from the first three months is not necessarily
indicative of the whole year. 5. The vertical axis changes
units; the graph continues through a break. 6. The number of
people who prefer Yummy Cereal is increasing sharply.
Albany
Atlanta
Boise
Houston
St. Louis
Seattle
R
6–10
125
100
75
50
25
0
Albany
Atlanta
Boise
Houston
St. Louis
Seattle
a
Lois’s Graph
Teachers’ Years of Service
Number of
Teachers
Number
80
70
60
50
40
30
20
10
M
1.
Number of
Teachers
U.S. Endangered Animals
page 47 Activity: Analyzing Graphs
Cities
18
Answers
Data Analysis and Probability Teacher’s Guide
© Pearson Education, Inc. All rights reserved.
page 43 Practice: Choosing an Appropriate Graph
1. private owners 2. about 25,000 people 3. line graph,
shows change over time 4. bar graph, compares quantities
5. scatter plot, shows a relationship between sets of data
6. circle graph, compares parts of a whole 7. bar graph: See
students’ graphs. Sample:
Data Analysis and Probability Workbook Answers
6. U.S. Union Membership
20
18
16
14
12
10
8
6
4
2
0
page 49 Using Graphs to Persuade 2
Union members (millions)
1.60
1.50
1.40
1.30
1.20
1.10
199
6
199
7
199
8
199
9
200
0
200
1
Year
7.
page 50 Practice: Using Graphs to Persuade
1. birds 2. no 3. the break in the vertical axis.
U.S. Endangered Species
Number of Species
© Pearson Education, Inc. All rights reserved.
4.
80
70
60
50
40
30
20
10
Mammals Birds
Type
Fish
5. The differences seem much less.
Data Analysis and Probability Teacher’s Guide
90
19
80
19
70
19
60
40
19
30
3. The first graph implies prices decreased rapidly from 1997
to 1998, then increased rapidly from 1999 to 2000. The second
graph implies slower changes.
20
18
16
14
12
10
8
6
4
2
0
19
Year
U.S. Union Membership
Union members (millions)
1
200
0
200
199
9
8
199
199
199
7
1.60
1.50
1.40
1.30
1.20
1.10
6
Price (dollars)
Gasoline Retail Prices
19
2.
19
3
19 0
4
19 0
50
19
6
19 0
70
19
8
19 0
90
Year
50
Price (dollars)
Gasoline Retail Prices
19
1.
Year
8. The horizontal scales are different.
page 51 Analyzing Games and Making Predictions
1. No; it suggests that Player A wins 3 and Player B wins 5.
2. HTHH, THTT, HTTH, HTHT, HHHT, THHT, TTHT,
TTHT, HTHT, TTTH; Player A wins 4 and B wins 6. 3. No:
Player A wins 7 and Player B wins 13. 4. Yes; in 16 rounds,
both players are likely to win 8 points.
page 52 Practice: Analyzing Games and Making Predictions
1. not fair 2. fair 3. fair 4. not fair 5. Possible answers:
0–1 for correct, 2–9 for incorrect 6. Check students’ work.
11
50 22% if answer shown here for Exercise 5 is used.
7. possible answer: 0–4 for correct, 5–9 for incorrect
29
8. Check students’ work. 50
58% if answer shown here for
Exercise 7 is used. 9. Possible answer: 0–7 for correct, 8–9
for incorrect 10. Check students’ work. 41
50 82%
if answer shown here for Exercise 9 is used.
Answers
19
Data Analysis and Probability Workbook Answers
2. 1st choice 2nd
buttons
small
snaps
buttons
medium
snaps
Lawns Mowed
0 1 2 3 4
Week Number
2. Answers will vary. Sample:
5
4
3
2
1
0
buttons
large
snaps
0
1
2
3
Week Number
Section 4 Counting Principles
page 54 Counting Outcomes
1. 12 2. 15 3. 16 4. 20 5. 12 6. 30
page 55 Practice: Counting Outcomes
3rd choice
choice
milk
pizza
juice
milk
spaghetti
juice
blue
beige
blue
beige
blue
beige
blue
beige
blue
beige
blue
beige
3. 45,697,600 license plates 4. 6,760,000 license plates
5. 24 types 6. 19,656 results 7. 1,000 codes
8. 358,800 passwords
4
3. Answers will vary. Graph A is good for giving the
impression that something is decreasing rapidly. Sample:
showing the change in a bank account over time to explain
why a bigger allowance is needed. 4. Answers will vary.
Graph B is good for giving the impression that something is
decreasing more slowly. Sample: showing the change in a bank
account over time to convince one’s self that a job is needed.
1. 1st choice 2nd
3rd choice
choice
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
pudding
apple
pudding
apple
pudding
apple
pudding
apple
page 56 Permutations
1. 24 2. 6 3. 4 4. 90 5. 1 6. 240 7. 120 8. 20
9. 1,320 10. 24 11. 210 12. 360 13. 6P2 30
14. 10P3 720
page 57 Practice: Permutations
1. 720 2. 479,001,600 3. 362,880 4. 336 5. 79,833,600
6. 15,120 7. 56 8. 1,814,400 9. 120 10. 3,603,600
11. 24 ways 12. 5,040 ways 13. 40,320 orders 14. 362,880
arrangements 15. 68,880 ways 16. 2,730 ways
17. 1,406 ways 18. 151,200 arrangements
19. 17,576 arrangements 20. 3,628,800 arrangements
page 58 Combinations
1. 20 2. 10 3. 21 4. 4 5. 28 6. 15 7. 126 8. 10 9. 6
10. 35 11. 56 12. 84 13. 10C3 120 14. 6C2 15
15. 7C2 21 16. 15C5 3,003 17. 10C5 252
18. 8C3 56
page 59 Practice 1: Combinations
1. 9 2. 70 3. 330 4. 330 5. 1 6. 84 7. 924 8. 28
9. 120 10. 6 11. 70 12. 21 13. 210 ways
14. 490,314 ways 15. 5,005 ways 16. 1,370,754 ways
17. 66 handshakes 18. 10,626 committees 19. B, J; B, E; B,
M; B, X; J, E; J, M; J, X; E, M; E, X; M, X 20. D
page 60 Practice 2: Permutations and Combinations
1. 42 2. 21 3. 336 4. 3,024 5. 3 6. 210 7. a. 24
b. 120 c. 24 d. 51 8. 10 9. 42,840 10. 30 11. 120
12. 360 13. 720 14. 720
page 61 Assessment 4
1. 5040 2. There would be 8 times as many (or 40,320).
3. 15 4. 15 or 20% 5. 60 ways 6. Answers will vary.
Sample: 6P3 could represent how many ways you could make
a 3-digit combination out of the numbers 1 to 6; 6C3 could
represent how many ways you select three students from a
group of six. 7. 24C6, 24P6, and 24!
20
Answers
Data Analysis and Probability Teacher’s Guide
© Pearson Education, Inc. All rights reserved.
Lawns Mowed
page 53 Assessment 3
1. Answers will vary. Sample:
Data Analysis and Probability Workbook Answers
page 67 Practice: Sample Spaces
Section 5 Theoretical Probability
page 62 Theoretical Probability
3
1. 1; 0.2; 20% 2. 5; 0.4; 40% 3. 12, 0.5, 50% 4. 10
, 0.3, 30%
9
7
10
8
3
22
2
5. 10
, 0.9, 90% 6. 11
7. 11
8. 11
9. 25
10. 25
11. 25
page 63 Practice: Theoretical Probability
1
1. 10
; 0.1; 10% 2. 12; 0.5; 50% 3. 1; 1.0; 100% 4. 15; 0.2; 20%
6
8
5. 23
or 34.8% 6. 23
or 26.1% 7. Add a marble that is not
blue. 8. 21 9. 14 10. 41 11. 43 12. 43 13. 21 14. 43 15. 34
997
3
16. 12 17. 21 18. 34 19. 43 20. 1000
, 0.3%, 0.003 21. 1000
,
99.7%, 0.997
page 64 Reteaching: Theoretical Probability
3
1
1
1
1
1. 51 2. 52 3. 51 4. 10
5. 10
6. 0 7. 10
8. 12
9. 12
1
2
3
4
5
6
7
8
9
10
A
A1
A2
A3
A4
A5
A6
A7
A8
A9
A10
B
B1
B2
B3
B4
B5
B6
B7
B8
B9
B10
C
C1
C2
C3
C4
C5
C6
C7
C8
C9
C10
D
D1
D2
D3
D4
D5
D6
D7
D8
D9
D10
E
E1
E2
E3
E4
E5
E6
E7
E8
E9
E10
F
F1
F2
F3
F4
F5
F6
F7
F8
F9
F10
G
G1
G2
G3
G4
G5
G6
G7
G8
G9
G10
H
H1
H2
H3
H4
H5
H6
H7
H8
H9
H10
2a.
10. 61 11. 23 12. 16 13. 0 14. 13
H
page 65 Activity: Finding Errors in Statistical Analysis
1. 2 Incorrect probability for each event
P(even) 5 P(2) 1 P(4) 1 P(6)
T
5
5
1
1
1
6 1 6 1 6
3
1
6 , or 2
2
5
5
1
8
2
10 2 10
6
3
10 , or 5
H
T
H
T
H
T
H
T
H
T
1
21
1
21
2b. 14 3. 12 kinds 4. 32
5. 64
6. 64
7. 64
page 68 Counting Outcomes and Probability
1. 18 possible outcomes 2. C1-S1-F1, C1-S1-F2, C1-S1-F3,
C1-S2-F1, C1-S2-F2, C1-S2-F3, C1-S3-F1, C1-S3-F2, C1-S3-F3,
C2-S1-F1, C2-S1-F2, C2-S1-F3, C2-S2-F1, C2-S2-F2, C2-S2-F3,
C2-S3-F1, C2-S3-F2, C2-S3-F3 3. 31 4. 91
P(,4) 1 P(odd) 2 P(, 4>odd)
5 (P(1) 1 P(2) 1 P(3) 1 P(1) 1 P(3) 1
P(5) 1 P(7) 1 P(9)) 2 P(1) 1 P(3))
1
1
1
1
1
1
1
1
5 Q10
1 10
1 10
1 10
1 10
1 10
1 10
1 10
R
1
Q10
H
T
2. 1 Cannot count the same event twice
© Pearson Education, Inc. All rights reserved.
1.
page 69 Practice: Counting Outcomes and Probability
1. m1A, m1B, m1C, m2A, m2B, m2C, m3A, m3B, m3C, m4A,
1
m4B, m4C 2. 31 3. 12
4. 6 choices: AM, AN, BM, BN, CM,
CN 5. 8 combinations; P1C1; P1C2; P2C1; P2C2; P3C1; P3C1;
P3C2; P4C1, P4C2 6. 140 routes 7. 468 combinations.
1
10 R
median is the middle number once the data set is in order.
9
Mean: 64
11 , or 5 11 Median: 6 Mode: 7
page 70 Independent and Dependent Events
9
3
6
4
1
1
1. 64
32
2. 64
3. 64
16
4. 91 5. 61 6. 12
7. independent 8. independent 9. independent
10. dependent
page 66 Sample Spaces
1. 12
page 71 Practice 1: Independent and Dependent Events
6
1
1
1
1. 51 2. 15
3. 10
4. 0 5. 41 6. 81 7. 41 8. 32
9. 11
3. Labels on the horizontal and vertical axes are transposed
4. 1 2 3 Terms are not written in numerical order; the
1
2
H
T
2. 10
H
3
T
H
T
m
a e i
4
H
5
T
H
T
H
a e i
T
second guest’s choice is limited by the first guest’s choice.
15. Independent; the second flip is not affected by the first.
1
1
16. 81
17. 72
n
o u
5
1
1
1
10. 22
11. 11
12. 33
13. 11
14. Dependent; the
6
page 72 Practice 2: Independent and Dependent Events
o u
3. 32 4. 54 5. 9 6. 24
Data Analysis and Probability Teacher’s Guide
9
6
6
6
4
2
4
1
1. 25
2. 25
3. 25
4. 25
5. 55
6. 55
7. 55
8. 221
9. 61
2
10a. 65
10b. 42
65
Answers
21
Data Analysis and Probability Workbook Answers
trash
> > /¶2> >
/¡``/3``£/
1¡/```/`£0/
brown
/¡``4/``£/
5/¶
/¡``/9``/£
black
laundry
brown
black
brown
black
6
43 17
11b. 91
12. 100
; 200
page 73 Reteaching: Independent and Dependent Events
9
3
3
1
1
1
1
1
1. 25
2. 50
3. 100
4. 100
5. 45
6. 15
7. 30
8. 10
page 74 Analyzing Events Not Equally Likely
7 3 2 1
1. 12, 12, 20
, 4, 5, 4 2. 1 3. 8 4. 6
page 75 Practice: Analyzing Events Not Equally Likely
1
1
1
1
1
1
1
1. 13
2. 13
3. 13
4. 13
5. 13
6. 13
7. 26
8. 41
9.
1
2
16.
10.
3
13
11.
1
52
12.
3
4
13.
51
52
14.
2
7
15.
8
19
7
16
page 76 Activity: Analyzing Events Not Equally Likely
9
19
2
1. 47 2. 0.77 3. 0.9999 4. 52
5. 13
6. 13
page 77 Puzzle: Analyzing Theoretical Probability
1. 53 2. 25 3. You are more likely to pick a 3. 4. The
incorrect assumption is that there is a 13 chance of getting a
5 with two number cubes. With two number cubes there are 36
possible outcomes. Of those, 11 contain at least one 5, so your
11
chances are 36
, which is less than 31. With 6 number cubes,
chances are 1Q56R < 67%.
6
page 82 Activity 2: Theoretical and Experimental
Probability
1–2. Check students’ work. 3. no 4. 16 5. no 6. For a
large number of rolls, you will likely roll a 3 one-sixth of the
time. 7. Since P (any number from 1 through 6) 1,
students should be able to predict that this result will occur
12 times. Since P(7) 0, students should be able to predict
that this result will occur 0 times. 8. Yes, because through
repeated trials the experimental results should approach the
theoretical probability. 9–10. Check students’ work.
page 83 Activity 3: Increasing the Number of Trials
Check students’ tables. 1. 40 2. Answer should be close to
24. 3. answer should be about 3.16. 4. Check students’
calculations.
page 84 Predictions Based on Experimental Probabilities
8
24
1. P(broken) 600
0.04 2. Yes; P(broken) 500
0.016
3. 0.04 18,000 720 crayons 4. 0.016 18,000 288
crayons 5. 98.4%
page 85 Practice: Predictions Based on Experimental
Probabilities
3
4
1. 200
2. yes; P(defective) now is 375
, which is less than
3
.
3.
175
g
4.
250
g,
100
g
5.
150
g 6. The mass must
200
be within 75 g of 175 g. 7. 87 cars 8. 8,000 bearings
page 86 Reteaching: Predictions Based on Experimental
Probabilities
9
x
x
8
1. 160 shirts, 400
2. 144 shirts, 500
5 8000
5 8,000
16
3. 480 games, 400
5
x
12,000
x
12,000
30
4. 450 games, 800
5
x
12,000
65
x
games, 1,700
5 12,000
page 78 Assessment 5
1
1. They are equally likely: 21 3 12 3 12 3 12 5 16
. 2. There
are 6 ways of getting two heads and two tails: HHTT; HTTH;
HTHT; THHT; THTH; TTHH, but only one way of getting
1
HEADS TAILS HEADS TAILS. 3. 21 or 50% 4. 270
, 725
19
1
1
5. 28, 561 6. 27 or 70% 7. 3 8. The weather for two days
in a row is not an independent event. If we get the storm it
may be more likely to rain both days. If the storm misses us, it
probably won’t rain either day.
19
5. 456 games, 500
5
Section 6 Experimental Probability
page 88 Random Samples and Biased Questions
1. C 2. B 3. No; you are more likely to interview
homeowners. 4. No; you are more likely to interview renters.
