Download Discovering Geometry An Investigative Approach

Answers to Exercises
CHAPTER 0 • CHAPTER
0
CHAPTER 0 • CHAPTER
LESSON 0.1
1. possible answers: snowflakes and crystals;
flowers and starfish
2. Answers might include shapes and patterns,
perspective, proportions, or optical illusions.
3. Answers will vary. Bilateral symmetry.
4. A, B, C, and F
5. A, B, D, and E
6. 4 of diamonds; none
7. The line of reflection is a line along the
surface of the lake.
8. One line of symmetry is a vertical line through
the middle of the Taj Mahal, and the other is a
horizontal line between the Taj Mahal and the
reflecting pool. The pool reflects the building,
giving the scene more reflectional symmetry than
the building itself has.
9. Designs will vary, but all should have 2-fold
rotational symmetry.
2 lines of reflectional
symmetry
It is impossible to have
2 lines of reflectional
symmetry without
rotational symmetry.
10. Answers will vary.
11. Answers will vary.
Answers to Exercises
ANSWERS TO EXERCISES
1
LESSON 0.2
1. compass and straightedge
2. Answer will be the Astrid or 8-Pointed Star,
possibly with variations.
3. Answers will be a design from among the three
choices.
4. The first design has 4-fold symmetry and four
lines of reflectional symmetry. The triangle has
3-fold rotational symmetry and three lines of
reflectional symmetry. The third design has 6-fold
rotational symmetry if you ignore color and 2-fold
rotational symmetry if you don’t.
5. possible answer:
Answers to Exercises
2 lines of
reflectional
symmetry
2
ANSWERS TO EXERCISES
6. possible answer:
7.
3 lines of reflectional
symmetry
3-fold rotational
symmetry: rotated
120⬚, 240⬚, 360⬚
LESSON 0.3
4. possible answer:
1. Design with reflectional symmetry is drawn on
this background.
2. Design should have rotational symmetry.
Possible answer:
5. Drawing should resemble the hexagons in the
given figure, except that each radius, and the side
of each hexagon, should measure 1 inch.
Answers to Exercises
3. possible answer:
ANSWERS TO EXERCISES
3
LESSON 0.4
Answers to Exercises
1. Possible answer: The designs appear to go in
and out of the page (they appear 3-D). The squares
appear to spiral (although there are no curves in
the drawing). The spiral appears to go down as if
into a hole. Lines where curves meet look wavy.
Bigger squares appear to bulge out.
2. Possible answer: When zebras group together,
their stripes make it hard for predators to see
individual zebras.
4
ANSWERS TO EXERCISES
3. Designs will vary.
4. Designs will vary.
5. Possible answers: 2-D: rectangle, triangle,
trapezoid; 3-D: cylinder, cone, prism. The palace
facade has one line of reflectional symmetry
(bilateral symmetry).
LESSON 0.5
5.
1. possible answers: Scotland, Nigeria
2. possible answer:
3. possible answer:
6. Answers will be a lusona from among the three
choices.
7. It means to solve a problem boldly and
decisively or in a creative way not considered by
others.
8. The square knot has reflectional symmetry
across a horizontal line. The less secure granny
knot has 2-fold rotational symmetry.
9. The result is a regular pentagon.
Answers to Exercises
4. possible answer:
Cut the middle ring.
ANSWERS TO EXERCISES
5
LESSON 0.6
Answers to Exercises
1. possible answers: Morocco, Iran, Spain,
Malaysia
2. Alhambra
3. 2.1 cm
4. in terms of the original square tile: 4-fold in the
center (center of orange) or corner (center of
white); 2-fold in the midpoint of the edge of the
tile (where two orange shapes meet)
6
ANSWERS TO EXERCISES
6. Designs will vary.
7. Designs will vary, but should contain a regular
hexagon.
8. Tessellations will vary.
CHAPTER 0 REVIEW
1. possible answer: Islamic, Hindu, Celtic
2. Sometimes the large hexagons appear as blocks
with a corner removed, and sometimes they appear
as corners with small cubes nestled in them.
3. Compass: A geometry tool used to construct
circles.
Straightedge: A geometry tool used to construct
straight lines.
4. possible answer:
8. Wheel A has four lines of reflectional symmetry;
Wheel C has five lines of reflectional symmetry.
Wheels B and D do not have reflectional symmetry.
9. Wheels B and D have only rotational symmetry.
Wheels A and B have 4-fold, Wheel C has 5-fold,
and Wheel D has 3-fold rotational symmetry.
10, 11. Drawing should contain concentric circles
and symmetry in some of the rings.
12. The mandala should contain all the required
elements.
13a. The flag of Puerto Rico is not symmetric
because of the star and the colors.
13b. The flag of Kenya does not have rotational
symmetry because of the spearheads.
13c. possible answers:
Japan
5.
Answers to Exercises
6. possible answers: hexagon: honeycomb,
snowflake; pentagon: starfish, flower
7. Answers will vary. Should be some form of an
interweaving design.
Nigeria
ANSWERS TO EXERCISES
7
Answers to Exercises
1
CHAPTER 1 • CHAPTER
CHAPTER 1 • CHAPTER
LESSON 1.1
1. sample answers: point: balls, where lines cross;
segment: lines of the parachute, court markings;
collinear: points along a line of the court structure;
coplanar: each boy’s hand, navel, and kneecap,
points along two strings
, TP
2. PT
, RA
, AT
, TA
, RT
, TR
3. any two of the following: AR
4. any two of the following: MA,MS,AS,AM,SA,SM
5.
6.
A
B
7.
K
, PN
22. PM
24. A
25.
26.
Y
M
N
27.
C
D
B
A
28. 10
29–31.
y
5
3
L
M
3
C
x
or CA
or QP
8. AC
9. PQ
or RT
, RI
or IR
, and TI
or IT
10. TR
y
11. A
12.
B
S
11
O
–5
–3 E
A (4, 0)
x
D
–5
B
32. Yes, P(6, 2), Q(5, 2), R(4, 6); the slope
between any two of the points is 14.
y
R
Q
R
P´
x
x
Q´
14.3 cm 14. mCD
6.7 cm
13. mAB
15–17. Check each length to see if it is correct.
Refer to text for measurements.
. X is the midpoint of
18. R is the midpoint of PQ
. No midpoints are
WY . Y is the midpoint of XZ
shown in ABC.
19. possible answers:
or
A
B
C
D
P
R´
33. possible answer:
y
Q
N
x
P
E
CE
AC
BC CD
20.
34.
P
R
A
T
G
Y
35. Answers will vary. Possible answers:
S
M
T
8 cm
8 cm
11 cm
11 cm
Q
8
B
X
y
D
–3
Answers to Exercises
, AC
21. AB
, XZ
23. XY
ANSWERS TO EXERCISES
USING YOUR ALGEBRA SKILLS 1
1. (3, 4)
2. (9, 1.5)
3. (5.5, 5.5)
4. (6, 44)
5. Yes. The coordinates of the midpoint of a
segment with endpoints (a, b) and (c, d ) are found
ac
by taking the average of the x-coordinates, 2 ,
bd
and the average of the y-coordinates, 2. Thus the
ac bd
midpoint is 2 , 2 .
6. (3, 2) and (6, 4). To get the first point of
trisection, sum the coordinates of points A and B
to get (9, 6), then multiply those coordinates by 31 to
get (3, 2). To get the second point of trisection, sum
the coordinates of points A and B to get (9, 6), then
multiply those coordinates by 32 to get (6, 4). This
works because the coordinates of the first point
are (0, 0).
7. Find the midpoint, then find the midpoint of
each half.
8a. Midpoints for Figure 1 are (5.5, 6.5); for
Figure 2, (16, 6.75); and for Figure 3, (29.75, 5.5).
8b. For these figures the midpoints of the two
diagonals are the same point.
Answers to Exercises
ANSWERS TO EXERCISES
9
LESSON 1.2
1. TEN, NET, E; FOU, UOF, 1;
ROU, UOR, 2
N
2.
A
24.
135
D
67.5
22
22
A
67.5
D
26. no
B
G
E
25.