5. Yes; you can’t tell if people own or rent.
page 79 Theoretical and Experimental Probability
9 1
3 1
3
1
4 1
4
2 2
1. 25
; 5 2. 25
; 15 3. 25
; 3 4. 17
75; 15 5. 15 ; 15 6. 10 7. 2
3
1
8. 20
9. 20
page 80 Practice: Theoretical and Experimental Probability
9
5
5
7
1. 31 2. 12
3. 41 4. 32 5. 12
6. 43 7. 14
8. 14
9. 92
8
4
10. 15
11. 31 12. 45
13. 4 to 11 14. 8 to 37 15. 32
1
1
16. 79 17a. 1200
; 1 to 1199 17b. 400
; 1 to 399
1
18. 0.55; 90 slices 19. 10,000
; 1 to 9,999
page 81 Activity 1: Theoretical and Experimental
Probability
1a-b. Check students’ work. 1c. Answers may vary; graphs
should approach 50% as number of tosses increases.
22
Answers
6. 459
page 87 Assessment 6
9
3 1
3 1
1
1. 100
, 1 to 99 2. 38, 52 3. 14, 40
4. 10
, 4 5. 40
,8
x
1
1
6. 10 or 10% 7. 100,000 disks 8. 10 disks 9. 25
;
5 500
20 students
Section 7 Statistical Investigations and
Simulations
page 89 Practice: Random Samples and Biased Questions
1.–10. Answers may vary. Samples are given. 1. random
sample 2. Not a random sample; students who use the
vending machine may not represent all types of students.
3. fair 4. Biased; Do you prefer hardwood floors in your
home? 5. fair 6. Biased; How many servings of fruits and
vegetables do you eat? 7. Biased; Do you prefer thick
carpeting? 8. fair 9. fair 10. Biased; Does TV news portray life accurately?
Data Analysis and Probability Teacher’s Guide
© Pearson Education, Inc. All rights reserved.
11a.
Data Analysis and Probability Workbook Answers
page 90 Planning A Survey
1. No; the sample includes only students interested in art, not
the whole population. 2. Yes; every student has an equal
chance of being selected. 3. Closed-option; Which sport do
you think is the most exciting? 4. Open-option; Did you
enjoy the book we just read in English? 5. Questioner is
suggesting a preferred answer. 6. Question assumes that the
movie was boring.
page 91 Practice 1: Planning a Survey
1. shoppers in a mall who are at least 16 years old 2. 966
never, 644 occasionally, 536 regularly 3. 2,146 people 4. no
5. open 6. The question makes gourmet meals seem better
than plain ones. 7. The question makes shortened school
days seem inferior. 8. No. People who work full-time are
more likely to shop after work than midday. 9. Answers may
vary. Some things to consider are using good questions, getting
a large enough random sample, and avoiding biased questions.
page 92 Practice 2: Random Samples and Surveys
1. 320 students 2. 352 students 3. 200 students
4. 192 students 5. The views of people coming out of a
computer store may not represent the views of other voters.
This is not a good sample because it is not random. 6. The
city telephone book may cover more than one school district.
It would also include people who do not vote. This is not a
good sample because it does not represent the population.
7. This is a good sample. It is selected at random from the
population you want to study.
© Pearson Education, Inc. All rights reserved.
page 93 Conducting a Statistical Investigation
1. Check students’ work.
page 94 Investigation 1: Happy Birthday
1–6. Check students’ work. EXTRA: James Polk (1796) and
Warren Harding (1865) were born on November 2. John
Adams (1826), Thomas Jefferson (1826), and James Monroe
(1831) all died on July 4.
page 95 Investigation 2: Hits and Misses (Geometric
Probability)
2
1. A 5 pr2 2. 2r ? 2r 5 4r2 3. pr2 5
6–8. Answers may vary.
4r
p
4
4. 0.785
page 96 Investigation 3: Sticky Dot Number Cubes
1
2
1–5. Answers may vary. 6. P(0) 5 36
, P(1) 5 36
,
3
3
5
4
P(2) 5 36
, P(3) 5 36
, P(4) 5 36
, P(5) 5 36
,
5
3
3
4
P(6) 5 36, P(7) 5 36, P(8) 5 36, P(9) 5 36,
2
1
P(10) 5 36
, P(11) 5 36
,
page 97 Planning a Simulation
1. the probability that out of 3 children, one will be a girl
2. a coin toss 3. three tosses of a coin 4. the probability
that out of 6 tees, there will be 2 red, 2 white, and 2 blue
5. a number cube with 2 numbers assigned to each color
6. 6 rolls of the number cube 7. a number cube with
2 numbers assigned to each color 8. 12 rolls of the number
cube
Data Analysis and Probability Teacher’s Guide
page 98 Simulation 1: A Multiple Choice Test
1. 15 2. Answers may vary. Theoretically, 8.4 rolls 3. Answers
51
may vary. The theoretical probability is 243
. 5. Answers may
vary. The theoretical value is 20. 6. Answers may vary. The
theoretical probability is 21. 7. Answers may vary. The
theoretical value is 7.5. 8. Answers may vary. The theoretical
3
probability is 16
. 9. Probability of getting 4 or more answers
correct.
page 99 Siumulation 2: Dyeing T-Shirts
1. 12 2. 52 3. 18 4. 35 5. Answers may vary. Theoretically,
11.25 shirts. 6. Answers may vary. The theoretical probability
is 38. 7. Answers may vary. The theoretical value is 18.75.
8. Answers may vary. The theoretical probability is 58.
9. twice. Answers may vary. Theoretically, 1.875 times.
1
1
10. 15
. Answers may vary. The theoretical probability is 16
.
page 100 Simulation 3: A Fair Game?
In game 1, there is 1 chance out of 3 to match. In game 2,
chances are 1 out of 2 that there will be a match.
page 101 Analyzing Simulations
5 3
3
1
4
1. m
n 2. 5 3. 5 4. 1 5. 15 6. 8 7. 25, 15 8. 8 , 8
3
9. 40
10. 12 11. No. The probability of each person winning
the lottery is much less than 12.
page 102 Reteaching 1: Simulations
1. Sample: Use a set of number cards 1–10 and draw one card
at a time from a bag. Then replace the card. 2. Sample:
Roll a number cube letting each of the numbers represent a
different shape. (Ignore 6.) 3. Sample: Use a spinner with
8 equal parts. 4. Sample: Roll a 12-sided number cube,
letting each of the numbers represent a different symbol.
page 103 Reteaching 2: Simulations
1. Sample: Use a spinner divided into four parts, one for each
symbol. 2. Sample: Roll two number cubes, letting each of
the 36 combinations represent a different saying.
3. Sample: Toss a number cube. 4. Sample: Use two sets of
number cards 1–10 and draw them from a bag.
page 104 Practice 1: Simulations
5
1a. 18 1b. 87 2a. 11
16 b. 16
3. Check students’ work. Answer should be about 12.5%.
4. Check students’ work. Answer should be about 37.5%.
5. Check students’ work. Answer should be about 37.5%.
6. Check students’ work. Answer should be about 12.5%.
7. Possible answer: Let 1 basket made, 2 – 6 basket
missed. 8. Check students’ work.
page 105 Practice 2: Simulations
1. Sample: Use a number cube with one number for W, I, E,
and R, and two numbers for N.
page 106 Assessment 7
1. Answers may vary. Sample: Registered drivers selected
randomly is best because that represents a fair sample of the
Answers
23
Data Analysis and Probability Workbook Answers
relevant population, while people at the restaurant, waiting
to buy tickets, or visiting the dealership probably do not.
2. Open and fair. The question lets people answer generally
rather than having them pick from limited choices, so it is
open. It asks their opinion neutrally without suggesting an
answer so it is fair. 3. 260 people 4. $20,000 5. The
simulation shows that even with a 50/50 chance, the outcomes
will not be exactly equal. In the simulation, the 75 coin flips
come out 41 heads and 34 tails. The number of students is
10 times the number of coin flips, so this is equivalent to
410 students preferring one beverage and 340 wanting the
other drink (rather than exactly 375 of each). Based on this
result, the beverage committee should buy 35 extras of each
drink (410 instead of 375) if they want a good chance of being
able to satisfy everyone.
pages 107–108 Cumulative Assessment
1.
Other
7%
Transp.
27%
Book/CD
33%
Food
33%
2. 86.8, 88, 88 3. The median, because there is an outlier and
there is no mode. 4. Any number greater than or equal to
27. 5. 44 6. One or both of the axes can be stretched or
© Pearson Education, Inc. All rights reserved.
compressed so that the rate of change looks bigger or smaller
than it really is. Or the y-axis can be broken so that the
differences in magnitude are exaggerated. 7. 35, 210
1
1
8. 13,440 9. 9,000 more 10. 256
11. 18 12. 17
4 1
1
1
;6
13. 221
14. 169
15. 8 16. 160 17. 25
18. Answers will vary. Sample answer: Do you prefer the
greasy pizza from Joe’s or the fancier pizza from Pistachios?
Re-write: Which pizza do you prefer, Joe’s or Pistachios?
19. 40,000 people
24
Answers
Data Analysis and Probability Teacher’s Guide
Answers
Activity 1
Activity 5
1. Check students’ work. 2. 1 3. 2
4.
Category
numbers
Possible
stores
for best buy
Store with best buy
so that each store
occurs exactly twice
Number of twists
Number of strips when
cut along the center line
1 and 2
A
A
1
1
1 and 3
C, D
C
2
2
1 and 4
A, C
A
3
1
1 and 5
D, E
E
4
2
2 and 3
B
B
2 and 4
A, B
B
2 and 5
D
D
3 and 4
C
C
3 and 5
D
D
4 and 5
E
E
5. Sample: A Möbius strip with 5 twists may produce 1 new
strip, and one with 6 twists may produce 2 new strips.
6. Sample: By examining the information from the experiments, it is likely that Möbius strips with an odd number of
twists result in 1 new strip when cut down the middle. However,
Möbius strips with an even number of twists result in 2 new
strips when cut down the middle.
Activity 6
Activity 2
1a. no 1b. There is an infinite number of ways.
2a. no 2b. There is an infinite number of ways. 3. 1 way
4a. Sample: The cardboard is not stable. 4b. There is an
infinite number of ways. 5a. Sample: Yes, it is possible.
© Pearson Education, Inc. All rights reserved.
However, if all four erasers are not the same length, they cannot all touch the cardboard. 5b. The cardboard is stable.
5c. The cardboard is not stable. 6. Sample: Three distinct
points are needed to determine a unique plane. However, if
one line and another point are present, a unique plane can be
determined.
Activity 3
1. Check students’ work. 2. Check students’ work.
3. ABCD, ABFE, EFGH, CDHG, ADHE, BCFG,
ABGH, CDEF, BCHE, ADGF, ACGE, BDHF 4. 12
Activity 4
1.–2.
Hypothesis (p)
T or F
Negation (Mp)
T or F
X Y (6 5)
T
X not greater
than Y (6 5)
F
X Y (6 5)
F
X not less than
Y (6 5)
T
X 2 Y (6 2 5)
T
X = Y (6 = 5)
F
Activity 7
1a. The 50-yd line is a transversal across the players’ paths.
The measures of the same-side interior s on the west side
of the 50-yd line are 65° for player #21 and 90° for player #13.
1b. Because the sum of the measures of the same-side
interior s is less than 180°, the two pathways meet beyond
the 50-yd line. Since player #13 has less distance to travel, he
should be able to overtake player #21. 2. Measurements
may vary, however, the alternate interior s will not be . 3.
No; the alternate
* ) interior s formed are not . 4. It was
not, because BX is not 6 to the 40-yd line.
Activity 8
Triangle
3. IF X > Y THEN PRINT Y, X; IF X < Y THEN PRINT X,Y;
IF X = Y THEN PRINT X = Y 4. IF 3 < 10 THEN PRINT 3 = 10
5. False; the hypothesis is true, and the conclusion is false.
Geometry Hands-On Activities
1. Sample: HCA and GCF 2. Sample: BEG and
IEG, CHG and BHG 3. Sample: DJA and
EBA, BAJ and EID 4. Sample: EGH and HGC,
DJF and FJA 5. Sample: DJF and HBA, FJA
and EBH 6. Sample: DJF and FJA, BAC and
CAJ 7. Sample: BEG and BHG 8. Sample:
GCH, DIE, and BCA 9. Sample: DGCF, IBAJ,
and EGAB 10. EBHG and DJBE
Classification
IMQ
acute, scalene
AHD
obtuse, scalene
BEJ
acute, equilateral
FLK
acute, isosceles
GRC
right, scalene
STP
obtuse, scalene
Answers
41
Answers
(continued)
Activity 9
Activity 13
1a.
1b.
1. the width of the straightedge 2. They are equal.
3. They are equal. 4. Point T lies on QS. 5. They are
equal. 6. QS is the bisector of PQR in PQR.
Activity 14
2.
3.
1.
B
A
4a.
C
2.
B
2
D
4b.
E
J
1
A
3.
Activity 10
1.–3. Check students’ work. 4. congruent intersecting
circles 5. congruent crescentlike shapes; congruent triangular
shapes; congruent oval shapes 6. two pairs of congruent
triangles 7. three pairs of congruent triangles 8. Check
students’ designs for requirements.
H
3K
F
4
G
C
3
1
2
4
4.
1. Because all pins are equidistant, each side length is the
sum of two equal distances. So AB JK, BC KL,
and AC JL . Therefore, ABC JKL by SSS.
2. TSR VUQ. The pairs of corresponding sides are
TS and VU, SR and UQ, and TR and VQ . By the previous
reasoning (each side is the sum of line segments), the pairs
of corresponding sides are . So the triangles are by SSS.
3a. Sample: A
B
C
D
E
F
G
H
I
© Pearson Education, Inc. All rights reserved.
Activity 11
5.
6.
J
3b. Sample: ABE, BFC, CDG, BEF, CFG,
EFH, FGI, HIJ, FHI
7.
Activity 12
1. The number of smaller triangles in an n-frequency dome is
n2. 2. As the frequency increases, the dome becomes more
spherical in shape.
42
Answers
Geometry Hands-On Activities
Answers
(continued)
Activity 15
Activity 17
Number of
Number of
differentNumber of
isosceles
Number of
shaped
Lengths equilateral triangles (nontoothpicks triangles of sides triangles
equilateral)
3
1
1, 1, 1
1
0
4
0
Not
possible
0
0
5
1
2, 2, 1
0
1
6
1
2, 2, 2
1
0
7
2
1, 3, 3 and
2, 2, 3
0
2
8
1
2, 3, 3
0
1
9
3
3, 3, 3;
2, 3, 4;
1, 4, 4
1
1
10
2
2, 4, 4;
3, 3, 4
0
2
11
4
1, 5, 5;
2, 4, 5;
3, 4, 4;
3, 3, 5
0
3
2, 5, 5;
3, 4, 5;
4, 4, 4
1
© Pearson Education, Inc. All rights reserved.
12
3
1
1. A triangle cannot be formed. 2. 5, 7, 8, 9, 10, 11, 12; there
does not seem to be a pattern. 3. 3, 6, 9, 12; yes; the pattern
seems to be multiples of 3; prediction: 15. 4. 1, 2, 3, 4;
prediction: 5; the prediction is correct: 1, 6, 6; 2, 5, 6; 3, 5, 5;
3, 4, 6; 4, 4, 5. 5. Prediction: 6; the prediction is not correct,
7 triangles can be formed: 1, 7, 7; 2, 6, 7; 3, 5, 7; 3, 6, 6; 4, 4, 7;
4, 5, 6; 5, 5, 5. 6. greater than; cannot
Number of
toothpicks (y)
Sketch of
arrangement(s)
Number of
squares (x)
4
1
7
2
8
1
10
3; 2
11
3
12
1; 5
1. 1 2. 4 3. Sample: One square can be made from any
multiple of
134 toothpicks.
4
1. 1 2. 4 3. Answers may vary. Sample: If the number of
toothpicks is divisible by 4, there can be 1 square formed.
Activity 16
1. All triangles, when combined with a congruent triangle,
form a parallelogram. This is true because any parallelogram
split in half on its diagonal produces two congruent triangles.
2. The two acute isosceles triangles yielded a rhombus.
Because of the two equal sides of each triangle, the two
triangles form four congruent sides when a pair of sides are
placed together. 3. The two right scalene triangles yielded
another triangle. This triangle is classified as isosceles. This
is possible because of the 90° angle in each triangle. When
joined, these two 90° angles form a straight angle. 4. The
two obtuse scalene triangles yielded a concave quadrilateral.
When these triangles are joined at their shorter sides, an
interior angle greater than 180° is formed. 5. All three pairs
of triangles yielded a kite. When the triangles are joined at
their longer sides, congruent sides are adjacent to each other,
which is the definition of a kite. 6. parallelogram, rhombus
7. parallelogram, rhombus
Activity 18
1. square, rectangle, rhombus, or parallelogram 2. square or
rectangle 3. square or rhombus
4.
quadrilateral
European
kite
European
trapezoid
European
isosceles
trapezoid
parallelogram
rhombus
rectangle
square
Geometry Hands-On Activities
Answers
43
Answers
(continued)
3b. Sample:
5.
Quadrilaterals with area 10 square units
general
quadrilateral
Classification
trapezoid
MutBis
MutBis
Dimensions
rectangle
b = 5, h = 2
parallelogram
b = 2, h = 5
isosceles trapezoid
b1 = 6, b2 = 4, h = 2
kite
d1 = 4, d2 = 5
trapezoid
b1 = 3, b2 = 1, h = 5
s = "10
square
4a. Sample:
MutBis
MutBis
Activity 19
1. Check students’ work. 2. Sample: The area of JKLM is
one-fifth the area of ABCD.
3.–5.
10
D
G
8
2
Classification
L
6
4
Quadrilaterals with area 12 square units
C
M
H
F
K
J
A
E
2
4
6
B
8 10 12
* )
* )
* )
* )
* ) * )
*ED) : y =* -)2x + 10; AF and CH have the same
* slopes;
)
BG and ED have the same slopes. The slopes of AF and
* )
* ) * )
CH are the negative reciprocals of the slopes of BG and ED .
* ) * )
6. AF : y = 12 x, CH : y = 12 x + 5, BG : y = -2x + 20,
7. They are parallel. BG and ED are also parallel. JKLM
is a parallelogram. 8. J(4, 2) 9. K(8, 4), L(6, 8), M(2, 6)
10. area of JKLM = 20 square units; area of ABCD = 100
square units; ratio = 15 11. Check students’ work.
Dimensions
rectangle
b = 6, h = 2
rectangle
b = 4, h = 3
isosceles trapezoid
b1 = 5, b2 = 3, h = 3
kite
d1 = 6, d2 = 4
parallelogram
b = 3, h = 4
trapezoid
b1 = 3, b2 = 1, h = 6
5a. Sample:
5b. Sample:
Quadrilaterals with area 15 square units
Classification
Dimensions
Activity 20
rectangle
b = 3, h = 5
1a. Check students’ work. 1b. Check students’ work.
2. rectangle: b = "18, h = "2; kite: d1 = 2, d2 = 6;
parallelogram
b = 5, h = 3
isosceles trapezoid
b1 = 4, b2 = 2, h = 5
rectangle
b = 15, h = 1
trapezoid
b1 = 7, b2 = 3, h = 3
parallelogram: b = 3, h = 2; trapezoid: b1 = 4,
b2 = 2, h = 2
3a. Sample:
6. Sample: Use factors of the given area. 7. Sample: Work
backward from area formulas of parallelograms, trapezoids,
and kites.
44
Answers
Geometry Hands-On Activities
© Pearson Education, Inc. All rights reserved.
12
4b. Sample:
Answers
(continued)
6.
Activity 21
1. 100 square units 2. PS = 6; SG = 8
3. YPS SGA by SAS, so SA YS by CPCTC.
4. 1 4 and 2 3 by CPCTC. 5. They are
complementary. 6. 90 7. Check students’ work; area:
c2 = 100. 8. original squares: 82 + 62 = 64 + 36 = 100;
new squares: 10 ? 10 = 100 9. Check students’ work.
10. X
U
R
a
Y
c
S
c
V
Z
7. and 8.
b
T
Using the figure above, the area of square XYZU is a2,
and the area of square RZST is b2. So the area of figure
XYTSRU is a2 + b2. If you cut out XYV, VST, and
figure XVSRU and then rearrange them to form a square
with sides c, the area of the square formed is c2. Therefore,
a2 + b2 = c2.
Activity 22
1.
9.
10.
2.
© Pearson Education, Inc. All rights reserved.
3.
Activity 23
1. Check students’ work. 2. Check students’ work.
3. Triangle A: hypotenuse = 5 cm; Triangle B: hypotenuse
4.
5.
= 13 cm; Triangle C: hypotenuse = 10 cm; Triangle D:
hypotenuse = 15 cm; Triangle E: hypotenuse = 17 cm
4. 3, 4, 5; 5, 12, 13; 6, 8, 10; 9, 12, 15; 8, 15, 17 5. 3, 4, 5;
6, 8, 10; 9, 12, 15 6. Sample: Each side of a bigger triangle is
a multiple of its corresponding side of a smaller triangle. For
triangles to be similar, they must have a common ratio among
their sides. 7. Sample: 10, 24, 26; 16, 30, 34 8. Sample:
similar to triangle B: 10, 24, 26; 15, 36, 39; similar to triangle E:
16, 30, 34; 24, 45, 51 9a. 72 + 242 = 252; 49 + 576 = 625;
625 = 625 9b. Sample: 14, 48, 50; 21, 72, 75
Activity 24
1. Check students’ work.
2.
ABC
Geometry Hands-On Activities
mA = 90
mB = 27
mC = 63
MNP mM = 90
mN = 27
mP = 63
RST
mR = 90
mS = 27
mT = 63
ABC
AB = 3 cm
BC = 3.4 cm
CA = 1.5 cm
MNP MN = 4 cm NP = 4.5 cm
PM = 2 cm
RST
TR = 2.5 cm
RS = 5 cm
ST = 5.6 cm
Answers
45
Answers
(continued)
RST, and the sides of MNP are proportional to the sides
of RST. 4. Yes; by Theorem 8-2, if the ratios are equal,
then the triangles are similar. 5a. They are parallel.
5b. MW is half the length of MP. MX is half the length
of MN. 6. The length of the part of the shorter leg is half
the length of the part of the longer leg.
Activity 25
1. Check students’ work. 2. Triangle A:
hypotenuse 3.6 cm; Triangle B: hypotenuse ≈ 7.2 cm;
Triangle C: hypotenuse 14.4 cm 3. The corresponding
sides are proportional. 4. The three triangles are similar.
5. The corresponding angles are congruent.
6.
Smaller
angle
side opposite
side adjacent
Triangle A
2
3
1
1.8
1
1.2
Triangle B
2
3
1
1.8
1
1.2
Triangle C
2
3
1
1.8
1
1.2
side opposite side adjacent
hypotenuse
hypotenuse
7. The ratios in each column are equal.
8.
Larger
angle
side opposite
side adjacent
Triangle A
3
2
1
1.2
1
1.8
Triangle B
3
2
1
1.2
1
1.8
Triangle C
3
2
1
1.2
1
1.8
side opposite side adjacent
hypotenuse
hypotenuse
The values in the second column are the reciprocals of
the values in the second column of the table in Exercise 6.
The values in the third and fourth columns are switched
with the values in the third and fourth columns of the table
in Exercise 6.
Activity 27
1. 180 2. 270 3. 315 4. 112.5
5. The paper moves faster than the speed of the force with
which you rolled it. Its actual speed is the sum of the force
with which you rolled the paper and the force with which you
blew on it.
Blow
Push
6. The paper moves slower than the speed of the force with
which you rolled it. Its actual speed is the difference between
the force with which you rolled it and the force with which
you blew on it.
Blow
Push
7.
Blow
Actual path
Push
8. The faster you roll the paper, the more closely it follows
its original direction. The harder you blow, the more the
paper deviates from its original direction. 9. 580 mi/h
10. 420 mi/h
11.
Actual
velocity
Wind
80 mi/h
Velocity with no wind
500 mi/h
12a. 506 mi/h 12b. This result is between the speed in
a headwind and the speed in a tailwind. This makes sense
because a headwind gives the greatest resistance, and a
tailwind gives the greatest assistance.
13a. 9° 13b. This makes sense when compared with the
vector diagram that illustrates the situation (see answer to
Exercise 11). The course of the plane is slightly north of east.
Activity 28* )
* )
* )
1. Sample: G–AB –D, A– CD
* –E,
) C– EG –A
* )
GD –E 4. A– EC –G
2. triangular
prism
3.
B–
* )
5. E– CD –A 6. AEC and BGD
Activity 26
1.–4. Check students’ work.
46
Answers
Geometry Hands-On Activities
© Pearson Education, Inc. All rights reserved.
3a. These sides are approximately proportional. 3b. yes
3c. Yes; the sides of ABC are proportional to the sides of
Answers
(continued)
1.–2. Check students’ work. 3. The angles appear
to be supplementary. 4. Yes; the angle formed by the
Activity 29
1. Check students’ work.
2.
Circle
1
2
3
Radius of
circle (R)
Take-out
angle (x)
Cone angle
(y)
A
10 cm
72
288
B
10 cm
90
270
C
10 cm
180
180
Activity 32
D
10 cm
216
144
1., 3.–6., and 8.
4
5
6
cone angle
360
Radius of
base of
cone (r)
Circle
Key
x
cone radius
circle radius
4
5
8 cm
B
3
4
15 cm
2
3
4
C
1
2
5 cm
1
2
D
2
5
4 cm
2
5
A
© Pearson Education, Inc. All rights reserved.
two tangents and the angle formed by the two radii are
supplementary. 5. APO BPO. Justifications will
vary, but the triangles must be congruent by the HL Theorem
(right angles by construction, PO is the hypotenuse of both
triangles, and the congruent radii OA and OB are the
corresponding legs). 6a. PO is the bisector of APB by
CPCTC. 6b. PO is the bisector of AOB by CPCTC.
4
5
3. The radius of the circle is the slant height of the cone.
4. Slant height of cone 5. The values are equal.
2 x
2 x
6. 2pr = 360360
(2pR) 7. r = (360360
)R
360 2 x
r
8. R = 360
9. circle A: h = 6 cm; circle B: h 6.6 cm;
circle C: h 8.7 cm; circle D: h 9.2 cm 10. Check
students’ work.
F
M
O
Altitudes
Medians
Perpendicular
bisectors
F
M
C2
x
x
C1
F
M
Euler line
C1 – Circumcenter
O – Orthocenter
C2 – Centroid
x – Euler Points
F – Feet of Altitudes
2. The distance from the centroid to the orthocenter is
twice the distance from the centroid to the circumcenter.
7. The center of the nine-point circle lies on the Euler line
and is midway between the circumcenter and the orthocenter.
9. The radius of the nine-point circle is equal to half the
radius of the circumscribed circle of the triangle.
Activity 33
Activity 30
1. Check students’ work. 2. 4 3. They are congruent.
4. 6 5. no 6. 2 7. The sum will always be greater than
180º. 8. Sample: Any two points on a sphere lie on the same
great circle. Therefore, the shortest distance between any
two locations on Earth will be along a great-circle route.
1. 90 2. 180 3. semicircle 4. Sample: AG = 42 mm;
BG = 42 mm; CG = 42 mm; DG = 42 mm 5. G is the
center of the circle. 6. Check students’ work.
Activity 34
1. Sample:
Activity 31
Steps 1.–5.
1
of base
2
P
C
2. Sample:
D
A
E
B
base
O
T
Geometry Hands-On Activities
Answers
47
Answers
(continued)
3. Sample:
8.
altitude
of k
endpoint
of base
endpoint
of base
4. Sample:
9. perpendicular bisector 10. Locate the perpendicular
2
of base
3
bisector of the base of each triangle, and then reflect, or
draw a mirror image of, the triangle on each side of this line.
11. Check students’ work.
Activity 35
1. Figures 4 and 6 2. Figures 3 and 1 3. Figure 5
4. Figure 2 5.–8. Check students’ work.
5.
side
Activity 36
1. Sample:
part of base
6.
base
part of side
2. 60 3. 40 4. 5 5. 2 6. 90; 65; 5; 2 7. Check
students’ work.
7.
vertex
48
Answers
© Pearson Education, Inc. All rights reserved.
endpoint
of base
Geometry Hands-On Activities
Answers
Screening Test
Quarter 1 Test, Form A
1. C 2. H 3. B 4. H 5. C 6. J
11. C 12. F 13. B 14. H 15. C
19. B 20. J 21. C 22. H 23. C
27. D 28. G 29. D 30. F 31. C
35. C 36. G 37. D 38. F 39. B
7. A 8. F 9. C 10. G
16. J 17. D 18. H
24. J 25. B 26. H
32. F 33. C 34. J
40. J 41. C 42. J
1. 13, 21 2. Sample: a tiger 3. right triangle
10 in.
4.
6 in.
Benchmark Test 1
All rights reserved.
1. D 2. H 3. B 4. F 5. C
10. F 11. B 12. H 13. D
17. B 18. G 19. C 20. F
24. G 25. A 26. H 27. D
31. B 32. F 33. C
6. H 7. D 8. J 9. A
14. G 15. D 16. F
21. C 22. G 23. C
28. G 29. C 30. J
2 in.
10 in.
2 in.
5. -27 6. x = 3 7. 25
8.
L
M
Benchmark Test 2
1. A 2. H 3. A 4. J 5. A 6. H 7. B 8. F 9. C
10. H 11. A 12. G 13. C 14. G 15. B 16. H
17. B 18. H 19. B 20. F 21. C 22. G 23. D
24. H 25. B 26. J 27. C 28. G 29. A 30. H
9. Subtraction Property
10.
Benchmark Test 3
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1. A 2. F 3.
10. G 11. A
17. B 18. H
24. H 25. B
31. B 32. H
D 4. J 5. B 6. F 7. B 8. F 9. D
12. G 13. C 14. F 15. B 16. G
19. D 20. H 21. C 22. H 23. D
26. J 27. C 28. H 29. C 30. G
33. B
Benchmark Test 4
1. B 2. F 3. C 4. J 5. A
10. G 11. B 12. H 13. C
17. B 18. J 19. D 20. G
24. F 25. D 26. F 27. A
6. F 7. C 8.
14. J 15. C
21. B 22. H
28. F 29. B
J 9. D
16. J
23. B
30. H
Benchmark Test 5
1. B 2. H 3. B 4. J 5. C 6. F 7. B 8. J 9. D
10. F 11. C 12. H 13. B 14. H 15. A 16. G
17. D 18. F 19. C 20. H 21. B 22. H 23. A
24. H 25. B 26. J 27. B 28. F 29. D 30. G
31. D 32. J 33. B 34. G 35. A 36. J 37. D
38. J 39. C
Answers
Fro
nt
ht
Rig
11. 70 12. 62 13. 29 14. Suppose that 2x + 4 = 6.
Subtract 4 from both sides of the equations by the
Subtraction Property. This gives an equation of 2x + 4 - 4
= 6 - 4, which reduces to 2x = 2. Now divide both sides of
the equation by 2 by using the Division Property. This gives an
* )
2
equation of 2x
2 = 2 , which simplifies to x = 1. 15. AC
16. 12p m 17. Sample: A student can drive if and only if he
or she is over the age of sixteen. If a student can drive, then
he or she is over the age of sixteen. If a student is over the
age of sixteen, then he or she can drive. 18. 144
19. (1, 2.5) 20. -23 21. 2 4 by the Converse of
the Corresponding Angles Theorem or 2 + 3 = 180
by the Converse of the Same-Side Interior Angles Theorem
22. 2 "17 23. no intersection or a point 24. Two lines
must not intersect and be in the same plane to be parallel.
25. 40 26. 50 27. 90 28. 75 29. hypothesis: if you are
not at school; conclusion: you are at home 30. If you are at
home, then you are not at school. 31. 20 in. 32. Transitive
Property 33. center (5, -5); radius = "5
Geometry
155
Answers
(continued)
Quarter 1 Test, Form B
Quarter 1 Test, Form D
1. 8, 5 2. Sample: obtuse triangle 3. isosceles triangle
3 cm
4.
14 cm
1. 21, 25 2. scalene triangle 3. 36p m2 4. A 5. 5 6. 9
7.
14 cm
6 cm
1
Right
2
3 cm
Front
8. 138 9. 46 10. C 11. 40 cm 12. BF 13. 10p m
14. 120 15. (-2, 4) 16. 12 17. B 18. 5 19. 45 20. 90
21. 60 22. 85 23. hypothesis: two angles have a sum of
5. -18 6. x = 6 7. 7
8.