T
3.
C
C
Y
D
I
4.
N
S
A
M
27.
L
Answers to Exercises
5. S, P, R, Q; none in the second figure
6. possible answer:
B
28.
C
A
D
7. 90°
8. 120°
9. 45°
10. 135°
11. 45°
12. 135°
13. 30°
14. 90°
15. Yes; mXQA mXQY 45° 90° 135°,
which equals mAQY.
16. 69°
17. 110°
18. 40°
19. 125°
20. 55°
21. SML has the greater measure because
mSML 30° and mBIG 20°.
22.
A
30. One possibility is 4:00.
H
31.
90
8
6
O
T
32.
T
H
R
33.
S
A
T
G
44
A
23.
I
90
10
29.
B
ANSWERS TO EXERCISES
N
34.
W
O
B
T
45. 180 km. The towns can be represented by
three collinear points, P, S, and G. Because S is
between P and G, PS SG PG.
12 cm
46.
6 cm
47.
I
35.
37.
39.
41.
43.
MY; CK; mI
x 54°
z 32°
242°
no
36.
38.
40.
42.
SEU; EUO; MO
y 102°
288°
They add to 360°.
6 cm
4.36 cm
2.18 cm
2.18 cm
48. MS DG means that the distance between
M and S equals the distance between D and G.
MS DG means that segment MS is congruent
to segment DG. The first statement equates two
numbers. The second statement concerns the
congruence between two geometric figures.
However, they convey the same information and
are marked the same way on a diagram.
49. Answers will vary. Possible answer.
C
70
4 cm
Answers to Exercises
40
A
B
F
44. You will miss the target because the incoming angle is too big.
40
D
70
6 cm
E
Target
Mirror
Laser light
source
ANSWERS TO EXERCISES
11
LESSON 1.3
1–3. possible answers:
D
1.
45
O
2.
G
T
E
R
3.
11. The measures of complementary angles sum
to 90°, whereas the measures of supplementary
angles sum to 180°.
12. No, supplementary angles can be unconnected, while a linear pair must share a vertex and
a common side.
13a. an angle; measures less than 90°
13b. angles; have measures that add to 90°
13c. a point; divides a segment into two congruent
segments
13d. a geometry tool; is used to measure the sizes
of angles in degrees
14.
E
B
P
C
B
I
G
4.
Answers to Exercises
A
G
and CD
intersect at point P so that P is
If AB
between A and B, and P is between C and D, then
APC and BPD are a pair of vertical angles.
15. true
D
S
M
5.
A
16. true
P
6.
E
R
D
A
17. true
B
E
C
7.
40
A
18. false
50
B
8.
19. false
140
40
C D
9. B is a Zoid. A Zoid is a creature that has in its
interior a small triangle with a large black dot
taking up most of its center.
10. A good definition places an object in a class
and also differentiates it from other objects in that
class. A good definition has no counterexamples.
12
D
ANSWERS TO EXERCISES
20. false
28. One possible answer is A(8, 8), B(4.5, 6.5),
C(11, 1).
50
y
A
B
50
80
6
S
C
21. true
–12
x
–6
T
22. false; Though the converse is true, a
counterexample is
C
T
R
29. 12 cm
30. 36°
31. possible answer:
A
23. false
Answers to Exercises
32.
140
0 pt.
C
1 pt.
D
40
24. false
2 pt.
A
3 pt.
33.
C
1 pt.
2 pt.
T
25. possible answers: (3, 0), (0, 2), or (6, 6)
26. possible answers: (2, 3) or (5, 1)
27. The reflected ray and the ray that passes
through (called the refracted ray) are mirror
images of each other. Or they form congruent
angles with the mirror.
A
Reflected
segment
B
Mirror
4 pt.
3 pt.
5 pt.
6 pt.
Infinitely
many points
120° 1 2
, 34a. 360° 3 3 left
60°
1 1
, 34b. 360° 6 6 missing
360°
34c. 9 40°
ANSWERS TO EXERCISES
13
LESSON 1.4
1.
FI
and IVE ANC
18. PA
19. possible answer:
130
2. possible answers:
130
130
5 cm
20. possible answer:
Answers to Exercises
3.
4. octagon
5. hexagon
6. heptagon
7. pentagon
8–10. One possible answer for each is shown.
8. pentagon FIVER
9. quadrilateral FOUR
10. equilateral quadrilateral BLOC
11a. a polygon; has eight sides
11b. a polygon; has at least one diagonal outside
of the polygon
11c. a polygon; has 20 sides
11d. a polygon; has all sides of equal length
12. One possibility is C and Y are consecutive
and YN
are consecutive sides.
angles; CY
13. 9; possible answer:
, AD
, BD
, BE
, CE
14. AC
15. TIN
16. WEN
17a. a 44, b 58, c 34
17b. mT 87° and mI 165°
14
ANSWERS TO EXERCISES
21. 84 in.
22. 5.25 cm
23. AB 14 m, CD 25 m
24. complementary angles: AOS and SOC;
vertical angles: OCT and ECR or TCE and
RCO
25.
E
A
B
C
F
D
26. possible answer:
5
20
27. All are possible except two points.
0 pts.
1 pt.
3 pts.
4 pts.
5 pts.
6 pts.
16. possible answers:
LESSON 1.5
1. D
3. C
5.
2. A
4. B
6. C
C
C
14 cm
10 cm
40
A
7.
T
S
M
L
A
B
A
F
10 cm
R
40
C
A
8. possible answer:
2a
2a
b – 2a
b
6 cm
B
40
6 cm
A
B
40
y
C
C
(–3, 5)
(2, 4)
10. possible answer:
C
(–4, 1)
(1, 1)
x
4 cm
22. (4,1) → (3,2) (1,1) → (2, 2) (2, 4) →
(3, 1) (3, 5) → (2, 2) Yes, the quadrilaterals are
congruent.
23. Find the midpoint of each rod. All the
midpoints lie on the same line; place the edge
of a ruler under this line.
24–26. Sample answers for 24–26
P
P
24.
25.
80
A
4 cm
B
11. possible answer:
A
4 cm
120
P
4 cm
Z
12. possible answers: (1, 1), (1, 0), or (4, 3)
13. possible answers: (3, 3), (3, 4), (11, 4),
(11, 3), (0.5, 0.5), or (7.5, 0.5)
14. possible answers: (3, 1), (1, 9), (2, 2),
(4, 8), (0, 1), or (1, 6)
15. possible answers:
C
E
A
N
26.
E
T
A
N
Q
U
D
A
T
C
7 cm
12 cm
45
A
9 cm
B
45
A
9 cm
B
ANSWERS TO EXERCISES
15
Answers to Exercises
9. possible answer:
A
E
17. true
18. true
19. False, a diagonal connects nonconsecutive
vertices.
20. False, an angle bisector divides an angle into
two congruent angles.
21. true
3a
3a
10 cm
LESSON 1.6
1. first figure: quadrilateral ABCD with two
congruent sides and one right angle; second
figure kite EFGH; third figure: trapezoid IJKL
with two right angles; fourth figure parallelogram
MNPQ
2. B
3. D
4. F
5. C
6. A, D, F
I
7. D
Z
8.
19b. Possible answer: Rotate one triangle so that
a congruent pair of sides forms a diagonal of a
parallelogram.
20. There are three possibilities: a rhombus, a
concave kite, or a parallelogram.
O
N
E
B
F
Answers to Exercises
9.
L
A
E
21a. right triangles
21b. isosceles right triangles
22.
U
Q
10.
T
H
23.
I
R
5 cm
5 cm
G
11.
24.
12. square
13. possible answers: (2, 6), (2, 4); (6, 2),
(2, 4); (3, 1), (1, 3)
14. 90 cm
15. (5, 6)
16. S(9, 0), I(4, 2)
17. S(3, 0), I(1, 5)
18. A(7, 6), N(5, 9) or A(5, 2), N(7, 1)
19a. Possible answer: Flip one triangle and align it
so that a congruent pair of sides forms a diagonal
of a kite.