3
A
B
180; conclusion: the angles are supplementary angles
9. Addition Property
Quarter 1 Test, Form E
10.
1. 26, 31 2. equilateral triangle 3. 100p m2 4. 5 5. 6
6.
2
Fro
nt
h
Rig
t
11. 112 12. 60 13. 44 14. Suppose that 3x + 5 = 11.
Subtract 5 from both sides of the equation by the Subtraction
Property. This gives an equation of 3x + 5 - 5 = 11 - 5,
which reduces to 3x = 6. Now divide both sides of the
equation by 3 by using the Division Property. This gives an
6
equation of 3x
3 = 3 , which simplifies to x = 2. 15. point R
16. 26p m 17. Sample: A student can drive if and only if he
or she is over the age of sixteen. If a student can drive, then he
or she is over the age of sixteen. If a student is over the age of
sixteen, then he or she can drive. 18. 135
19. (-6, 11) 20. 32 21. Sample: 1 3 or 2 4 by
the Converse of the Corresponding Angles Postulate;
2 + 3 = 180 by the Converse of the Same-Side Interior
Angles Theorem 22. 5 23. no intersection or a line
24. A triangle cannot have two sides and no sides equal at the
same time. 25. 60 26. 90 27. 70 28. 45
29. hypothesis: if you are not at work; conclusion: you are at
school 30. If you are at school, then you are not at work.
31. 96 in. 32. Substitution Property 33. center (2, 1); radius
= "117
156
Geometry
2
2
Right
1
Front
*
)
7. 110 8. 70 9. B 10. G 11. 60 in. 12. AB 13. 14p m
14. 135 15. (1, 5) 16. -13 17. D 18. 10 19. 72 20. 55
21. 50 22. 75 23. G 24. Hypothesis: two lines form right
angles; conclusion: the lines are perpendicular. 25. C
Quarter 2 Test, Form A
1. RS 2. 20 3. trapezoid 4. 50 5. 40; 82 6. SSS
Postulate 7. SAS Postulate 8. (c - a, b) 9. Sample: (0, 4),
(4, 0), (-4, 0), (0, -4) 10. R, P, Q 11. AAS, ABC
ABD 12. 133; 47 13. Sample: 4 ft, 3 ft, and 8 ft
14. 4 15. Sample: The diagonals must be perpendicular
bisectors of each other, or all sides must be congruent.
16. The orthocenter is the intersection of the three altitudes of
a triangle. 17. an altitude 18. 45 19. Hypotenuse-Leg
Theorem 20. centroid 21. It is the median and altitude of
the triangle. 22. Suppose that a triangle had more than one
right angle. If this were true, then two of the angle measures
alone would add up to 180, and the third angle would have a
measure that would contradict the Triangle Angle-Sum
Theorem. 23. AB , BC , AC 24. 90; 30 25. rhombus
26. 65 27. c, e, b, a, d or e, c, b, a, d 28. No; the midsegment
is not half the length of the third side. 29. a kite
30. 31 31. They must be perpendicular bisectors of each
other. 32. AC
Answers
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
C
All rights reserved.
24. G 25. D
Answers
(continued)
Quarter 2 Test, Form B
All rights reserved.
1. B 2. 10 3. parallelogram 4. 58 5. 41; 75 6. ASA
Postulate 7. AAS Theorem 8. (c + a, b) 9. Sample: (0, 4),
(8, 0), (-4, 0), (0, -4) 10. L, N, M 11. SAS,
ACD CAB 12. 142; 38 13. Sample: 4 ft, 3 ft, and
8 ft 14. 4 15. Sample: The diagonals must be perpendicular
bisectors of each other and all sides must be congruent, or
all sides and angles must be congruent. 16. The incenter
is the intersection of the three angle bisectors of a triangle.
17. a median 18. 41 19. Hypotenuse-Leg Theorem
20. centroid 21. It is the median and altitude of the
triangle. 22. Suppose that a triangle had more than one
obtuse angle. If this were true, then two of the angle measures
alone would add up to more than 180, and the third angle
would have a measure that would contradict the Triangle
Angle-Sum Theorem. 23. AB , BC , AC 24. 90; 50
25. rectangle 26. 70 27. a, c, e, b, d or c, a, e, b, d
28. Yes; the midsegment is half the length of the third side.
29. a parallelogram 30. 31 31. They must be equal in
measure. 32. AC + BC
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Quarter 2 Test, Form D
1. XZ 2. 4 3. D 4. 55 5. x = 85, y = 60 6. J
7. C, A, B 8. m1 = 54.5, m2 = 54.5, m3 = 71
9. D 10. C 11. B 12. E 13. 55 14. 56 15. H
16. m1 = 55, m2 = 45, m3 = 45, 17. m1 = 25,
m2 = 25, m3 = 130 18. A 19. parallelogram,
rectangle, square, rhombus 20. 164 21. B D
22. altitude 23. 33
25. Sample:
6
4
6
Quarter 3 Test, Form A
1. 14 2. ABC JKL by SSS ~ Theorem
3. TUV WXY by AA ~ Postulate 4. obtuse
5. 56.3 6. 10.7 7. reflectional symmetry and 180º rotational
8
symmetry (point symmetry) 8. 2 "26 9. sin A = 17
;
8
15
cos A = 17 ; tan A = 15 10. A(7, 2), B(11, 2), C(10, -1),
D(6, -1) 11. A(-5, 4), B(-9, 4), C(-8, 1), D(-4, 1)
12. A(-3, -4), B(-7, -4), C(-6, -1), D(-2, -1)
13. 150 in.2 14. 53.9 mi/h; 21.8° south of east 15. A
16. (3, 3) 17. 4 "3 18. 38 19. 6 20. 18.4 21. Check
students’ work. 22. No; there will be gaps when the pattern
is repeated. 23. A(-2, 6), B(0, 14), C(8, 4) 24. 48 ft
Answers
9
6
6
9
26. x 5 3, y 5 3"2 27. x = 2, y = 4 28. 3, -9
29. 12 in.
Quarter 3 Test, Form B
1. 14 2. not similar 3. ABC LPR by
SAS ~ Theorem 4. obtuse 5. 14.9 6. 55.0 7. reflectional
symmetry 8. "66 9. sin A = 53; cos A = 54; tan A = 34
10. A(-6, -3), B(-2, -3), C(-2, 0), D(-5, 0)
11. A(-18, 15), B(-6, 15), C(-6, 6), D(-15, 6)
12. A(-9, 6), B(-5, 6), C(-5, 3), D(-8, 3)
13. 18 cm2 14. 40.3 mi/h; 29.7º north of west 15. D
16. (-4, 1) 17. 2 "34 18. 4 19. 63 20. 33.7
21. Check students’ work.
22. yes
23. X(0, -3), Y(3.5, -2), Z(3, 1) 24. 265 ft
25. Sample:
3
Quarter 2 Test, Form E
1. R 2. 5 3. C 4. 50, 65 5. x = 100, y = 80 6. F
7. C, B, A 8. m1 = 122, m2 = 29, m3 = 29
9. E 10. A 11. C 12. B 13. 78 14. 58 15. G
16. m1 = 44, m2 = 84, m3 = 40 17. m1 = 65,
m2 = 65, m3 = 50 18. D 19. parallelogram,
rectangle, square, rhombus 20. 187 21. BX DX
22. perpendicular bisector 23. 40
4
3
6
6
26. x 5 5, y 5 5"2 27. x = 3, y = 6 28. 2, 9
29. 35 in.
Quarter 3 Test, Form D
1. BRG NDK by SAS Theorem
2. AYM XQH by AA Postulate 3. 12.8
5
4. 32.2 5. 13.5 6. D 7. 4 "5 8. sin A = 13
; cos A =
5
2
tan A = 12 ; 9. (10, 0) 10. G 11. 112 in. 12. A
12
13 ;
13. 5"7 14. x = 9, y = 15 15. y = 14 16. X(4, 6),
Y(5, 10), Z(8, 7) 17. X(2, -1), Y(3, -5), Z(6, -2)
18. X(-2, -1), Y(-3, -5), Z(-6, -2) 19. 6"3
20. 88 ft 21. 12 in. 22. J
Quarter 3 Test, Form E
1. WLB HER by AA Postulate
2. CMJ FKP by SAS Theorem 3. 24 4. 12.2
15
8
5. 63.0 6. C 7. 6"3 8. sin A = 15
17; cos A = 17; tan A = 8 ;
Geometry
157
(continued)
9. (-6, 9) 10. J 11. 588 in.2 12. A 13. 8"10
14. x = 3, y = 25 15. x = 10 16. A(0, 1), B(-3, 7),
C(-4, 2) 17. A(4, -2), B(7, 4), C(8, -1) 18. A(-1, 0),
B(-7, 3), C(-2, 4) 19. 9"3 20. 124 ft 21. 3 in. 22. F
Quarter 4 Test, Form A
1. 268.1 cm3 2. 66.3 cm2 3. 25 4. 32 5. 144
6. Sample: Draw d1, a diagonal of the kite that divides it into
two congruent triangles. Let d1 represent the base of each
triangle. The area of one triangle is 21 d1h1. The area of the
14. 122 15. 91.1 16. 75.9 cm3 17. 2 cans of paint
18. 55 19. 35 20. 270 21. 11 22. 93.6 in.2 23. b, c, a, d
24. (x + 3)2 + (y + 2)2 = 9 25. 423.2 cm 2 26. 73 p ft
0
0
27. Sample: RS is an arc of Q. The measure of RS is
equal to the measure of RQS. 28. 7162.8 mm2
Quarter 4 Test, Form D
1. 904.8 in.2 2. 4 : 7, 16 : 49 3. 90 4. 75 5. B 6. 96 cm2
7. 8 8. 245.0 ft2 9. 40 cm2 10. 5 11. 9
12. center: (0, 0); radius: 5
y
other triangle is 12 d1h2. Because the triangles are congruent,
4
h1 = h2. The other diagonal, d2, is the sum of h1 and h2.
Therefore, the area of a kite is 12 d1d2. 7. 14.14 cm2 8. 12.5
9. a circle with center (-2, -3) and radius 5 units
10. 113.1 m2 11. 81 m2 12. 6
13. center: (5, 6); radius: 4
2
2
4
x
4 2
2
4
2
4
y
13. 104 14. 90 15. 128 cm2 16. 184 in.2 17. 180 cm2
18. 201.1 ft2 19. 133 20. 180 21. 192.5 in.2 22. 150.80 m 2
23. G 24. 188.5 cm2
x
2 4 6 8 10
Quarter 4 Test, Form E
ft3
14. 73.7 15. 92 16. 400
17. 15 rolls of wallpaper
18. 30 19. 60 20. 270 21. 8 22. 310.4 in.2 23. b, c, a, d
24. (x - 2)2 + (y - 2)2 = 4 25. 188.1 in.2 26. 2p in.
* )
* )
27. Sample: QR is tangent to P. QR is perpendicular to a
radius of P. 28. 284.7 cm2
1. 1436.8 cm3 2. 3 : 8, 9 : 64 3. 65 4. 140 5. B 6. 255 in.2
7. 16 8. 326.7 in.2 9. 95 m2 10. 10 11. 15
12. center: (0, 0); radius: 4
6
y
2
Quarter 4 Test, Form B
1. 1436.8 in.3 2. 63.4 in.2 3. 39.5 4. 25.5 5. 108
6. Sample: Draw d1, a diagonal of the rhombus that divides
it into two congruent triangles. Let d1 represent the base of
each triangle. The area of one triangle is 21 d1h1. The area of
the other triangle is 12 d1h2. Because the triangles are
congruent, h1 = h2. The other diagonal, d2, is the sum of h1
and h1. Therefore, the area of a rhombus is 12 d1d2.
7. 19.63 in.2 8. 6 9. a circle with center (-4, 6) and radius
8 units 10. 466.5 ft2 11. 42 in.2 12. 12
13. center: (-3, 5); radius: 3
y
8
6
4
2
64 2
2
4
6
158
x
2 4 6
Geometry
x
2
2
2
13. 90 14. 87 15. 125 cm3 16. 804.2 m2 17. 483.8 in.2
18. 262 ft2 19. 59 20. 239 21. 173.8 in.2 22. 94.25 cm2
23. H 24. 55.0 in.2
Mid-Course Test, Form A
1. 63; 127 2. 86 3. 12 4. 7 5. rectangle 6. 12
7. Sample: (2, 0), (8, 0), (10, 5), (4, 5) 8. B, A, C
9. 36, 36 10a. If it is summer, then it is sunny.
10b. If it is not sunny, then it is not summer. 11. C
12. 26 13. 68 cm 14. 60 15. CAB BDC by SAS.
16. (0, b) 17a. 3 and 5 or 4 and 6 17b. 1 and
5, 2 and 6, 3 and 7, or 4 and 8 18. 90; 20
Answers
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
10
8
6
4
2
2
All rights reserved.
Answers
Answers
(continued)
20. G 21. m1 = 105, m2 = 75, m3 = 105, m4 = 75
22. parallelogram, rectangle, rhombus, square 23. B
24. Hypothesis: a transversal intersects two parallel lines;
conclusion: alternate interior angles are congruent. 25. J
26. A 27. 166
28.
y
19.
3
1
Right
2
Front
4
All rights reserved.
20. 115; 65 21. always 22. never 23. never
24. sometimes 25. x = 34; y = 24; z = 88
26. Sample: A rectangle always has opposite sides parallel,
making it a parallelogram. A parallelogram doesn’t always
have four right angles, so it is not always a rectangle.
27. C, A, B 28. rectangle 29. rhombus 30. square
31. trapezoid 32. rectangle 33. rhombus 34. 17 35. 48;
90; 42; 57; 33 36. x = 25; y = 31 37. (2.5, 1) 38. J
39. 118; 34 40. 7.2 41. area = 452.4 in.2; circumference
= 75.4 in. 42. 74 43. HL 44. not possible
45. SSS 46. AAS 47. not possible 48. SAS
49. ASA or AAS 50. SSS or SAS
x
4
2
2
4
2
4
2
4
29.
4
y
2
Mid-Course Test, Form B
1. 22, 29 2. 120 3. 17 4. 12 5. parallelogram 6. 7
7. Sample: (0, 4), (-6, 0), (0, -4) and (6, 0) 8. B, C, A
9. x = 120; y = 30 10a. If the ground is wet, then it is
raining. 10b. If it is not raining, then the ground is not wet.
11. D 12. 12 13. 120 m 14. 36 15. WXY XWZ
by AAS. 16. (a - b, c) 17a. 2 and 3 or 6 and 7
17b. 1 and 3, 2 and 4, 5 and 7, or 6 and 8
17c. 2 and 6 or 3 and 7 18. 34; 68
19.
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
2
2
2
2
x
4
2
2
4
30. 42 31. (1, 2) 32. 120 cm 33. 13 34. G 35. 24
36. 95 37. 157 38. DX JX 39. m1 = 40,
m2 = 95, m3 = 45 40. m1 = 54, m2 = 63
Right
1
Front
20. 63; 117 21. always 22. never 23. sometimes
24. sometimes 25. x = 30; y = 36; z = 84 26. Sample:
Because m1 + m2 = 135 and m2 + m3 = 135,
m1 + m2 = m2 + m3 by substitution. By the
Subtraction Prop. of Equality, m1 = m3, and therefore
1 3. 27. YZ, XZ, XY 28. quadrilateral 29. kite
30. trapezoid 31. parallelogram 32. rhombus 33. square
34. 5 35. x = 49; y = 90; z = 18 36. w = 40; x = 30;
y = 30; z = 90 37. (1, -6) 38. J 39. 71; 34; 75
40. 14.4 41. area = 254.5 cm2; circumference = 56.5 cm
42. 54 43. ASA 44. SSS 45. SAS and ASA 46. not
possible 47. HL 48. not possible 49. AAS 50. SAS
Mid-Course Test, Form E
1. -19, -23 2. 13 3. 12 4. RT , RS, TS 5. B 6. 24
7. F 8. 60 9. C 94.2 in., A 706.9 in.2 10. 13
11. m1 = 47, m2 = 43 12. AAS 13. ASA 14. SAS
15. not possible 16. HL 17. SSS 18. 50 19. D 20. J
21. m1 = 80, m2 = 80, m3 = 100, m4 = 80
22. square, rhombus 23. A 24. Hypothesis: a transversal
intersects two parallel lines; conclusion: corresponding angles
are congruent. 25. H 26. B 27. 216
28.
4
y
2
x
4
2
2
4
2
Mid-Course Test, Form D
1. -9, -11 2. 11 3. 16 4. T, R, S 5. D 6. 14
7. G 8. 45 9. C 100.5 in., A 804.2 in.2 10. 10
11. m1 = 38, m2 = 52 12. SAS 13. HL 14. not
possible 15. SSS 16. AAS 17. ASA 18. 19 19. A
Answers
Geometry
159
Answers
29.
4
(continued)
27. A, C, B 28. 17.9 29. never 30. sometimes
31. always 32. sometimes 33. never 34. 18 35. 34
36. 92.3 in.2 37. 100 38. 25 : 4 39. 199.0 cm2 40. 9.7
41.
y
2
x
2
2
4
2
42. 59 55.6% 43a. 1809.6 cm2 43b. 7238.2 cm3 44a. If
4
30. 52 31. (2, 1) 32. 160 cm 33. 23 34. J 35. 47
36. 80 37. 67 38. MZ TR or WZ WR
39. m1 = 52, m2 = 30, m3 = 98
40. m1 = 59, m2 = 62
Final Test, Form A
1. 36 2. (2.5, 1) 3. B 4. G 5. (6, -5) 6. 128 cm2
7. 48 cm2 8. reflectional and rotational symmetry 9. D
12 cm
10.
two angles have the same measure, then they are congruent.
44b. If two angles are not congruent, then they do not have
the same measure. 45a. PQ 45b. F 46. 13.0
47. AAS 48. rhombus 49. 113 50. y = -13 x + 10
Final Test, Form B
1. 45 2. (9, 2) 3. C 4. J 5. (-1, -1) 6. 216 "3 cm2
7. 96 in.2 8. rotational and reflectional symmetry 9. B
10. surface area: 84 cm2
5 cm
6 cm
5 cm
5 cm
All rights reserved.
4
10p
p cm
11. 13.6 ft
12.
Fro
nt
Rig
ht
13. 47.0 cm2 14. 1280 m3 15. The locus would consist of
Front
Right
13. 120 cm2 14. 1296 cm2 15. two parallel, horizontal
lines, one 3 units above and one 3 units below y = -2
y
5
4 2
2
4
x
a congruent segment 5 cm above and another 5 cm below
the given segment. The endpoints of these two segments
would be connected with semicircles of radius 5 cm.
16. 144 "3 in.2 17. 8.9 18. H 19. 6.4 20. 10.5
21. 175 cm2 22. 63 m 23. 44 m
24. 53.2 mi/h; 41.2° west of north 25. 11 26. (-1, -6)
27. C, B, A 28. 10.8 29. always 30. sometimes
31. never 32. sometimes 33. sometimes 34. 5.5 35. 32.5
36. 24.6 cm2 37. 80 38. 81 : 25 39. 182.8 cm2 40. 6
41.
y 2 2
4
16. 40 m2 17. 15.0 18. F 19. 6.5 20. 7.5 21. 54 cm2
22. 110 ft 23. 64 ft 24. 9, 1 25. 7 26. (-2, -2)
160
Geometry
Answers
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
11. 14.0 in.
12.
Answers
(continued)
42. 41 = 25% 43a. 282.7 m2 43b. 314.2 m3
44a. If two angles are congruent, then they are vertical.
44b. If two angles are not vertical, then they are not
congruent. 45a. AE 45b. B 46. 17.3 47. HL
48. square 49. 125 50. y = -12 x - 2
Final Test, Form D
triangle are congruent; conclusion: the angles opposite those
sides are also congruent 4. H 5. A 6. 62 7. 74 8. 91
9. G 10. B 11. C 12. A 13. D 14. A 380.1 in.2,
C 69.1 in. 15. (-3, -4) 16. y = -x + 1 17. AAS
Theorem 18. ASA Postulate 19. not possible 20. SAS
Postulate 21. HL Theorem 22. x = 8, y = 4"3
23. F 24. 200p cm3 25. 40 26. B 27. 90 28. 100 m2
29. 11 ft 30. 17 31. H 32. 28 33. 71 34. 180 35. acute
36. center: (-4, 3); radius: 7 37. AC 41.3, CD 33.4
38. 288 square units 39. 8 40. 43.30 cm2
1. x = 18, y = 147 2. D 3. hypothesis: if a point is on the
perpendicular bisector of a segment; conclusion: it is
equidistant from the endpoints of the segment 4. J
5. A 6. 150 7. 78 8. 76 9. G 10. A 11. D 12. C
13. B 14. A 907.9 m2, C 106.8 m 15. (6, -2)
16. y = -x - 8 17. SAS Postulate 18. not possible
19. SSS Postulate 20. HL Theorem 21. AAS Theorem
22. x = 12, y = 6"3 23. F 24. 540p ft3 25. 8 26. D
27. 180 28. 128 cm2 29. 14 ft 30. 25 31. G 32. 66
33. 68 34. 174 35. obtuse 36. center: (-6, 5); radius: 8
37. MW 43.5, MX 34.3 38. 80 39. 15 40. 173.21 in.2
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
1. x = 19, y = 135 2. D 3. hypothesis: if two sides of a
Final Test, Form E
Answers
Geometry
161
(continued)
Test-Taking Strategies
Interpreting Data
1. C 2. G 3. C 4. J
Writing Gridded Responses
1. 6 2. 5 3. 11.4 4. 326.7 5. 615.75 6. 2.5 7. 12
8. 51 9. 87 10. 690 11. 552.9 12. 13.30
Using a Variable
is found,all work is shown, and a diagram is complete and
correct.2c. 1 point; correct answer, but did not complete a
drawing.3a. 0 points; incomplete and incorrect response
3b. 1 point; work is shown, however it contains a math error
Writing Extended Responses
1. No steps are shown and no justification is made about why
m/1 = m/2.
2.
B
4x 30
A
C
F
Answers may vary: Sample: BD bisects /ABC, therefore
m/CBD = m/ABD.
2x = 4x - 30
x = 15
m/CBD = 30
m/BDC = 90
The measure of an exterior angle is the sum of
the two remote interior angles therefore the
m/BCF = m/CBD + m/BDC
= 30 + 90
= 120.
3. Answers may vary. Sample: Find the slope of the line.
m 5 1 2 3 5 2 2 5 21 .
6
3
3 2 (23)
Parallel lines have the same slope.
y 1 1 5 21 (x 1 1) Substitute m = -13 and (-1, -1)
3
in the point-slope formula.
y 1 1 5 21 x 2 1
3
3
4
1
Solve for y to write the equation in
y52 x2
3
3
slope-intercept form.
(3, 3) y
2
(3, 1)
x
2 O
(1, 1)
2
162
Finding Multiple Correct Answers
1. C 2. J 3. C 4. F 5. B 6. G 7. C
Testing Multiple Choices
1. B 2. G 3. C 4. F 5. B 6. J 7. C 8. G 9. B 10. F
Eliminating Answers
1a. Answers may vary. Sample: D can be eliminated because
2x
D
Drawing a Diagram
1. rhombus 2. BE = ED = 8 and AE = EC = 24
3. Two side are 22 and two sides are 14 4. Y(-8, 2); Z(2, 6)
5. 35, 35 6. 79, 79 7. (0, 1), (3, 0) 8. (5, 5) 9. (-6, -6),
m = 1 10. 12 1 6"2 or 20.49 11. x = 1; each diagonal
is 8 12. 120° 13. Two sides are 12 and two sides are 32
All rights reserved.
Writing Short Responses
1a. 2 points; all parts are solved, work is shown, and answers
are correct. 1b. 0 points; no work shown and answer is
incorrect. 1c. 1 point; error in procedure, but problem is set
up correctly. incomplete answer 2a. 0 points; incorrect and
incomplete response 2b. 2 points; the measure of each angle
7 , 0b
1. 108, 78, and 94 2. 40, 60, and 80 3. (6.5, 0) 4. a2 19
5. 8 or 12 6. 40 7. 60 8. 2212 9. 4 in. 10. 12
11. 3 12. 7 13. 4 14. 108°
2
Geometry
142 = 196 and the length of the side will be less than 14.
A can be eliminated because the length of the side will be
longer than 7. 1b. C 2a. The angle is just a bit larger than a
30° angle, so using the special right triangle relationships you
can eliminate F. The hypotenuse is the longest side so you can
eliminate J. 2b. G 3a. The angle is just a little bit smaller
than a 30° angle, so using the special right triangle
relationships you can eliminate A and D. 3b. B 4a. H and
J are obtuse angles. 4b. G 5a. A and B are too small since
the area will be greater than 12 ? 7 ? 48.
5b. C 6a. The first coordinate has to be -9 so choices G
and J can be eliminated. 6b. F
Choosing “Cannot Be Determined”
1. C 2. G 3. D; need lateral surface area or total surface
area. 4. G 5. B 6. J; need the dimensions of the garden.
7. A 8. J; need the radius. 9. C 10. G 11. D; need the
length of the edge of the base. 12. G
Using Estimation
1. A 2. F 3. C 4. G 5. A 6. H
Answering the Question Asked
1. the sum of the x- and the y-coordinates of the reflected
point; D 2. the y-coordinate of the reflected point; J 3. the
distance of the translated point to the x-axis; B 4. the vector
of the translation; G 5. the sum of the x- and y-coordinates
of the rotated point; D 6. the difference between the
y-coordinate and the x-coordinate of the image after
translation; H 7. The length A'B'; B 8. lines of symmetry in
the figure; H 9. a false property of a reg. hexagon; C
10. image of (8, -5); G 11. the equation of the line after
reflection; B 12. the image after the dilation; H
Answers
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
Answers
Answers: SAT/ACT Practice Test
All rights reserved.
Multiple Choice
1.
A
B
C
D
E
12.
A
B
C
D
E
23.
A
B
C
D
E
2.
A
B
C
D
E
13.
A
B
C
D
E
24.
A
B
C
D
E
3.
A
B
C
D
E
14.
A
B
C
D
E
25.
A
B
C
D
E
4.
A
B
C
D
E
15.
A
B
C
D
E
26.
A
B
C
D
E
5.
A
B
C
D
E
16.
A
B
C
D
E
27.
A
B
C
D
E
6.
A
B
C
D
E
17.
A
B
C
D
E
28.
A
B
C
D
E
7.
A
B
C
D
E
18.
A
B
C
D
E
29.
A
B
C
D
E
8.
A
B
C
D
E
19.
A
B
C
D
E
30.
A
B
C
D
E
9.
A
B
C
D
E
20.
A
B
C
D
E
31.
A
B
C
D
E
10.
A
B
C
D
E
21.
A
B
C
D
E
32.
A
B
C
D
E
11.
A
B
C
D
E
22.
A
B
C
D
E
Student-Produced Responses
1.
2 / 5
/
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
/
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
4
4
4
3.
5
/
.
.
2.
/
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
.
7
/
1 0 8
/
/
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
.
4.
3
/
/
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
.
5.
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
3
4
4
4
4
4
.
.
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
6.
7 2 0
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
4
4
4
.
7.
2 0
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
8.
. 7 8
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
9.
64
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
10.
6
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
3
4
4
4
4
4
.
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
Answers
Geometry
163
Name
Class
Date
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
All rights reserved.
Answer Sheet
1.
A
B
C
D
27.
A
B
C
D
2.
F
G
H
J
28.
F
G
H
J
3.
A
B
C
D
29.
A
B
C
D
4.
F
G
H
J
30.
F
G
H
J
5.
A
B
C
D
31.
A
B
C
D
6.
F
G
H
J
32.
F
G
H
J
7.
A
B
C
D
33.
A
B
C
D
8.
F
G
H
J
34.
F
G
H
J
9.
A
B
C
D
35.
A
B
C
D
10.
F
G
H
J
36.
F
G
H
J
11.
A
B
C
D
37.
A
B
C
D
12.
F
G
H
J
38.
F
G
H
J
13.
A
B
C
D
39.
A
B
C
D
14.
F
G
H
J
40.
F
G
H
J
15.
A
B
C
D
41.
A
B
C
D
16.
F
G
H
J
42.
F
G
H
J
17.
A
B
C
D
43.
A
B
C
D
18.
F
G
H
J
44.
F
G
H
J
19.
A
B
C
D
45.
A
B
C
D
20.
F
G
H
J
46.
F
G
H
J
21.
A
B
C
D
47.
A
B
C
D
22.
F
G
H
J
48.
F
G
H
J
23.
A
B
C
D
49.
A
B
C
D
24.
F
G
H
J
50.
F
G
H
J
25.
A
B
C
D
51.
A
B
C
D
26.
F
G
H
J
52.
F
G
H
J
Answer Sheet
Geometry
165
Blank Sheet of Grids
2.
4.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
6.
7.
All rights reserved.
.
5.
8.
.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
9.
166
3.
10.
11.
12.
.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
.
/
.
/
.
/
.
.
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
0
1
2
3
4
5
6
7
8
9
Geometry
Answer Sheet
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
1.
Student Answer Sheet: SAT/ACT Practice Test
All rights reserved.
Multiple Choice
1.
A
B
C
D
E
12.
A
B
C
D
E
23.
A
B
C
D
E
2.
A
B
C
D
E
13.
A
B
C
D
E
24.
A
B
C
D
E
3.
A
B
C
D
E
14.
A
B
C
D
E
25.
A
B
C
D
E
4.
A
B
C
D
E
15.
A
B
C
D
E
26.
A
B
C
D
E
5.
A
B
C
D
E
16.
A
B
C
D
E
27.
A
B
C
D
E
6.
A
B
C
D
E
17.
A
B
C
D
E
28.
A
B
C
D
E
7.
A
B
C
D
E
18.
A
B
C
D
E
29.
A
B
C
D
E
8.
A
B
C
D
E
19.
A
B
C
D
E
30.
A
B
C
D
E
9.
A
B
C
D
E
20.
A
B
C
D
E
31.
A
B
C
D
E
10.
A
B
C
D
E
21.
A
B
C
D
E
32.
A
B
C
D
E
11.
A
B
C
D
E
22.
A
B
C
D
E
Student-Produced Responses
1.
2.
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
4
4
4
© Pearson Education, Inc., publishing as Pearson Prentice Hall.
.
3.
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
4.
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
5.
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
3
4
4
4
4
4
.
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
/
/
.
.
.
.
0
0
0
6.
7.
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
4
4
4
.
8.
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
9.
/
/
.
.
.
0
0
0
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
.
10.
/
/
.
.
.
0
0
0
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
4
4
4
4
4
4
4
4
4
.
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
5
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
7
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
8
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
9
Answer Sheet
Geometry
167
Answers
Activity 1
1. a. 12 ft; 2 s
b. 8 ft; 3 s
2. Check student’s work.
3. Walk at a constant speed away from the motion
detector.
4. Check student’s work.
5. Divide the change in distance by the change in time; ft/s
6. Check student’s work.
7. The speed equals the slope of the line;
vertical change
slope =
.
horizontal change
8. a line with a steeper slope
9. Students should sketch a line with a steeper slope
10. Students should sketch a line with a less steep slope.
11–12. Check student’s work.
13. Walk toward the motion detector.
14. Stand still; impossible, because the person would have
to be in two places at the same time.
15. the initial distance from the motion detector
16a. stood still, then walked at a constant speed away from
the detector, then stood still again
16b. walked away from the detector at a constant speed,
stood still, walked toward the detector at a faster,
constant speed
16c. walked away from the detector at a constant speed,
stood still, walked away from the detector at a much
faster, constant speed
17. Yes; the distance at the beginning and end of the graph
is the same.
18. Toward the detector; the steepness is greater so the
speed is greater.
Take a Walk, Part 2
1.
2.
3.
4.
5.
6–8.
9–10.
11.
12–14.
15.
16.
104
Activity 2
negative; positive
the line through A; 6
yes; y = 6x
Check students’ work.
Check students’ work; initial distance from the motion
detector
Check students’ work.