16
ANSWERS TO EXERCISES
72 72
72 72
72
25. no
y
C (0, 5)
B (4, 4)
x
A (5, 0)
LESSON 1.7
1. Answers will vary. Sample answers:The green
area in the irrigation photo is a circle,the water is a
radius,and a path on the far side of the circle appears
to be tangent to the circle.The wood bridge is an arc of
a circle,and the railings are arcs of concentric circles.
The horizontal support beam under the bridge is
a chord.
, BD
, EC
, EF
2. three of the following: AB
3. EC
, EP
, FP
, BP
, CP
4. AP
, AE
, AB
, BC
, CD
, DF
,
5. five of the following: EF
EB , ED , FC , AC , DB , AF , AD , BF
or EFC
, EBC
or EAC
6. EDC
, FEC
, DEC
, . . .
7. two of the following: ECD , EDF
B
P
Q
A
17. equilateral; 3 to 1
s
18.
19. yes; yes
Answers to Exercises
, HB
8. FG
9. either F or B
10. possible answers: cars, trains, motorcycles;
washing machines, dishwashers, vacuum cleaners;
compact disc players, record players, car racing,
Ferris wheel
110°; mPRQ
250°
11. mPQ
12.
16. Equilateral quadrilateral. (The figure is
actually a rhombus, but students have not yet
learned the properties needed to conclude that the
sides are parallel.)
y
5
5
–5
x
65
–5
215
20. yes; no
13. possible answers: concentric rings on cross
sections of trees (annual rings),bull’s-eye or target,
ripples from a rock falling into a pond.
14.
y
5
–5
5
x
–5
21. no; no
15.
y
8
P
Q
8
16
x
ANSWERS TO EXERCISES
17
22. 80°. The slices of pizza do not overlap, so
60° x 140°, where x is the measure in
degrees of the angle of the second slice.
23.
24. not possible
25.
C
60
60
32. not possible
33. R
G
T
34.
A
B
27. about 0.986°
28. 15°
29.
4a + 2b
35.
T
100
50
36.
Answers to Exercises
120
B
C
70
I
P
E
A
37.
50 70
2p
2p
F
30.
120
60
120
A
26.
A
31.
U
2p
55
2p
Q
E
120
T
R
K
T
I
18
ANSWERS TO EXERCISES
120
60
22. False. The two lines are not necessarily in the
same plane, so they might be skew.
LESSON 1.8
1.
2.
3.
4.
23. true
5.
6.
7.
8.
24. true
25. true
2
9. 60 boxes
Answers to Exercises
3
4
10.
26. False. They divide space into seven or eight
parts.
5
3
4
11.
12.
27. true
13. B, D
15. C
17. A
18.
19.
20. true
21. true
28. (3, 1)
14. B
16. D
29. perimeter 20.5 cm; m(largest angle) 100°
30.
8 cm
120
13 cm
ANSWERS TO EXERCISES
19
18. pyramid with hexagonal base
LESSON 1.9
1. Sample answer: Furniture movers might
visualize how to rotate a couch to get it up a
narrow staircase.
2. yes
3. W
O M E N ; Nadine is ahead.
4. 28 posts
5. 28 days
6. 0 ft (The poles must be touching!)
B
7.
20. pyramid with square base
A
B
A
21. x 15, y 27
22. x 12, y 4
23.
8.
Answers to Exercises
19. prism with hexagonal base
A
9.
A
3 cm
B
3 cm
24.
3 cm
A
B
10.
Polygons
Quadrilaterals
Trapezoids
11.
12.
13.
14.
15.
16.
Triangles
Obtuse
Isosceles
no
C(2, 3), A(0, 0), B(0, 5), D(2, 1)
Y(4, 1), C(3, 1), N(0, 3)
(x, y) → (x 3, y 2)
CP
, EF
GH
; i k, j k
AB
perimeter 34 cm
25.
26.
28.
30.
32.
34.
point, line, plane
AB
vertex
CD
AB
ABC
congruent to
20
ANSWERS TO EXERCISES
AB
AB
protractor
CD
AB
35. The distance is two times the radius.
r
r
P
Q
PQ = 2r
36. They bisect each other and are perpendicular.
A
17. Left photo: Three points determine a plane.
Middle photo: Two intersecting lines determine a
plane. Right photo: A line and a point not on the
line determine a plane.
27.
29.
31.
33.
P
Q
B
CHAPTER 1 REVIEW
1.
2.
3.
4.
5.
6.
7.
8.
true
.
False; it is written as QP
true
False; the vertex is point D.
true
true
False; its measure is less than 90°.
false; two possible counterexamples:
T
Y
E
A
P
30.
31.
A
D
A
P
R
C
D
A
29.
P
P
B
C
T
APD and APC
are a linear pair.
APD and APC
are the same angle.
E
Y
A
C
B
Answers to Exercises
9. true
10. true
11. False; they are supplementary.
12. true
13. true
14. true
15. False; it has five diagonals.
16. true
17. F
18. G
19. L
20. J
21. C
22. I
23. no match
24. A
25. no match
T
26.
32.
D
33.
34.
2 in.
5 in.
3 in.
35.
125
36.
K
40
27.
S
5
3
T
7
28.
O
P
37.
A
N
R
I
E
G
ANSWERS TO EXERCISES
21
38.
39.
40.
41.
42.
43.
44.
45.
46.
47.
114°
x 2, y 1
x 12, y 4
x 4, y 2.5
x 10, y 8
AB 16 cm
96°
105°
30°
51. He will get home at 5:46 (assuming he goes
inside before he gets blown back again).
52. (2, 3)
53.
Quadrilateral
54.
Rectangle Square Rhombus
55.
Answers to Exercises
Trapezoid
48. 66 inches
49. 3 feet
50. The path taken by the midpoint of the ladder is
an arc of a circle or a quarter-circle if the ladder
slides all the way from the vertical to the horizontal.
57.
Shed
Path of flashlight
22
56.
ANSWERS TO EXERCISES
Answers to Exercises
CHAPTER 2 • CHAPTER
2
CHAPTER 2 • CHAPTER
LESSON 2.1
13.
14.
28.
29. possible answers:
30. sample answer:
31.
33.
35.
37.
39.
41.
collinear
dodecagon
protractor
diagonal
90°
sample answer:
32.
34.
36.
38.
40.
42.
Answers to Exercises
1. “All rocks sink.” Stony needs to find one rock
that will not sink.
2. If two angles are formed by drawing a ray from a
line, then their measures add up to 180°.
3. 10,000, 100,000, . . . . Each term is 10 times the
previous term.
4. 65, 1, . . . . (Written with the common
denominator 6, the pattern becomes
1 2 3 4 5 6
, , , , , , . . . .)
6 6 6 6 6 6
5. 17, 21
6. 28, 36
7. 21, 34
8. 49, 64
9. 10, 24
10. 64, 128
11.
12.
27.
isosceles
parallel
radius
regular
perpendicular
sample answer:
I
15.
16.
A
17. 1, 4, 7, 10, 13
18. 15, 21, 28, 36, 45
19. Answers will vary.
20. sample answers: 3, 6, 12, 24, 48, . . . and 4, 8, 12,
16, 20, . . .
21. Answers will vary.
22. 7th term: 56; 10th term: 110; 25th term: 650
23. Conjecture is false; 142 196 but 412 1681.
24. 11,111 11,111 123,454,321. For the sixth
line, if 111,111 is multiplied by itself, then the
product will be 12,345,654,321. But 1,111,111,111 1,111,111,111 1,234,567,900,987,654,321.
25.
26.
G
N
T
43. sample answer:
44. possible answer:
A
C
Clearly AC
does not
bisect A
or C.
45. Methods will vary. It is not possible to draw a
second triangle with the same angle measures and
side length that is not congruent to the first.
C
40
A
60
9 cm
B
ANSWERS TO EXERCISES
23
4.
5.
6.
7.
LESSON 2.2
1. See table below.
2. See table below.
3. See table below.
See table below.
See table below.
See table below.
See table below.
1. (Lesson 2.2)
n
1
2
3
4
5
6
…
n
…
20
f (n)
3
9
15
21
27
33
… 6n 3 …
117
…
20
…
56
…
20
…
148
Answers to Exercises
2. (Lesson 2.2)
n
1
2
3
4
5
6
…
f (n)
1
2
5
8
11
14
…
n
3n 4
3. (Lesson 2.2)
n
f (n)
1
2
3
4
5
6
…
4
4
12
20
28
36
…
n
8n 12
4. (Lesson 2.2)
Number of sides
3
4
5
6
…
n
…
35
Number of triangles formed
1
2
3
4
…
n2
…
33
5. (Lesson 2.2)
Figure number
1
2
3
4
5
6
…
n
…
200
Number of tiles
8
16
24
32
40
48
…
8n
…
1600
Figure number
1
2
3
4
5
6
…
n
…
200
Number of tiles
1
5
9
13
17
21
… 4n 3 …
797
6. (Lesson 2.2)
7. (Lesson 2.2)
24
Figure number
1
2
3
4
5
6
…
Number of matchsticks
5
9
13
17
21
25
…
Number of matchsticks
in perimeter of figure
5
8
11
14
17
20
ANSWERS TO EXERCISES
n
…
200
4n 1 …
801
… 3n 2 …
602
8. 8n is steeper; the coefficient of n.
13.
R
E
f (n)
T
8n
C
50
40
30
4n 1
4n 3
20
3n 2
14.
2
1
2
1
3
3
10
1
2
4
6
1
n
8
H
H
H
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
Octane (C8H18)
3
10. y 2x 3
11.
1
2
1
2
9. CnH2n2
H
1
1
L
H
15. Márisol should respond by noticing that all
the triangles José drew were isosceles, but it is
possible to draw a triangle with no two sides
congruent. In other words, she should show him a
counterexample.
16. Methods will vary. It is not possible to draw
another triangle with the given side lengths and
angle measure that is not congruent to the first.
C
Answers to Exercises
9 cm
T
45
A
E
12.
Q
O
M
L
8 cm
B
17. She could try eating one food at a time;
inductive.
18. 2600
Y
ANSWERS TO EXERCISES
25
n(n 1)
8. Use 2 to get 780 direct lines. Use a central
hub with a line to each house to get 40 lines. The
art shows the direct-line solution and the practical
solution for six houses.
LESSON 2.3
1.
See table below.
2. n 1, 36
9. Use points for the 10 teams and segments
connecting them to represent four games played
n(n 1)
between them. So, use 4 2 to get 180 games
played.
n(n 1)
10. If 2 66, then n(n 1) 132. What
two consecutive numbers multiply to equal 132?
12 times 11. Thus, there were 12 people at the
party.
11. true
12. true
13. False; an isosceles right triangle has two sides
congruent.
14. false
3. You can draw a diagonal from one vertex to all
the other vertices except three: the two adjacent
vertices and the vertex itself.
Answers to Exercises
n(n 3)
4. 2, 560
n(n 1)
5. 2, 595
E
B
C
A
AED and BED are
not a linear pair.
n(n 1)
6. 2, 595
15. False; they are parallel.
16. true
17. False; a rectangle is a parallelogram with all of
its angles congruent.
18. False; a diagonal is a segment in a polygon
connecting any two nonconsecutive vertices.
19. true
20. 5049
7. Answers will include relationships between
points in Exercises 5 and 6 and vertices of a polygon.
The total number of segments connecting n random
points is the number of diagonals of the n-sided
n(n 3)
polygon, 2, plus the number of sides, n.
Thus, the total number of segments connecting n
n(n 3)
2n
n(n 3) 2n
random points is 2 2 2 n2 3n 2n
n2 n
2
2
D
n(n 1)
2 .
1. (Lesson 2.3)
26
Lines
1
2
3
4
5
…
n
…
35
Regions
2
4
6
8
10
…
2n
…
70
ANSWERS TO EXERCISES
LESSON 2.4
1. inductive; deductive
2. mB 65°; deductive
3. inductive
25
18.
19.
36
4. DG 258 cm; deductive
5. 180°; 180°; The sum of the five angles is 180°;
inductive.
20.
20
32
64
27
37
I
E
L
33.
D
B
140
O
G
34.
C
T
O
D
35.
K
D
N
ANSWERS TO EXERCISES
27
Answers to Exercises
6. LNDA is a parallelogram; deductive.
7. Possible answer: AB CD, so AB BC CD BC. Because AB BC AC and CD BD
.
BC BD, AC BD, so AC
8. 3; 10; 7
9. Just over 45°; if m 90°, then 21m 45°.
10. 48°; 17°; 62°; mCPB
11. Possible answers: Because mAPC mBPD, add the same measure to both sides to get
mAPC mCPD mBPD mCPD.By
angle addition, mAPC mCPD mAPD
and mBPD mCPD mCPB. Therefore,
mAPD mCPB.
12. Answers will vary.
13. The pattern cannot be generalized because
once the river is straight, it cannot get any shorter.
14. 900, 1080
15. 75, 91
4
16. 5, 12
17.
21. Sample answer: There were three sunny days
in a row, so I assumed it would be sunny the fourth
day, but it rained.
22. L
23. M
24. A
25. B
26. E
27. C
28. G
29. D
30. H
31. I
W
32.
15.
Answers to Exercises
LESSON 2.5
1. a 60°, b 120°, c 120°
2. a 90°, b 90°, c 50°
3. a 77°, b 52°, c 77°, d 51°
4. a 60°, b c 120°, d f 115°, e 65°,
g i 125°, h 55°
5. a 90°, b 163°, c 17°, d 110°, e 70°
6. The measures of the linear pair of angles add up
to 170°, not 180°.
7. The angles at which he should cut measure 45°.
8. Greatest: 120°. Smallest: 60°. One possible
explanation: The tree is perpendicular to the
horizontal. The angle of the hill measures 30°.
The smaller angle and the angle between the hill
and the horizontal form a pair of complementary
angles, so the smaller angle equals 90° 30° 60°.
The smaller angle and larger angle form a linear
pair, so the larger angle equals 180° 60° 120°.
9. The converse is not true. ; sample counterexample:
A
A
O
T
16.
17.
6
3
4
18. Possible answer: All the cards look exactly as
they did, so it must be the 4 of diamonds, because
it has rotational symmetry while the others do not.
19. 22.5°
20.
55
125
B
10. each must be a right angle
21. CnH2n
H
H
11. Let the measures of the congruent angles be x.
They are supplementary, so x x 180°, 2x 180°,
x 90°. Thus each angle is a right angle.
12. The ratio is 1. The ratio does not change as
long as the lines don’t coincide. Because the
demonstration does not explain why, it is not
a proof.
13. P
14.
5
H
H
H
H
H
C
C
C
C
C
C
C
C
H
H
H
H
H
H
H
H
22. See table below.
n(n 1) 80(79)
23. handshake problem: 2 ; 2 3160
pieces of string
24. 3160 intersections
n(n 2)
25. 2 yields 760 handshakes.
26. 21; 252
n(n 3)
27. 2 560; there are 35 vertices.
; deductive.
28. M is the midpoint of AY
3
A
H
T
22. (Lesson 2.5)
Rectangle
1
2
3
4
5
6
…
n
…
200
Perimeter of rectangle
10
14
18
22
26
30
… 4n 6 …
806
Number of squares
6
12
20
30
42
56
…
… 40,602
(n 1)(n 2)
28
ANSWERS TO EXERCISES
13. No, tomorrow could be a holiday. Converse:“If
tomorrow is a school day, then yesterday was part
of the weekend;” false.
14. 42°
15. 20°
16. No, the lines are not parallel.
17. isosceles triangle
18. a parallelogram that is not also a rectangle or a
rhombus
19. 18 cm
20. 39°
21. 3486
22. 30 squares (one 4-by-4, four 3-by-3, nine
2-by-2, and sixteen 1-by-1)
23. The triangle moved to the left 1 unit.Yes,
congruent to original.