Check students’ work. If an equation models the data
well, the line should overlap the plot. If an equation
does not model the data well, the line may either have
a different slope or an incorrect y–intercept.
The model will predict reasonable values for x 0
and x the final time measured by the CBL 2 unit.
Other values of x predict values before or after the
student was walking.
Check students’ work.
No; the y–intercept will be negative and does not represent the initial distance from the motion detector to
the student, because the student stood still for 2 s.
Yes; the slope is the speed of the student after he or
she began walking.
Answers
Synchronized Strut
Activity 3
1. Lines have different positive slopes and have the same
y–intercept.
2. parallel with different y-intercepts
3. Same line
4–5. Check students’ work.
6. y–intercept; represents initial distance from the motion
detector
7–9. Check students’ work.
10. Speed each student walks
11. They walk at different speeds.
12. Check students’ work.
13. the student whose graph has a steeper slope
14–15. Check students’ work.
16. The slopes should be about the same because they
walked at the same speed.
17. They have the same slope.
18. m
19. The difference is constant because they are walking at
the same speed.
20. They are the same line.
Coming and Going
Activity 4
1. No, a motion detector is like a function machine
because there is only one distance for any given time.
2. Students should sketch a line of positive slope with a
y–intercept near zero. They should also sketch a line of
negative slope with a y-intercept near the top of the
screen. The negative slope indicates a person walking
toward the detector. The positive slope indicates a person walking away from the detector. The y–intercepts
show how far from the detectors each person started.
3. Not necessarily. It can be approximated from the data,
however.
4–7. Check students’ work.
8. Check students’ work. If an equation models the data
well, the line should overlap the plot. If an equation
does not model the data well, the line may either have
a different slope or an incorrect y–intercept.
9–10. Check students’ work.
11. The measurements should agree closely with the
calculations. Discrepancies might be caused by delays
due to the reaction times of the marker and timer.
12. A motion detector records only the distance to the
closest object in front of it. A motion detector cannot
record two objects at once. A function has only one
y–value for each x–value.
13. Check students’ work.
Keep Your Eye on the Ball
Activity 5
1. The student should sketch a series of parabolas
opening downward with maximum values that
gradually decrease.
2. decrease
3. Yes, each rebound is shorter than the previous bounce.
4–5. Check students’ work.
Technology Activities
© Pearson Education, Inc. All rights reserved.
Take a Walk, Part 1
Answers
(continued)
6. vary
7. decreasing
8. Multiply a height value by 75% or 80% and see if the
result is the height of the previous bounce.
9. yes; constant
10. Check students’ work.
11. b = -2ax; check students’ work.
12–16. Check students’ work.
© Pearson Education, Inc. All rights reserved.
Race Cars
Activity 6
1. Yes; the battery-powered car gives a linear graph
because it travels at a constant speed. The speed of the
wind-up car changes, so its graph will show a curve as
it speeds up and a curve as it slows down.
2. Students should sketch a line for the battery-powered
car and a curve showing the acceleration and deceleration of the wind-up car.
3. The battery-powered car can be modeled by a line. The
wind-up car can be modeled by two parabolas, one for
acceleration and one for deceleration.
4. Check students’ work; a line
5. Check students’ work; the slope is the speed and the
y–intercept is the initial distance from the motion
detector.
6. Check students’ work. The equations may not model
the data after the car initially starts driving because
there is some acceleration.
7. Check students’ work.
8. The values should be very close. The slope might be
higher than the average speed from the graph because
the slope ignores the brief acceleration of the car.
9. 0 x 1.5; students may want the domain to extend
beyond x = 1.5 because the car continues to drive.
Any domain that they can justify is reasonable, but it
should not be very long, since the car may stop.
10. The collected data would span a longer period of time.
Everything else should remain constant.
11. Check students’ work. Should be approximately the
same as the previous model.
12. Check students’ work; yes.
13. 0 x 6
14. Check students’ work.
15. The data will appear linear but you would expect
some type of curve. You will see a line because you are
recording data for a very short time. But the springloaded car must be either speeding up or slowing down
since it does not go at a constant speed as does the
battery-powered car. Thus you would expect to see a
curve.
16. A longer time interval will give a better look at the
car’s motion. You will probably see s-shaped data. You
will see a parabola opening upward for when the car
starts and is speeding up and a parabola opening
downward for when it is slowing at the end.
Technology Activities
Back and Forth It Goes
Activity 7
1. Sketch should resemble a cosine curve with negative
amplitude.
2. decrease; increase
3. Check students’ work.
4. Students should label the minimums and maximums
of the graph. The motion detector records points at
regular intervals. It probably does not record the actual
minimum and maximum but records points near the
maximum and minimum.
5–9. Check students’ work.
10. Check students’ work; yes; non-linear
2
11. y = g a x b ; check students’ work; it should be a
2p
good model.
12. 1.74; the two methods should give close answers but
there will be some experimental discrepancy.
13. l = 0.64; the answer is reasonable, although it does
not agree exactly with the measurement from the
activity.
l ; independent; as you change the length of
Äg
the pendulum, the period changes.
15. Check students’ work. The model gives T = 2.84 s.
The model and the experiment should give similar
answers.
16. You should shorten it.
14. t = 2p
Charge It!
Activity 8
1. f(x); g(x)
2. a is the starting amount or the y–intercept. b is the
base, the growth factor, or the decay factor.
3. b 1; 0 b 1
4. 1; 2; y = 2x
5. 30; 0.75; y = 30(0.75)x
6–7. Check students’ work.
8. Check students’ work; it is the voltage of the battery.
9. The two answers should be close. The y–intercept may
be lower because the capacitor may still have been
collecting charge.
10. They are the same.
11. exponential decay
12–15. Check students’ work.
16. They should be very close.
17. No, exponential functions never reach their asymptotes. No, the voltage of the capacitor will eventually
reach zero.
18. 0.9782; they should be close.
Falling Objects
Activity 9
1. Students should graph a horizontal line with a
positive y–intercept that changes to the right side of
a downward-opening parabola. There should be a
horizontal line once the parabola reaches the x–axis.
Answers
105
(continued)
2. quadratic
3. Check students’ work. First the book was held over the
motion detector. Next it fell. Finally it lies on top of the
motion detector.
4–11. Check students’ work.
12. The vertex represents the time at which the student
dropped the book and the book’s initial height above
the motion detector.
13. Check students’ work. Only the part of the parabola
that corresponds to the actual time the book was
falling should be included in the domain. At the other
times, the book was actually stationary.
14. Check students’ work.
15. The model should not agree exactly. The physics model
assumes that there is no air resistance. The book is
affected by air resistance. The physics model assumes
that the book starts to fall when t = 0. In the activity,
the book was held briefly so that it started to fall after
t = 0.
Bouncing Ball, Part 1
Activity 10
1. The student should sketch a series of parabolas that
open upward that all stop about 5 units above the
x–axis.
2. The student should sketch a series of parabolas opening downward with maximum values that gradually
decrease. Students who understand that the motion
detector measures distance from itself should understand that the final result will be a graph of the distance from the floor instead of the distance from the
motion detector.
3. The ball will bounce to a maximum height that is a
percent of the previous height. The percent should be
constant.
4–5. Check students’ work.
6. One method is to substitute the coordinates of the
vertex for h and k; next substitute the coordinates of a
point on the parabola for x and y; then solve for a.
Another method is to find three points on the parabola
and substitute their coordinates into the equation
y = ax2 + bx + c; then solve for a, b, and c.
7. Check students’ work. The curve is too wide and opens
up instead of down.
8. A parabola with a negative a opens down. A parabola
with a positive a opens up; negative
9. The curve is narrower. A larger a means a narrower
curve. A smaller a means a wider curve.
10. Check students’ work, but a should be near -16.
11. No; if students use an actual data point, their models
may be off in height or not centered on the bounce
horizontally.
12. Check students’ work. They should be similar.
13. a is the acceleration due to gravity, h is the time when
the ball is at the maximum height, k is the maximum
height of the ball for the bounce. a should remain
constant because gravity is constant.
106
Answers
Bouncing Ball, Part 2
Activity 11
1. The student should sketch a series of parabolas
opening downward with maximum values that
gradually decrease.
2. eventually comes to a stop
3. decrease
4. remains constant
5–8. Check students’ work.
9. exponential or possibly quadratic
10. A quadratic equation does not have an asymptote.
After an infinite amount of time, a quadratic equation
predicts that the ball is an infinite distance below the
floor. After the same amount of time, an exponential
equation predicts that the ball is at 0 ft, or resting on
the floor.
11–13. Check students’ work.
14. Check students’ work. A well-fitting model should
approximately touch the vertex of each bounce.
15–16. Check students’ work.
17. coefficient of restitution
When’s the Tea Ready?
Activity 12
1. 100°C; room temperature. A liquid cannot cool below
room temperature.
2. Students should draw an exponential function with an
asymptote at room temperature.
3. The graphs will be similar, but the probe should cool
more quickly because it has less mass.
4–5. Check students’ work.
6. exponential
7. the x–axis
8. Check students’ work; no.
9. By adding a constant c, the graph of the exponential
equation is translated vertically and the asymptote is
no longer the x–axis.
10. c should be given the value found in Exercise 8; it is
the temperature of the classroom.
11–13. Check students’ work.
14. Check students’ work. The model should fit well
because it overlaps most of the data points. The model
may not fit the first few points.
15. Check students’ work.
16. No, a is the difference between the initial value and the
room temperature, c.
17. Check students’ work. This model is a vertical translation of the previous model. The translation is necessary
because the room temperature is greater than zero.
18. yes; yes
19. yes; yes
20–21. Check students’ work.
22. a is the difference between room temperature and the
initial temperature of the water. b is the rate at which
the water cools; it does not change, c is room temperature; it does not change. Check students’ work.
Technology Activities
© Pearson Education, Inc. All rights reserved.
Answers
Answers
(continued)
Full Speed Ahead
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
© Pearson Education, Inc. All rights reserved.
14–15.
16.
Activity 13
Students’ should sketch a line.
Students should sketch a curve opening upward.
Students should sketch a curve opening downward.
Check students’ work; yes.
Check students’ work; the slope is the speed and the
y–intercept is the initial distance from the motion
detector.
The model should fit reasonably well, the residual plot
should show a definite parabolic pattern.
quadratic because the speed is increasing
Check students’ work.
The model should fit better. The residuals should
confirm the quality of the model.
The speed is increasing because as x increases the
change in y divided by the change in x is getting larger.
This means that the car is traveling farther during each
x-unit.
The sketch should show a parabola opening upward
representing an increase in speed joined to a parabola
opening downward. The data shows the car speeding
up at the start and then slowing down.
The model is reasonable only over the domain
0 x 1.5. The time interval was too short to see
the car slow down.
Check students’ work. Students should label when the
car is accelerating and decelerating.
Check students’ work. Models should fit well.
Students can use the intersect option on the graphing
calculator or solve the system of quadratic equations
formed by the two models. Check students’ work.
In the Swing of Things
Activity 14
1. The graph should be a periodic function with a maximum or minimum at the y–axis.
2. the length of the string
3. cosine or sine
4. a = amplitude; b = 2p divided by the period;
c = horizontal translation; d = vertical translation
5–7. Check students’ work.
8. Take half the difference between the y–value of a peak
and the y–value of a trough. Check students’ work.
9. Take the difference between the x–values of two peaks
or two trough. Check students’ work.
10. Find the x–value of the first peak. Check students’ work.
11. Take the average of the y–values of one peak and one
trough. Check students’ work.
12–13. Check students’ work.
14. The amplitude is greater, because the pendulum is
pulled back farther. The graph has a greater vertical
translation because the motion detector sill needs to be
1.5 m away from the closest swing of the pendulum.
15. The period decreases because the string is longer. The
amplitude and vertical translation increase because a
longer string means that the pendulum starts farther
away from the center of its swing.
Technology Activities
16. The magnitude of a is determined by the angle at
which the string is pulled back and by the length of
the string. b decreases with the length of the string.
c increases when the motion detector is started after
the pendulum is swinging. d increases when the string
is pulled back farther and when the string is longer.
Playing With Numbers
Activity 15
Investigate
1. See screen
2. The digits are the same but in a different order.
3. The digits are the same but in a different order.
Exercises 1–3
4. Yes
5. Same digits with zero included
Exercise 4
Exercise 5
6. Multiples of 3 have been omitted from the list.
7. 22A, 23A, 25A, 26A
8. The same patterns occur.
Investigate
1. See screen
Exercise 1
Exercise 4
2. 111112 = 123454321
3. The paper and pencil calculation of 1112 is as follows:
111
3111
111
111
111
12321
Notice that the number of one’s in each column
increases (then decreases) by 1 from right to left.
4. It can be seen that the digits increase (then decrease)
by 1 from left to right.
5. The last 7 in the number has been rounded up from 6.
Answers
107
(continued)
6. Using paper and pencil, it can be seen that there is a
column of 10 one’s, whose sum is 10. This results in a
“carry-over” which causes the pattern to break down.
Investigate
1.
2.
3.
4.
Answers may vary. Sample: 82471
Answers may vary. Sample: 27814
Answers may vary. Sample: 82471 - 27814 = 54657
Answers may vary. Sample:
5 + 4 + 6 + 5 + 7 = 27; 72 - 27 = 45;
4+5=9
5. A sum of 9 always results.
6. Variables can take on any numerical value, therefore
proving a result for any number.
Exercises
1. The result contains five copies of the two-digit number
(i.e. 48 ? 101010101 = 4848484848).
2. The number is repeated as a decimal (i.e. 4 = 4).
9
They do not terminate; they keep repeating.
3. Yes. In this case 9 = 1 and .9 = 1 as well.
9
4. The number is repeated as a decimal (i.e. 36 = .36).
99
5. Yes. In this case 99 = 1 and .99 = 1 as well.
99
Finding Real Roots
of Equations
Activity 16
Investigate
1–3. Check student’s work.
4. .8241323123
5. -.8241323123
Investigate
Investigate
1. Check student’s work.
2. .8241323123
3. -.8241323123
Exercises
1. They are approximations.
2. .3862368706; 1.961569035
3. The (x, y) coordinates of the graphs are real numbers,
so graphical methods can only reveal real answers.
Activity 17
Investigate
1–3. Check student’s work.
Investigate
4. Check student’s work.
5. All values of x where the graph is below the x-axis.
108
Answers
Investigate
7. Check student’s work.
8. The inequality returns a value of -1 for 1 x 3
and 0 everywhere else. The points on the graph with
y-coordinate 0 are hidden by the x-axis.
Exercises
1. -2 x 1; methods of solution may vary
2. - 17 x 57
3. 43 x 3
4. 1 x 2
Magic Pricing Numbers
Activity 18
Investigate
1. 1 or 24–1
24
2. 1.16
3. .9
4. 1.028
5. 1 ? 1.16 ? .9 ? 1.028 = .044718
24
6. y = .044718x
7. Here are two examples to describe the rounding
process. The number 31.243 typically rounds down to
31.24 (to the nearest penny). By adding .005, you get
31.243 + .005 = 31.248 which rounds up to 31.25.
The number 31.247 typically rounds up to 31.25. By
adding .005, you get 31.247 + .005 = 31.252 which
still rounds to 31.25.
8. Yes
9. $28.81
Exercises
1. $21.62
1 ? 1.16 ? .9 ? 1.028 = .029812;
2. Multiply 857.98 by 36
y = .029812(857.98) = $25.58
3. w = (.044718)–1 = 22.36236x, where w = wholesale
price and x = retail price
6. Check student’s work.
7. -.8241323123
Solving Absolute Value
Inequalities Graphically
6. All values of x where the graph is on or above the x-axis.
Mrs. Murphy’s
Algebra Test Scores
Activity 19
Investigate
1. first line: 30 units; second line: 20 units
2. x - 42
42
3. x 2
30
y 2 78
20
y 2 78
x
2
42
5. 30 =
20
6. The graph is a straight line.
4. y - 78;
Exercises
1. 60
Technology Activities
© Pearson Education, Inc. All rights reserved.
Answers
Answers
(continued)
y 2 78
2. x 2 32 =
or y = 1 (x - 32) + 78
2
40
20
4–6. Check student’s work.