LESSON 2.6
1. 63°
2. 90°
3. no
4. 57°
5. yes
6. 113°
7. a d 64°, b c 116°, e g i j k 108°, f h s 72°, m 105°, n 79°,
p 90°, q 116°, t 119°; Possible explanation:
Using the Vertical Angles Conjecture, n 79°.
Using the Linear Pair Conjecture, p 90°. Using the
Corresponding Angles Conjecture and b 116°,
q 116°.
8. Possible answer: In the
k
diagram, lines and m are parallel
1
and intersected by transversal k.
m
Using the Corresponding Angles
2
3
Conjecture, 1 2. Using the
Vertical Angles Conjecture,
2 3. Because 1 and 3 are both congruent
to 2, they must be congruent to each other. So
1 3. Therefore,if two parallel lines are cut by a
transversal, then alternate exterior angles are
congruent.
9. 56° 114° 170° 180°. Thus, the lines
marked as parallel cannot really be parallel.
10. Alternate interior angles measure 55°, but
55° 45° 180°.
11. The incoming and
45
90
outgoing angles measure
45
45°. Possible explanation:
Yes, the alternate interior
angles are congruent, and
thus, by the Converse of the
45
Parallel Lines Conjecture,
90
the mirrors are parallel.
45
24. The quadrilateral was reflected across both
axes or rotated 180° about the origin.Yes, congruent
to original.
y
4
x
4
25. The pentagon was reflected across the line
x y.Yes, congruent to original.
y
N
5
O'
L
M'
Q
P
x
5
O
N'
x
E
B
A
6
E'
M
L'
26. See table below.
26. (Lesson 2.6)
Figure number
1
2
3
4
5
6
…
n
…
35
Number of yellow squares
2
3
4
5
6
7
…
n1
…
36
Number of blue squares
3
5
7
9
11
13
… 2n 1 …
71
Total number of squares
5
8
11
14
17
20
… 3n 2 …
107
ANSWERS TO EXERCISES
29
Answers to Exercises
12. Explanations will vary.
Sample explanation:“I used the
protractor to make corresponding angles congruent when I
drew line PQ.”
y
5
USING YOUR ALGEBRA SKILLS 2
Answers to Exercises
1. 2
12
2. 1
3
97
3. 4
6 2.1
4. y 23
5. x 4
6. Any point of the form (3p, 4p).Possible answer:
(6,8),(9,12),and (12,16).To find another
point go right 3 and down 4.
30
ANSWERS TO EXERCISES
7. 66.7 mi/h
8. At 6 m/s, Skater 1 is 4 m/s faster than Skater 2 at
2 m/s.
9. A 100% grade has a slope of 1. It has an
inclination of 45°, so you probably could not
drive up it.You might be able to walk up it. Grades
higher than 100% are possible, but the angle of
inclination would be greater than 45°.
10. The slope of the adobe house flat roof is
approximately 0. The Connecticut roof is steeper,
with a slope of about 2, and will shed the snow.
19d. possible answers: AFG and FGD or
CFG and BGF
20. possible answers: GFC by the Vertical
Angles Conjecture, BGF by the Corresponding
Angles Conjecture, and DGH, using the
Alternate Exterior Angles Conjecture
21. True. Converse:“If two polygons have the
same number of sides, then the two polygons are
congruent”; false. Counterexamples may vary;
students might draw a concave quadrilateral and a
convex quadrilateral.
22.
CHAPTER 2 REVIEW
1. poor inductive reasoning, but Diana was
probably just being funny
2. Answers will vary.
3. Answers will vary.
4. 19, 30
5. S, 36
6. 2, 5, 10, 17, 26, 37
7. 1, 2, 4, 8, 16, 32
8.
9.
56
124
7 cm
56
4.5 cm
4.5 cm
56
124 56
7 cm
900
930
See table below.
See table below.
n2, 302 900
n(n 1) 100(101)
15. 2, 2 5050
n(n 1)
16. 2 741; therefore, n 39
n(n 1)
17. 2 2926; therefore, n 77
18. n 2, 54 2 52
19a. possible answers: EFC and AFG, AFE
and CFG, FGD and BGH, or BGF and
HGD
19b. possible answers: AFE and EFC, AFG
and CFG, BGF and FGD, or BGH and
HGD (there are four other pairs)
19c. possible answers: EFC and FGD, AFE
and BGF, CFG and DGH, or AFG and
BGH
RX
and SU
VX
; explanations will vary.
23. PV
24. The bisected angle measures 50° because of
AIA. So each half measures 25°. The bisector is a
transversal, so the measure of the other acute angle
in the triangle is also 25° by AIA. However, this
angle forms a linearpairwiththeanglemeasuring
165°,and 25° 165° 180°, which contradicts the
Linear Pair Conjecture.
25. a 38°, b 38°, c 142°, d 38°, e 50°,
f 65°, g 106°, h 74°; The angle with measure e
forms a linear pair with an angle with measure 130°
because of the CorrespondingAngles Conjecture.So
e measures 50° because of the Linear Pair Conjecture.
The angle with measure f is half of the angle with
measure 130°, so f 65°.The angle with measure g is
congruent to the angle with measure 106° by the
CorrespondingAngles Conjecture,so g 106°.
12. (Chapter 2 Review)
n
1
2
3
4
5
6
…
f(n)
2
1
4
7
10
13
…
n
…
20
…
55
…
20
…
210
3n 5
13. (Chapter 2 Review)
n
1
2
3
4
5
6
…
f(n)
1
3
6
10
15
21
…
n
n(n 1)
2
ANSWERS TO EXERCISES
31
Answers to Exercises
10.
11.
12.
13.
14.
Answers to Exercises
CHAPTER 3 • CHAPTER
3
CHAPTER 3 • CHAPTER
LESSON 3.1
1.
B
A
D
C
F
E
2.
AB
9. For Exercise 7, trace the triangle. For Exercise 8,
trace the segment onto three separate pieces of
patty paper. Lay them on top of each other, and
slide them around until the segments join at the
endpoints and form a triangle.
. Copy Q and
10. One method: Draw DU
construct COY QUD. Duplicate DUA
at point O. Construct OYP UDA.
CD
U
O
A
AB ⫹ CD
3.
AB
EF
EF
Q
AB ⫹ 2EF ⫺ CD
CD
Answers to Exercises
D
C
Y
11. One construction method is to create congruent
circles that pass through each other’s center. One
side of the triangle is between the centers of the
circles; the other sides meet where the circles
intersect.
12. a 50°, b 130°, c 50°, d 130°, e 50°,
f 50°, g 130°, h 130°, k 155°, l 90°,
m 115°, n 65°
13. west
14. An isosceles triangle is a triangle that has at
least one line of reflectional symmetry.Yes, all
equilateral triangles are isosceles.
15.
4.
5. possible answer:
L
G
P
E
copy
6. m3 m1 m2; possible answer:
⬔2
16. new coordinates: A(0, 0), Y(4, 0), D(0, 2)
y
1
3
2
⬔1
6
Y
7. possible answer:
D'
Y'
C
–6
AC
D A' A
6
x
BC
–6
A
B
AB
8.
17. Methods will vary. It isn’t possible to draw a
second triangle with the same side lengths that is
not congruent to the first.
11 cm
8 cm
10 cm
32
ANSWERS TO EXERCISES
LESSON 3.2
1.
A
B
2.
Q
3.
D
Edge of the paper
Original segment
4.
D
1 CD
2
1 CD
2
AB
AB
I
1 CD
2
2AB – 1 CD
2
5.
A
AB
M
L
CD
N
MN = 1 (AB + CD)
2
8. The medians all intersect in one point.
C
N
6. Exercises 1–5 with patty paper:
Exercise 1 This is the same as Investigation 1.
Exercise 2
.
Step 1 Draw a segment on patty paper. Label it QD
Step 2 Fold your patty paper so that endpoints Q
and D coincide. Crease along the fold.
Step 3 Unfold and draw a line in the crease.
Step 4 Label the point of intersection A.
Step 5 Fold your patty paper so that endpoints Q
and A coincide. Crease along the fold.
Step 6 Unfold and draw a line in the crease.
Step 7 Label the point of intersection B.
Step 8 Fold your patty paper so that endpoints A
and D coincide. Crease along the fold.
Step 9 Unfold and draw a line in the crease.
Step 10 Label the point of intersection C.
A
M
L
B
appears to be parallel to EF
, and its length
9. GH
is half the length of EF .
F
G
D
H
E
ANSWERS TO EXERCISES
33
Answers to Exercises
C
Exercise 3 This is the same as Investigation 1.
Exercise 4
Step 1 Do Investigation 1 to get 12CD.
Step 2 On a second piece of patty paper, trace AB
two times so that the two segments form a segment
of length 2AB.
Step 3 Lay the first piece of patty paper on top of
the second so that the endpoints coincide and the
shorter segment is on top of the longer segment.
Step 4 Trace the rest of the longer segment with a
different colored writing utensil. That will be the
answer.
Exercise 5
Step 1 Trace segments AB and CD so that the two
segments form a segment of length AB CD.
Step 2 Fold your patty paper so that points A and
D coincide. Crease along the fold.
Step 3 Unfold and draw a line in the crease.
7. The perpendicular bisectors all intersect in one
point.
10. The quadrilateral appears to be a rhombus.
V
E
O
C
R
D
14. One way to balance it is along the median. The
two halves weigh the same.
sample figure:
C
S
I
11.
D
Ness Station
Umsar
Station
A
Answers to Exercises
Any point on the perpendicular bisector of the
segment connecting the two offices would be
equidistant from the two post offices. Therefore,
any point on one side of the perpendicular bisector
would be closer to the post office on that side.
12. It is a parallelogram.
F
L
Area CDB = 158 in2
Area DAB = 158 in2
B
Ruler
15. F
16. E
17. B
18. A
19. D
20. C
21. B, C, D, E, H, I, O, X (K in some fonts, though
not this one)
22. Methods will vary.
C
T
10 cm
A
40⬚
13. The triangles are not necessarily congruent,
but their areas are equal. A cardboard triangle
would balance on its median.
34
ANSWERS TO EXERCISES
A
70⬚
B
It is not possible to draw a second triangle with the
same angle and side measures that is not congruent
to the first.
LESSON 3.3
1.
6. The two folds are parallel.
P
B
P
Q
I
G
The answer depends on the angle drawn and where
P is placed.
C
2.
D
A
B
7. Fold the patty paper through the point so that
two perpendiculars coincide to see the side closest
to the point. Fold again using the perpendicular
of the side closest to the point and the third
perpendicular; compare those sides.
8. Draw a line. Mark two points on it, and label
them A and C. Construct a perpendicular at C.
congruent to CA
. The altitude CD
is
Mark off CB
also the angle bisector, median, and perpendicular
bisector.
A
D
C
B
9.
U
O
B
T
O
T
Answers to Exercises
3.
B
U
10.
E
Two altitudes fall outside the triangle, and one falls
inside.
4. From the point, swing arcs on the line to
construct a segment whose midpoint is that point.
Then construct the perpendicular bisector of the
segment.
L
A
B
11.
12.
Complement
of ⬔A
⬔A
5. Construct perpendiculars from point Q and
and RE
congruent to QR
.
point R. Mark off QS
Connect points S and E.
Q
R
S
E
ANSWERS TO EXERCISES
35
13. See table below.
14.
18. not congruent
6 cm
6 cm
40⬚
40⬚
8 cm
15.
F
8 cm
19. possible answer:
Y
T
I
16.
A
E
20. possible answer:
C
D
5 cm
F
9 cm
B
5 cm
Answers to Exercises
17.
9 cm
13. (Lesson 3.3)
Rectangular pattern with triangles
Rectangle
1
2
3
4
5
6
…
Number of shaded
triangles
2
9
20
35
54
77
…
n
…
35
…
2484
(2n 1)(n 1)
36
ANSWERS TO EXERCISES
LESSON 3.4
1.
2.
3.
4.
5.
6.
D
F
A
C
E
11.
12. The angle bisectors are perpendicular. The
sum of the measures of the linear pair is 180°. The
sum of half of each must be 90°.
z
z
z
7.
A
B
M
R
P
8.
P
A
P
E
M
S
M
M
E
U
M
O
S
9a, b.
18.
G
45⬚
90⬚ 45⬚
N
9c.
I
A
S
L
135⬚
O
T
45⬚
10.
Altitude
Angle
bisector
Median
ANSWERS TO EXERCISES
37
Answers to Exercises
A
R
R
13. If a point is equally distant from the sides of
an angle, then it is on the bisector of an angle. This
is true for points in the same plane as the angle.
Mosaic answers: Square pattern constructions:
perpendiculars, equal segments, and midpoints;
The triangles are not identical, as the downward
ones have longer bases.
14. y 110°
15. mR 46°
16. Angle A is the largest; mA 66°,
mB 64°, mC 50°.
17. STOP
19.
22a. A web of lines fills most of the plane, except
a U-shaped region and a V-shaped region. (The
U-shaped region is actually bounded by a section
were
of a parabola and straight lines. If AB
extended to AB , the U would be a complete
parabola.)
B
5.6 cm
130⬚
A 3.5 cm C
20.
A
A
B
6.5 cm
C
21. No, the triangles don’t look the same.
60⬚
40⬚
60⬚
40⬚
8 cm
Answers to Exercises
8 cm
38
ANSWERS TO EXERCISES
D
B C
Parabola
and half the
22b. a line segment parallel to AB
length (The segment is actually the midsegment
of ABD.)
LESSON 3.5
1.
2.
3. Construct a segment with length z. Bisect the
segment to get length 2z. Bisect again to get a
segment with length 4z. Construct a square with
each side of length 4z.
8. 1 S and 2 U by of the Alternate
Interior Angles Conjecture
9. The ratios appear to be the same.
10. 1 3 and 2 4 by the
Corresponding Angles Conjecture
11. A parallelogram
12. Using the Converse of the Parallel Lines
Conjecture, the angle bisectors are parallel:
BC
.
DAB ABC, so AD
D
B
A
1z
4
13. Construct the perpendicular bisector of each
of the three segments connecting the fire stations.
1z
2
z
x
A
x
x
A
Eliminate the rays beyond where the bisectors
intersect. A point within any region will be closest
to the fire station in that region.
O
B
M
14. Z
15.
x
x
5. sample construction:
R
P
T
D
A
16.
I
R
R
60⬚
O
E
W
T
K
R
perpendicular to
6. Draw a line and construct ML
it. Swing an arc from point M to point G so that
MG RA. From point G, swing an arc to construct
. Finish the parallelogram by swinging an arc of
RG
length RA from R and swinging an arc of length GR
from M. There is only one possible parallelogram.
M
G
L
A
R
RC = KE = 8 cm
C
17. a 72°, b 108°, c 108°, d 108°, e 72°,
f 108°, g 108°, h 72°, j 90°, k 18°, l 90°,
m 54°, n 62°, p 62°, q 59°, r 118°;
Explanations will vary. Sample explanation:
c is 108° because of the Corresponding Angles
Conjecture. Using the Vertical Angles Conjecture,
2m 108°, so m 54°. p n because of the
Corresponding Angles Conjecture. Using the
Linear Pair Conjecture, n 62°, so p 62°.
Using the Linear Pair Conjecture, r p 180°.
Because p 62°, r 118°.
7. 1 S, 2 U
ANSWERS TO EXERCISES
39
Answers to Exercises
4.
C
USING YOUR ALGEBRA SKILLS 3
Answers to Exercises
1. perpendicular
2. neither
3. perpendicular
4. parallel
5. possible answer: (2, 5) and (7, 11)
6. possible answer: (1, 5) and (2, 12)
7. Ordinary; no two slopes are the same, so no
EM
because their
sides are parallel, although TE
slopes are opposite reciprocals.
8. Ordinary, for the same reason as in Exercise 7—
none of the sides are quite parallel.
RO
9. trapezoid: KC
61;
10a. Slope HA slope ND
slope NA
6. Quadrilateral HAND
slope HD
is a rectangle because opposite sides are parallel
and adjacent sides are perpendicular.