Exercises
1.
y 2 72
3
or y = (x - 40) + 72
3. x 2 40 =
36
4
27
Exploring Point-Slope Form
Activity 20
Investigate
1. y - 1 = 3(x - 2)
2.
2–4. Check student’s work.
5. L3 = {-2, -1, -.5, 0, .5, 1, 2}
Exercises
1.
3.
2.
4.
3.
Burgers cost $2.95; Fries cost $1.80; Soft Drinks
cost $1.10
5. If it costs $25.04 for 4 burgers, 4 fries, and 4 soft drinks,
then it should cost 25.04 2 = $12.52 for 2 burgers,
2 fries, and 2 soft drinks. The screen at right shows the
error message that results when [A]–1[C] is executed.
© Pearson Education, Inc. All rights reserved.
4.
Wheat on a Chessboard
Activity 22
Investigate
1.
5. Use L3 = {-2, -.5, 0, .5, 2, 200} and
Y1 = L3 * (X - 12) + 12
Fast Food Follies
Activity 21
Investigate
1. 3x + 4y + 4z = 16.45
5x + 2y + 5z = 19.30
4x + 6y + 6z = 23.50
2. The product of matrix A and matrix X will produce
matrix B.
3. If AX = B, then X = A–1B.
Technology Activities
Square
1
2
3
4
5
6
7
8
9
10
Num. of
Grains
1
2
4
8
16
32
64
128
256
512
2.
Num. of Total Num.
Squares of Grains
1
1
2
3
3
7
4
15
5
31
6
63
7
127
8
255
9
511
10
1023
Answers
109
(continued)
3. Raise 2 to the number of squares and subtract 1.
4. T(n) = 2n - 1 where n = number of squares and
T(n) = total number of grains; about 1.8 1019
grains of wheat
5. about 1.7 1013 or 17 trillion bushels
6. 6728.3 years
Exercises
1.
2.
3.
4.
5.
6.
1,075,000
21
20
255
4,294,967,295 (which is 232 - 1)
The wheat covering 51 squares provides a close match
to the 1998 U.S. wheat production.
7. about 2.6 inches
Manipulating a Polynomial
Activity 23
Investigate
1. Check student’s work.
2. It moves up.
3. Evaluate f(0) = 12 to find the y-intercept, (0, 12),
and 12 10.
4. The y-values range from -20 to 20 and the scale is 1,
which results in 40 tick marks on the y-axis.
5. Check student’s work.
Investigate
1. Because the polynomial has a value of 0 when x = -1.
2. Check student’s work.
3–4. Answers may vary. Sample:
f(x) = (5x4 - 17x3 + 13x2 - 8x + 12)(x + 1)(x + 3);
graph shown with Xmin = -4.7, Xmax = 4.7,
Xscl = 1, Ymin = -440, Ymax = 100, Yscl = 0
3.
Writing a Simple Program
Activity 24
Investigate
1–8. Check student’s work.
Exercises
1. Answers may vary. Sample: A = 1, B = 1, and
C = -20 has roots -5 and 4.
2. Answers may vary. Sample: A = 4, B = -12, and
C = 9 has root 3 .
2
3. Answers may vary. Sample: A = 2, B = -5, and
C = 7 has imaginary roots 5 - 1.392i and 5 + 1.392i.
4
4
4. Answers may vary. Sample: A = 6, B = 7, and
C = -5 has rational roots - 5 and 1 .
3
2
5. Answers may vary. Sample: A = 3, B = 8, and
24 4 !37
C = -7 has irrational roots
3
(-3.361 and .694).
6. a. $625.72
b. $2236.58
Repeated Radicals
Activity 25
Exercises
1. 9
2.
Exercises
1. Answers may vary. Sample:
f(x) = (x + 3)(x - 1)(x - 3)(x - 5); graph shown
with Xmin = -4.7, Xmax = 6, Xscl = 1,
Ymin = -120, Ymax = 20, Yscl = 0
2. No portion of the graph can be seen in the ZDecimal
window.
110
Answers
x = 6 + $6 1 #6 1 "6 1 !6 1 %
(x - 6)2 = a $6 1 #6 1 "6 1 !6 1 %b
2
x2 - 12x + 36 = 6 + #6 1 "6 1 !6 1 %
x2 - 12x + 36 = x
x2 - 13x + 36 = 0
(x - 4)(x - 9) = 0
x = 4 and x = 9
The solution x = 4 is extraneous, since x is clearly
greater than 6.
3. 5.302775638; the number is an approximation
7 1 !13
4.
; irrational
2
5. The calculator can only give an approximate answer,
whereas algebraic techniques give an exact answer.
Technology Activities
© Pearson Education, Inc. All rights reserved.
Answers
Answers
(continued)
Asymptotes and Holes
Activity 26
Investigate
1. The function is undefined at x = 42. This makes
sense algebraically since the denominator is equal to
zero at these two values of x.
2. It disappears.
3. The function approaches 4 on either side of the
singularity. This makes sense because the denominator
approaches zero as x approaches the singularity.
4. Vertical lines appear at the singularities.
5. The y–value ‘jumps’ from negative to positive (or vice
versa) as you trace through a singularity.
6. The calculator was drawing a line between two points
on either side of the singularity.
7. Yes; the singularity at x = -2 appears as a hole in the
graph.
8. The function is undefined when the denominator is
equal to zero.
9. As you approach the singularity at x = -2 from the
left or right, the corresponding y–values approach the
same number. As you approach the singularity at
x = 2 from the left, the y–values approach -. As
you approach the singularity at x = 2 from the right,
the y–values approach +.
10. No; the simplified function is defined at x = -2 and
the hole disappears.
11. When graphing Y1, the graphing ball ‘breaks’ when it
reaches x = -2. This does not happen when you
graph Y2.
Exercises
© Pearson Education, Inc. All rights reserved.
1. The first graph has a hole at x = 1 whereas the second
x2 2 1
is all real
graph does not. The domain of y =
x21
numbers except 1 and the domain of y = x + 1 is all
real numbers.
2
2. Answers may vary. Sample: f(x) =
(x 1 1) (x 2 1)
3
2
3. Answers may vary. Sample: f(x) = x 2 x or
x
2
1
x 2 (x 2 1)
f(x) =
x21
4. Check student’s work.
5. Type ‘.05 S 䉭X’ on the home screen to eliminate false
asymptotes.
Stepping Out
Activity 27
Investigate
1. The graph resembles a series of steps.
2. The range values are {1, 2, 4, 3, 4}. Answers may vary.
Sample: ‘Int’ rounds numbers down to the nearest
integer.
3. The range values are {-2, -2, -5, -4, -5}. Answers
may vary. Sample: ‘Int’ rounds numbers to the nearest
integer that is less than or equal to the number.
Technology Activities
4. The calculator will connect the endpoints of each step
to the next step.
5.
6. $6; $11; the function C(x) = 3.50 + 2.50x models a
cost that varies continuously as a function of mileage,
whereas the actual cost increases in steps.
7. The function rounds the mileage down, whereas the
cab company rounds the mileage up. For example, a
4.4 mile trip is charged at the same rate as a 4 mile trip
instead of a 5 mile trip.
8. C(x) = 3.50 + 2.50int(x + 1)
Exercises
1. C(x) = .34 + .23int(x + 1)
2. The two functions are identical except at the integer
values. A careful graph of y = int(x + 1) would show
a closed dot on the left of each step and an open dot
on the right, while a careful graph of y = -int(-x)
would show an open dot on the left of each step and a
closed dot on the right.
3. Answers may vary. Sample: Y1 = 3.5 - 2.5int(-x);
Answers may vary. Sample: Y1 = .34 - .23int(-x)
Systems With Many
Solutions
Activity 28
Investigate
1. The error message ‘ERR:SINGULAR MAT’ appears.
2. Yes
3. 2x + 3y + z = 6.85
4x + 3y + 3z = 11.65
3x + 3y + 2z = 9.25;
Yes; the prices given in the previous question solve
the system.
4. Because matrix A does not have an inverse.
5. $6.85; $11.65; $9.25
6. $6.85; $11.65; $9.25
7. There can only be one matrix [A]–1[B], so the calculator
could not produce multiple correct answers in this way.
Answers
111
(continued)
Exercises
1. Check students’ work.
2. Any values of x, y, and z satisfy the equation.
3. 3 ay 2 1 zb = 3 a 41 b S 3y - z = 41 S
3
20
60
3y - z = 2.05
4. x + z = 12
5
5. x = 2.2; z = .2
6. 2(2.2) + 3(.75) + 2 = 6.85 S 6.85 = 6.85
4(2.2) + 3(.75) + 3(2) = 11.65 S 11.65 = 11.65
3(2.2) + 3(.75) + 2(2) = 9.25 S 9.25 = 9.25
Extend
7. the cost of a burger is too low; some of the numbers
are not rounded to the nearest penny.
8. Answers may vary. Sample: $1.30, $1.05, $1.10; $1.60,
$0.95, $0.80; $1.90, $0.85, $0.50
Parabolic Reflectors
Activity 29
Investigate
tional. This explains why no graph results.
6. The calculator only evaluates y = (-1)x for values of
x that only have real answers (such as x = 2 , 3 , 4 , c).
5 5 5
These values cause y to alternate between -1 and 1.
7. The calculator only evaluates y = (-1)x for values of
x whose fractions reduce such that the numerators are
always even numbers (such as x = 2 , 4 , 6 , c). These
5 5 5
x–values always produce y–values of 1.
8. The calculator only evaluates y = (-1)x for values of
x whose fractions reduce such that the numerators are
always odd numbers (such as x = 1 , 3 , 5 , c). These
5 5 5
x–values always produce y–values of -1.
Exercises
1. y = (-2)x; y = (-2)–x; y = 2(-1)x
2. p is irrational—it cannot be written as the quotient of
two integers; use a rational approximation for p such
as 314159 .
100000
Bull’s-eyes
1–4. Check students’ work.
5. They are equal.
6. Check students’ work.
Activity 31
Investigate
Exercises
1. (0, 4)
2. closer
3. By placing the lamp at the focus of a parabolic
reflector, all rays that strike the reflector are directed
outward parallel to the axis. This helps ‘concentrate’
the light in a specific direction.
Exploring Powers of –1
1. Check students’ work.
2. Check students’ work.
3.
Activity 30
Investigate
1. Check students’ work.
2
1
2. (21) 6 = -1; f (21) 2 g 6 = 1; The calculator
2
1
evaluates (21) 6 as (21) 3 = -1, whereas the
law of exponents would seem to imply that
2
1
1
(21) 6 = f (21) 2 g 6 = 16 = 1. The answer depends
on the fraction being reduced to lowest terms before
the laws of exponents are applied.
3. It reduces 2 to 1 , which is the same as taking the cube
6 3
root.
4. Values of x are considered as fractions in lowest terms
before being used as exponents. For example, f(.4) is
4. the second and third one
5. type the function shown below
evaluated as f a 2 b = (21) 5 = 1. f(.5) is evaluated
5
2
as f a 1 b = (21) 2 which is undefined and f(.6) is
2
1
evaluated as f a 3 b = (21) 5 = -1.
5
5. Using this window, the calculator attempts (but fails)
to evaluate y = (-1)x for values of x that are irra3
112
Answers
Technology Activities
© Pearson Education, Inc. All rights reserved.
Answers
Answers
(continued)
Exercises
Exercises
1. D: -2 x 2
R: 0 y 2
D: -4 x 4
R: 0 y 4
D: -4 x 4
R: 0 y 2
D: -2 x 2
R: 0 y 4
(x 2 2 1) (x 2 2 4)
x6 1 1
2. The graph crosses its own asymptote an infinite number of times.
3. Yes, because the function gets arbitrarily close to
y = 2 as x approaches 4. The fact that the function
actually equals 2 is technically not an issue.
4. Answers may vary. Sample: x2 if x 0
1. Answers may vary. Sample: y =
Making Ellipses
out of Circles
Activity 34
Investigate
1–3. Check students’ work.
4. It looks identical to the graph shown in the introduction.
5. A horizontal ellipse with major axis of length 6 and
minor axis of length 4 is a circle of radius 1 that is
stretched horizontally by a factor of 3 and vertically by
a factor of 2. These scale factors can be produced by
dividing the x–axis window values by 3 and the y–axis
window values by 2.
2.
Exercises
1. Divide Xmin, Xmax, and Xscl by 2; divide Ymin,
Ymax, and Yscl by 3.
3. Check students’ work.
The Natural Logarithm
Activity 32
© Pearson Education, Inc. All rights reserved.
Investigate
1.
2.
3.
4.
Check students’ work.
1
Check students’ work.
Upper limit = e
2. Divide Xmin, Xmax, and Xscl by 4; divide Ymin,
Ymax, and Yscl by 3.
Exercises
1. .6931471806; ln(2) = .6931471806; the two numbers
are equal
2. they are equal
3. they are equal
4. it is negative
5. For any a 0, ln(a) = the area under the graph of
1 between x = 1 and x = a, with “area” being
y= x
negative for 0 a 1.
Can a Graph Cross Its
Own Asymptote?
3. Divide the X–Window values by 6 and the Y–Window
values by 2. The graph does not fit in the window.
Activity 33
Investigate
1. The x–axis
2. no
3. yes; Y2 =
4. twice
5. three times
6. yes
x
(x 2 1 1)
Technology Activities
Answers
113
Answers
(continued)
Exercises
4. First, multiply Xmin, Xmax, Ymin, and Ymax by 2.
Then follow the steps in the answer to Exercise #3
above.
1.
2.
3.
4.
Check students’ work.
5.187377518; ln(100) 艐 4.605, 1 + ln(100) 艐 5.605
6.79282343; ln(500) 艐 6.215, 1 + ln(500) 艐 7.215
No; it’s between ln(109) = 20.723 and
1 + ln(109) = 21.723.
5. e100 艐 2.688 1043 terms
Monte Carlo π
5. First, multiply Xmin, Xmax, Ymin, and Ymax by 2.
Then divide Xmin, Xmax, and Xscl by 2 and divide
Ymin, Ymax, and Yscl by 6.
6. First, multiply Xmin, Xmax, Ymin, and Ymax by 2.
Then divide Xmin, Xmax, and Xscl by 2 and divide
Ymin, Ymax, and Yscl by 6. Finally, add 3/6 to Ymin
and Ymax to re-center the screen vertically. The circle
command will be ‘Circle(2/2, 3/6, 1).’
The Harmonic Series
Activity 35
Investigate
1. The graph rises (slowly) without bound.
2. y = ex; all reals
3. 1, 1 , 1 , c ; 1, 1 , 1 , c
2 3
2 3
4. In both figures, the area of the rectangles represents
the terms of the harmonic sequence. The expression
ln(a) represents the area under the curve from x = 1
to x = a. Since the rectangles are all above the curve,
1 ln(a) must be
the inequality 1 1 1 1 1 1 % 1 a
2
3
true. In the second inequality, since the rectangles are
entirely contained within the region below the curve,
1 1 ln(a),
the inequality, 1 1 1 1 1 1 % 1 a
2
3
must be true.
5. between 9.21 and 10.21
114
Answers
Activity 36
Investigate
1. The probability that a randomly selected point lies
within the shaded region is equal to the ratio of the
area of the shaded region to the area of the square,
p
that is, 4 5 p
4
1
2. Check students’ work.
3. Because the coordinates, A and B, will always be less
than or equal to 1.
4. Answers may vary. Sample: The ratio will most likely
fall between 2 and 5, with most results clustered near 3.2.
5. Answers may vary. Sample: The ratio should
consistently be very close to 3.14.
Exercises
1. Line 1: Clears any drawings
Line 2: Turns off all plots
Line 3: Turns off any functions located in the ‘Y=’ editor
Line 4: Sets the viewing window, graphs square, graphs
circle
Line 5: Set the value of H equal to 0
Line 6: Prompts the user to input the number of points
for the simulation
Line 7: Sets up a loop that will execute the next 4 lines
N times; K is a counter
Line 8: Stores random numbers between 0 and 1 as
A and B
Line 9: Plots the ordered pairs (A, B) on the graph
Line 10: Calculates the distance from the origin to point
(A, B) and tests if this distance is less than or equal to 1
(determines if the point is within the circle)
Line 11: Adds 1 to the value of H if the point falls
within the circle.