40
ANSWERS TO EXERCISES
midpoint AD
21, 3. The
10b. Midpoint HN
diagonals of a rectangle bisect each other.
11a. Yes, the diagonals are perpendicular.
1; slope VR
1.
Slope OE
midpoint OE
(2, 4).
11b. Midpoint VR
The diagonals of OVER bisect each other.
11c. OVER appears to be a rhombus. Slope
slope RE
51 and slope OR
slope
OV
VE 5, so opposite sides are parallel. Also, all of
the sides appear to have the same length.
12a. Both slopes equal 21.
12b. The segments are not parallel because they
are coincident.
12c. distinct
13. (3, 6)
LESSON 3.6
1. Sample description: Construct one of the
segments, and mark arcs of the correct length from
the endpoints. Draw sides to where those arcs meet.
M
angle at each end of that segment congruent to one
of the angles in the book. Where they meet is the
third vertex of the triangle.
5.
C
A
A
B
C
A
M
A
S
S
A
M
S
2.
O
D
B
Sample description: Construct A and mark off
the distance AB. From B swing an arc of length BC
to intersect the other side of A at two points.
Each gives a different triangle.
C
6.
A
y ⫺x
____
2
O
O
O
T
x
T
y
T
Sample description: Construct O. Mark off
distances OD and OT on the sides of the angle.
Connect D and T.
I
3.
Sample description: Mark the distance y, mark
back the distance x, and bisect the remaining
length of y x. Using an arc of that length, mark
arcs on the ends of segment x. The point where
they intersect is the vertex angle of the triangle.
7.
I
Y
G
Y
I
Y
. Construct I at
Sample description: Construct IY
I and Y at Y. Label the intersection of the rays
point G.
4. Yes, students’ constructions must be either
larger or smaller than the triangle in the book.
Sample description: Draw an angle. Mark off equal
segments on the sides of the angle. Use a different
compass setting to draw intersecting arcs from the
ends of those segments.
8. Sample description: Draw an angle and mark
off unequal distances on the sides. At the endpoint
of the longer segment (not the angle vertex), swing
an arc with the length of the shorter segment. From
the endpoint of the shorter segment, swing an arc
the length of the longer segment. Connect the
endpoints of the segments to the intersection
points of the arcs to form a quadrilateral.
Sample description: Draw one side with a different
length than the lengths in the book. Duplicate an
ANSWERS TO EXERCISES
41
Answers to Exercises
D
y ⫺x
____
2
x
12. new coordinates: E(4, 6), A(7, 0), T(1, 2)
9.
y
–5
Sample description: Draw a segment and draw an
angle at one end of the segment. Mark off a
distance equal to that segment on the other side of
the angle. Draw an angle at that point and mark off
the same distance. Connect that point to the other
end of the original segment.
10.
E
T'
A'
–5
A
–5 T
E'
13.
Reflectional
symmetries
Rotational
symmetries
Trapezoid
0
0
Kite
1
0
Parallelogram
0
2
Rhombus
2
2
Rectangle
2
2
Answers to Exercises
Figure
Sample description: Draw an angle and mark off
equal lengths on the two sides. Use that length to
determine another point that distance from the
points on the sides. Connect that point with the
two points on the side of the angle.
11. Answers will vary. The angle bisector lies
between the median and the altitude. The order of
the points is either M, R, S or S, R, M. One possible
conjecture: In a scalene obtuse triangle the angle
bisector is always between the median and the
altitude.
m⬔ABC = 111⬚
14. half a cylinder
15. 503
16.
110⬚
E
R
B
3.2 cm
R
A
S
Altitude
42
Median
M
Angle
bisector
ANSWERS TO EXERCISES
110⬚
C
x
A
5.5 cm
110⬚
C
LESSON 3.7
1. incenter
Because the station needs to be equidistant from
the paths, it will need to be on each of the angle
bisectors.
2. circumcenter
3. incenter
8. Yes, any circle with a larger radius would not
fit within the triangle. To get a circle with a larger
radius tangent to two of the sides would force the
circle to pass through the third side twice.
9. No, on an obtuse triangle the circle with the
largest side of the triangle as the diameter of the
circle creates the smallest circular region that
contains the triangle. The circumscribed circle of
an acute triangle does create the smallest circular
region that contains the triangle.
Stove
Fridge
Sink
To find the point equidistant from three points,
find the circumcenter of the triangle with those
points as vertices.
5. Circumcenter. Find the perpendicular bisectors
of two of the sides of the triangle formed by the
classes. Locate the pie table where these two lines
intersect.
6.
10. For an acute triangle, the circumcenter is
inside the triangle; for an obtuse triangle, the
circumcenter is outside the triangle. The
circumcenter of a right triangle lies on the
midpoint of the hypotenuse.
11. For an acute triangle, the orthocenter is inside
the triangle; for an obtuse triangle, the orthocenter
is outside the triangle. The orthocenter of a right
triangle lies on the vertex of the right angle.
12. The midsegment appears parallel to side MA
and half the length.
A
S
M
7.
H
T
13. The base angles of the isosceles trapezoid
appear congruent.
A
T
O
M
ANSWERS TO EXERCISES
43
Answers to Exercises
The center of the circular sink must be equidistant
from the three counter edges, that is, the incenter of
the triangle.
4. circumcenter
14. The measure of A is 90°. The angle inscribed
in a semicircle appears to be a right angle.
21.
Y
40⬚ 40⬚
A
4.8 cm
6.4 cm
M
T
K
15. The two diagonals appear to be perpendicular
bisectors of each other.
A
T
E
22. construction of an angle bisector
T
16.
y
9
23. construction of a perpendicular line through a
point on a line
Answers to Exercises
x+y=9
9
x
17.
Construct the incenter by bisecting the two angles
shown. Any other point on the angle bisector of the
third angle must be equidistant from the two
unfinished sides. From the incenter, make congruent
arcs that intersect the unfinished sides. The
intersection points are equidistant from the incenter.
Use two congruent arcs to find another point that
is equidistant from the two points you just
constructed. The line that connects this point and
the incenter is the angle bisector of the third angle.
18. Answers should describe the process of
discovering that the midpoints of the altitudes are
collinear for an isosceles right triangle.
19. a triangle
M
20.
6.0 cm
R
6.0 cm
60⬚ 6.0 cm 60⬚
O
60⬚
60⬚
6.0 cm
6.0 cm
H
44
ANSWERS TO EXERCISES
24. construction of a line parallel to a given line
through a point not on the line
25. construction of an equilateral triangle
26. construction of a perpendicular bisector
LESSON 3.8
1. The center of gravity is the centroid. She needs
to locate the incenter to create the largest circle
within the triangle.
2. AM 20; SM 7; TM 14; UM 8
3. BG 24; IG 12 4. RH 42; TE 45
5. The points of concurrency are the same point
for equilateral triangles because the segments are
the same.
9. circumcenter
10. The shortest chord through P is a segment
perpendicular to the diameter through P, which is
the longest chord through P.
O
P
11.
A
B'
C
6.
B
12. rule: 2n 2, possible answer:
H
Centroid
Incenter
7. ortho-/in-/centroid/circum-; the order changes
when the angle becomes larger than 60°. The
points become one when the triangle is equilateral.
Orthocenter
Centroid
Circumcenter
8. Start by constructing a quadrilateral, then make
a copy of it. Draw a diagonal in one, and draw a
different diagonal in the second. Find the centroid
of each of the four triangles. Construct a segment
connecting the two centroids in each quadrilateral.
Place the two quadrilaterals on top of each other
matching the congruent segments and angles.Where
the two segments connecting centroids intersect is
the centroid of the quadrilateral.
C
D
H
C
C
H
H
H
C
C
H
H
H
C
C
C
H
H
H
H
14. a 128°, b 52°, c 128°, d 128°,
e 52°, f 128°, g 52°, h 38°,
k 52°, m 38°, n 71°, p 38°
15. Construct altitudes from the two accessible
vertices to construct the orthocenter. Through the
orthocenter, construct a line perpendicular to the
southern boundary of the property. This method
will divide the property equally only if the southern
boundary is the base of an isosceles triangle.