Line 12: Specifies where the loop ends, which happens
when K is equal to N
Lines 13 and 14: Displays “4 * RATIO IS:” along with
the value of 4H/N (which should be approximately
equal to p)
2. Check students’ work.
The Way the Ball Bounces
Activity 37
Investigate
1–2. Check students’ work.
3. A ball bouncing on a spring.
Technology Activities
© Pearson Education, Inc. All rights reserved.
Extend
Answers
(continued)
4. The ball moves quicker as it passes through the origin
because its motion is mostly in the vertical direction
(keeping in mind that that ball is really traveling along
a unit circle). The ball moves slower at the top and
bottom of its path because its motion is mostly in the
horizontal direction.
5. Yes
1.
2.
3.
4.
The ball moves twice as fast.
Change Tmax to 18p.
3 cycles of a sine curve.
Nothing; the scale along the x–axis is so large that the
graph is entirely contained along the y–axis and cannot
be seen.
5. The wave has been compressed horizontally (because
of the large x–axis scale), giving a result identical to
step 3 of the investigation.
6. The vertical position of a point traveling along a unit
circle at a constant speed can be used to model harmonic motion. If the vertical motion is graphed as a
function of time, it generates a sine wave.
Graphing Ellipses and
Hyperbolas
Activity 38
Investigate
© Pearson Education, Inc. All rights reserved.
1–2. Check students’ work.
2
(3cosT) 2
(2sinT) 2
y2
5
1
3. x 1
=
9
4
9
4
2
2
9cos T 1 4sin T 5 cos2T + sin2T = 1; horizontal
9
4
stretch factor = a, vertical stretch factor = b
1 2 sin2T 5 1 2 sin2T 5
cos2T
cos2T
cos2T
cos2T 5 1; a hyperbola
cos2T
5. Check students’ work.
4. x2 - y2 =
Exercises
3 ; Y1T = 2sinT 5 2tanT
cosT
cosT
Vertical Angles
They are congruent.
m⬔EAD + m⬔EAB = 180
m⬔EAD + m⬔BAC = 180
(⬔EAD + ⬔EAB) - (⬔EAB + ⬔BAC) = 0
(⬔EAD + ⬔EAB) - (⬔EAB + ⬔BAC) = 0
simplifies to ⬔EAD - ⬔BAC = 0 which means
⬔EAD = ⬔BAC
Technology Activities
Activity 40
Investigate
1. The measure of an exterior angle of a triangle is equal
to the sum of the remote interior angles.
2. supplementary; ⬔A + ⬔B + ⬔ACB = 180 and
⬔ACB + ⬔BCD = 180; rewrite both equations as
⬔A + ⬔B = 180 - ⬔ACB and
⬔BCD = 180 - ⬔ACB; by substitution
⬔A + ⬔B = ⬔BCD
Extend
3. their sum is 90; they must be complementary since the
angles of a triangle have a sum of 90 and ⬔B = 90
4. their sum is 270;
⬔DAC + ⬔ACE = (⬔BAC + ⬔B) +
(⬔ACB + ⬔B) = (⬔BAC + ⬔ACB) +
(⬔B + ⬔B) = 90 + 180 = 270
Investigate Further
5. ⬔ADE - (⬔A + ⬔B + ⬔C) = 180;
⬔ADE + 180 = ⬔A + ⬔B + ⬔C;
⬔A + ⬔B + ⬔C + ⬔ADC = 360 and
⬔ADC + ⬔ADE = 180;
⬔A + ⬔B + ⬔C = 360 - ⬔ADC and
⬔ADE = 180 - ⬔ADC; by substitution
⬔ADE + 180 = ⬔A + ⬔B + ⬔C
6. the sum of the remote interior angles of a pentagon is
equal to the sum of the exterior angle and 360.
7. Answers may vary. Sample:
No. of
Sides
Ext. ⬔
Sum of
Remote Int. >
3
4
5
6
80
80
80
80
80
260
440
620
Ext. ⬔ Sum of
Remote Int. >
0
180
360
540
Activity 39
Investigate
1.
2.
3.
4.
5.
6. Check students’ work.
7. They are perpendicular.
8. Answers may vary. Sample: Adjacent angles are supplementary. The angle bisectors of these angles create
a pair of angles that are complementary.
Exterior Angles of Triangles
Exercises
1. tanT
2. 3; 2
3. X1T =
Extend
The sum of the remote interior angles of any convex
polygon is equal to E + (n - 3)180 where E is the
measure of the exterior angle and n is the number of
sides of the polygon.
Relationships in Triangles
Activity 41
Investigate
1. They are congruent.
2. The triangles are congruent by SAS.
3. BE ⬵ AD by CPCTC
Answers
115
(continued)
Extend
4. They are congruent.
5. 䉭FAC ⬵ 䉭BAE by SAS so CF ⬵ BE by CPCTC
Investigate Further
6.
7.
8.
9.
they are equal to 120
Each angle of the triangle must be less than 120.
yes
It is equal to the sum of the lengths of the two shortest
sides of the triangle.
Isosceles Triangles
Activity 42
Investigate
1. isosceles
2. ⬔DBC and ⬔DCB are congruent therefore 䉭BCD
*remains
) isosceles
3. AD is the perpendicular bisector of BC
Investigate
4. The area of 䉭ABC is twice the area of ADFE.
5. the height is half that of 䉭ABC.
6. DE = 1 BC.
2
7. Each of the smaller triangles, 䉭ADE and 䉭DEF, have
base and height that are half those of 䉭ABC.
Therefore, the areas of each smaller triangle is
1 ? 1 5 1 䉭ABC. Together, the areas of 䉭ADE and
2 2
4
䉭DEF must be half the area of 䉭ABC.
8. They have the same area.
9. The slopes of 䉭DEF are equal to the corresponding
slopes of 䉭ABC. The measures of the sides of 䉭DEF
are one-half the measure of the corresponding sides
of 䉭ABC. This is a demonstration of the midsegment
theorem.
10. yes
Orthocenters
Activity 43
Investigate
1. a. when the triangle is acute;
b. when the triangle is right;
c. when the triangle is obtuse
2. the orthocenter of 䉭ABD is C; the orthocenter of
䉭BCD is A; the orthocenter of 䉭ACD is B
Investigate
3. The circumferences and areas of all four circles are
equal.
Quadrilaterals
Activity 44
Investigate
Using Different Menus to
Construct Special
Quadrilaterals
Activity 45
Investigate
1–2. kite; two pairs of congruent, adjacent sides
Investigate Further
3.
4.
5.
6.
rectangle
Check students’ work.
rhombus
rhombus
Extend
rectangle; congruent diagonals bisect each other
Parallelograms and
Triangles
Activity 46
Investigate
1. Check students’ work.
2. ⬔BAD is twice ⬔AFD; ⬔ABC is twice ⬔BEC
3. ⬔BAD = ⬔AFD + ⬔ADF by the Exterior Angle
Theorem; ⬔AFD ⬵ ⬔ADF by the 3.
Isosceles Triangle Theorem; therefore, by substitution,
⬔BAD = 2⬔AFD or 1 ⬔BAD = ⬔AFD; same rea2
soning used to show 1 ⬔ABC = ⬔BEC
2
4. ⬔BEC + ⬔AFD = 90, therefore ⬔CHD = 90
5. ⬔HCD ⬵ ⬔BEC and ⬔CDH ⬵ ⬔AFD
6. Corresponding Angles Postulate
Regular Polygons
Activity 47
Investigate
1. parallelogram
2. They are equal; They are parallel to BD.
3. One pair of opposite sides of EHGF are congruent
and parallel.
116
4. parallelogram
5. rectangle
6. The diagonals of a rhombus are perpendicular,
therefore the sides of EHGF are perpendicular.
7. rhombus
8. square
9. rhombus
10. rectangle
11. If a quadrilateral has congruent diagonals, then the
figure formed by joining consecutive midpoints is a
rhombus.
12. If a quadrilateral has perpendicular diagonals, then the
figure formed by joining consecutive midpoints is a
rectangle.
13. Opposite sides of a kite are not parallel. Both pairs of
opposite sides of an isosceles trapezoid are not parallel.
Answers
1.
2.
3.
4.
rhombus
108
isosceles
36
Technology Activities
© Pearson Education, Inc. All rights reserved.
Answers
Answers
(continued)
5. m⬔GDE = m⬔GFE = 72; m⬔E = 108
6. 108, since the measures of a quadrilateral have a sum
of 360
7. DE ⬵ EF
8. Since opposite angles are congruent, DGFE is a parallelogram. Because a pair of adjacent sides is congruent,
DGFE is a rhombus.
(n 2 1 1)
(n2 2 1)
n
2
2
3
= (A2 ? A2 – 1)/2 = A2 = (A2 ? A2 + 1)/2
= A2 + 2 = (A3 ? A3 – 1)/2 = A3 = (A3 ? A3 + 1)/2
= A3 + 2 = (A4 ? A4 – 1)/2 = A4 = (A4 ? A4 + 1)/2
n
which produces the following triples
Extend
9.
10.
11.
12.
2. Pythagoras’ method uses the formulas
rhombus
Check students’ work.
Check students’ work.
Draw the first diagonal such that it is parallel to one of
the sides of the regular polygon. The second diagonal
will join vertices that are adjacent to the first diagonal
such that the diagonals intersect.
3
5
7
Investigate
area of AECF =
=
areas
are equal
Check students’ work.
E and G are midpoints of BC and CD, respectively;
F coincides with C
Using x to represent the length of one side of ABCD,
2
each triangle has measure 1
2 ? x ? 2x = x . The areas
are equal.
The sides to which the segments are drawn are divided
into segments in a ratio of 2:2:1.
The sides to which the segments are drawn are divided
into segments in a ratio of 1:1:1.
The sides to which the segments are drawn are divided
into equal segments.
The sides to which the segments are drawn are divided
into segments in a ratio of 2:2: . . . :2:1.
9x2
3.
4.
5.
© Pearson Education, Inc. All rights reserved.
6.
7.
8.
9.
3x2
3x2
Pythagorean Triples
Investigate
1. Plato’s method uses the formulas
n2 - 1
4
4
= A2 ? A2/4 – 1
= A2 + 2 = A3 ? A3/4 – 1
= A3 + 2 = A4 ? A4/4 – 1
n
n
= A2
= A3
= A4
n2 + 1
4
= A2 ? A2/4 + 1
= A3 ? A3/4 + 1
= A4 ? A4/4 + 1
which produces the following triples
n
4
6
n2 - 1
4
3
8
Technology Activities
n
4
6
2n + 1
2n2 + 2n
2n2 + 2n + 1
produce the following triples
n
1
2
3
3x2; the
Activity 49
3
5
7
(n 2 1 1)
2
5
13
25
1
= 2?A2+1 = 2?A2?A2+2?A2 = 2?A2?A2+2?A2+1
= A2+1 = 2?A3+1 = 2?A3?A3+2?A3 = 2?A3?A3+2?A3+1
= A3+1 = 2?A4+1 = 2?A4?A4+2?A4 = 2?A4?A4+2?A4+1
Activity 48
1. both ratios are 2:1
2. the area of ABCD is 3x ? 3x = 9x2; the area of
2
䉭ABE = area 䉭ADF = 1
2 ? 2x ? 3x = 3x ; the
n
3. The formulas
n
Area
(n2 2 1)
2
4
12
24
n
2n + 1
3
5
7
2n2 + 2n
4
12
24
2n2 + 2n + 1
5
13
25
4. Plato’s method and Pythagoras’ method generate
triples from the same ‘family’ of integers, but not
necessarily the exact same numbers. For example,
the Pythagoras method produces the triple 5, 12, 13
whereas Plato’s method produces 10, 24, 26, which is
a multiple of 5, 12, 13. The unattributed method
produces the same triples as the Pythagoras method.
Similarity
Activity 50
Investigate
1. parallelogram
2. The sides of PQRS are midsegments, and therefore
parallel, to the sides of ABCD.
3. corresponding angles are congruent; corresponding
sides are in proportion
4. 2:1; 4:1
5. yes
Extend
6. corresponding sides are in a ratio of 2:1; corresponding
angles are congruent
7. the ratio of the areas is 4:1
8. Check student’s work.
Investigate Further
n2 + 1
4
5
10
9. ILMJ ⬃ KOPN; KIJN ⬃ OLMP
10. Answers may vary. Sample: IL : KO (for
ILMJ ⬃ KOPN); KI : OL (for KIJN ⬃ OLMP)
11. Check students’ work.
Answers
117
Answers
(continued)
Volume
Activity 51
Investigate
1. 12 - 4x
2. V(x) = x ? x ? (12 - 4x) = 12x2 - 4x3
Rotations
Investigate
1.
2.
3.
4.
3. 2
4. 16
5. S(x) = 2 ? x ? x + 4 ? x ? (12 - 4x) =
-14x2 + 48x
6. 1.714
7. no
8. (2.606, 30); maximum volume 10.695
each angle = 120; hexagon
each angle 90; square
100 is not a factor of 360
Any number that is a factor of 360 can used with a
whole number amount of rotations to produce a
regular polygon.
Extend
5. It is supplementary.
6. Yes
7. The angles that have a vertex at A are equal to the size
of rotation. Any other pair of angles formed will be
supplementary to the angle of rotation.
8. A square can be used to form a tessellation.
9. Their angles must be factors of 360.
10. The sum of the angles around any point equals 360.
11. Regular pentagons and rhombuses can be used to form
a semi-pure tessellation.
Extend
9. V(x) = px2(12 - 2px) where x is the radius of the
container
10. r = 1.273 and h = 4; 20.372
11. S(x) = 2px2 + 2px(12 - 2px)
12. r = 1.136 and h = 4.864
13. The shape of a package, in addition to its weight, is an
important consideration—Can the package be stacked
on other packages? Can it fit into certain spaces on a
truck or plane? Is it difficult to carry? etc. . . .
Web–Based Activities
Activity 52
Algebra 1
Using the Web to Study Mathematics
Check students’ work.
Mining Data Found on the Internet
year
Time
52
45.9
56
44.5
60
44.5
64
43.6
68
42.8
72
42.81
76
42.55
80
41.6
84
41.65
88
41.98
92
42.11
Step 2 and 3: There is a moderate linear relationship.
Line of best fit: y = -0.0950681818x + 230.5653636
Step 4: Actual 1996 Winning time: 41.95; Line of best fit prediction = 40.80927273; The data appears to be leveling off
from 1980–1992. This would seem reasonable since winning times cannot continue to decrease at the same rate.
118
Answers
Technology Activities
© Pearson Education, Inc. All rights reserved.
Check students’ work.
Step 1:
Winning Times for the Olympics Women’s 4 3 100–meter Relay
Answers
(continued)
Generating Quizzes on the Web
Check students’ work.
Graphing the Solar System
Investigate
1. 9.538
2. 0.5341
2
x2 1 y
51
2
9.538
9.5232
(x 1 0.5341) 2
y2
4.
1
51
2
9.538
9.5232
5.
3. b = 9.523;
6. An eccentricity close to zero, such as 0.056, gives the
appearance of a circle. Notice that the major axis,
2 ? 9.538 = 19.076, is slightly larger than the minor
axis, 2 ? 9.523 = 19.046.
© Pearson Education, Inc. All rights reserved.
Exercises
1. The average distances to the Sun for the terrestrial
planets are much closer to the Sun than the Jovian
planets. A scale drawing of all nine planets would
make it difficult to see the terrestrial planets.
2. Answers may vary: Pluto’s average distance to the Sun
is about 100 times greater than Mercury’s average
distance to the Sun. A sketch would need to be about
25 feet wide in order to reasonably see the orbits of all
nine planets.
y2
x2
3.
1
51
2
39.785
38.5422
4. Pluto’s orbit is much more elliptical (e = 0.248) than
Neptune’s (which is nearly circular with e = 0.009).
Because of this, there are times when Pluto’s orbit
brings the planet closer to the sun than Neptune’s.
5. The earth is tilted more towards the Sun during the
summer months, thereby creating a warmer climate.
6. There would be extreme temperature changes if the
Earth’s orbit was more elliptical.
Building a Web Page About Buildings
Check students’ work.
Creating Dynamic Images for the Web
Check students’ work.
Technology Activities
Answers
119