Altitude to
missing vertex
16. 1580 greetings
C⬘
D⬘
M4
M2
M3
M1
A
H
C
13.
Orthocenter
Incenter
H
B A⬘
B⬘
ANSWERS TO EXERCISES
45
Answers to Exercises
Circumcenter
CHAPTER 3 REVIEW
1. False; a geometric construction uses a straightedge and a compass.
2. False; a diagonal connects two non-consecutive
vertices.
3. true
4. true
5. false
B
A
28.
_1 z
2
y
y
Segment
29.
5x
P
C
D
R
3x
6. False; the lines can’t be a given distance from a
segment because the segment has finite length and
the lines are infinite.
7. false
C
x
_1 z
2
4x
Q
30. mA mD.You must first find B.
mB 180° 2(mA).
2y
B
Answers to Exercises
A
D
B
8. true
9. true
10. False; the orthocenter does not always lie
inside the triangle.
11. A
12. B or K 13. I
14. H
15. G
16. D
17. J
18. C
19.
20.
2y
⬔D
⬔B
⬔A
A
31.
21.
4x
A
Copy
B
y
22.
y
D
F
32.
23. Construct a 90° angle and bisect it twice.
24.
I
y
25. incenter
26. Dakota Davis should locate the circumcenter
of the triangular region formed by the three stones,
which is the location equidistant from the stones.
B
27.
A
46
z
ANSWERS TO EXERCISES
C
R
x
T
33. rotational symmetry
34. neither
35. both
36. reflectional symmetry
37. D
38. A
39. C
40. B
41. False; an isosceles triangle has two congruent
sides.
42. true
43. False; any non-acute triangle is a
counterexample.
44. False; possible explanation: The orthocenter is
the point of intersection of the three altitudes.
45. true
46. False; any linear pair of angles is a
counterexample.
47. False; each side is adjacent to one congruent
side and one noncongruent side, so two consecutive
sides may not be congruent.
48. false;
58c. no
59.
Q
Q'
49. False; the measure of an arc is equal to the
measure of its central angle.
50. false; TD 2DR
51. False; a radius is not a chord.
52. true
53. False; inductive reasoning is the process of
observing data, recognizing patterns, and making
generalizations about those patterns.
54. paradox
55a. 2 and 6 or 3 and 5
55b. 1 and 5
55c. 138°
56. 55
57. possible answer:
58a. yes
58b. If the month has 31 days, then the month is
October.
60. (Chapter 3 Review)
n
f(n)
1
2
3
4
5
6
…
1
2
5
8
11
14
…
n
…
20
…
56
f(n) 3n 4
61. (Chapter 3 Review)
n
1
2
3
4
5
6
…
f(n)
0
3
8
15
24
35
…
n
f(n) n2
…
20
…
399
1
ANSWERS TO EXERCISES
47
Answers to Exercises
60. See table below.
61. See table below.
62. a 38°, b 38°, c 142°, d 38°, e 50°,
f 65°, g 106°, h 74°.
Possible explanation: The angle with measure c is
congruent to an angle with measure 142° because
of the Corresponding Angles Conjecture, so
c 142°. The angle with measure 130° is
congruent to the bisected angle by the
Corresponding Angles Conjecture. The angle with
measure f has half the measure of the bisected
angle, so f 65°.
63. Triangles will vary. Check that the triangle is
scalene and that at least two angle bisectors have
been constructed.
64. mFAD 30° so mADC 30°, but
its vertical angle has measure 26°. This is a
contradiction.
65. minimum: 101 regions by 100 parallel lines;
maximum: 5051 regions by 100 intersecting,
noncurrent lines
Answers to Exercises
CHAPTER 4 • CHAPTER
4
CHAPTER 4 • CHAPTER
angles P and D can be drawn at each endpoint
using the protractor.
LESSON 4.1
Q
1. The angle measures change, but the sum
remains 180°.
2. 73°
3. 60°
4. 110°
5. 24°
6. 3 360° 180° 900°
7. 3 180° 180° 360°
8. 69°; 47°; 116°; 93°; 86°
9. 30°; 50°; 82°; 28°; 32°; 78°; 118°; 50°
55⬚
P
10.
Answers to Exercises
⬔R
⬔M
11.
⬔A
⬔G
⬔L
12. First construct E, using the method used in
Exercise 10.
85⬚
40⬚
7 cm
17. The third angles of the triangles also have the
same measures; are equal in measure
18. You know from the Triangle Sum Conjecture
that mA mB mC 180°, and mD mE mF 180°. By the transitive property,
mA mB mC mD mE mF.
You also know that mA mD, and mB mE. You can substitute for mD and mE in the
longer equation to get mA mB mC mA mB mF. Subtracting equal terms
from both sides, you are left with mC mF.
19. For any triangle, the sum of the angle measures
is 180°, by the Triangle Sum Conjecture. Since the
triangle is equiangular, each angle has the same
measure, say x. So x x x 180°, and x 60°.
20. false
E
R
A
13.
Fold
⬔M ⬔A
21. false
E
⬔R
⬔G
⬔L
A
R
22. false
14. From the Triangle Sum Conjecture
mA mS mM 180°. Because M is a
right angle, mM 90°. By substitution,
mA mS 90° 180°. By subtraction,
mA mS 90°. So two wrongs make a right!
15. Answers will vary. See the proof on page 202.
To prove the Triangle Sum Conjecture, the Linear
Pair Conjecture and the Alternate Interior Angles
Conjecture must be accepted as true.
16. It is easier to draw PDQ if the Triangle Sum
Conjecture is used to find that the measure of
can be drawn to be 7 cm, and
D is 85°. Then PD
48
ANSWERS TO EXERCISES
D
23. false
24. true
25. eight; 100
LESSON 4.2
F
D
E
C
A
B
16.
P
M
N
K
G
H
17. possible answer:
Fold 1
Fold 3
Fold 2
Fold 4
105⬚ 60⬚
45⬚
18.
19.
20.
21.
22.
23.
0
perpendicular
parallel
parallel
neither
parallelogram
40
8
16
24
32
40
24. New: (6, 3), (2, 5), (3, 0). Triangles are
congruent.
25. New: (3, 3), (3, 1), (1, 5). Triangles
are congruent.
ANSWERS TO EXERCISES
49
Answers to Exercises
1. 79°
2. 54°
3. 107.5°
4. 44°; 35 cm
5. 76°; 3.5 cm
6. 72°; 36°; 8.6 cm
7. 78°; 93 cm
8. 75 m; 81°
9. 160 in.; 126°
10. a 124°, b 56°, c 56°, d 38°, e 38°,
f 76°, g 66°, h 104°, k 76°, n 86°,
p 38°; Possible explanation: The angles with
measures 66° and d form a triangle with the angle
with measure e and its adjacent angle. Because d,
e, and the adjacent angle are all congruent,
3d 66° 180°. Solve to get d 38°. This is
also the measure of one of the base angles of the
isosceles triangle with vertex angle measure h.
Using the Isosceles Triangle Conjecture, the other
base angle measures d, so 2d h 180°, or
76° h 180°. Therefore, h 104°.
11. a 36°, b 36°, c 72°, d 108°, e 36°;
none
12a. Yes. Two sides are radii of a circle. Radii must
be congruent; therefore, each triangle must be
isosceles.
12b. 60°
13. NCA
14. IEC
15.
USING YOUR ALGEBRA SKILLS 4
1.
3.
5.
7.
false
not a solution
not a solution
y 4
true
solution
x7
x 8
1
10. n 2
9. x 4.2
Answers to Exercises
2.
4.
6.
8.
11. x 2
12. t 18
9
2
14a. x 4
13. n 5
14b. x 94; The two methods produce identical
results. Multiplying by the lowest common
denominator (which is comprised of the factors of
both denominators) and then reducing common
factors (which clears the denominators on either
side) is the same as simply multiplying each numerator
by the opposite denominator (or cross multiplying).
Algebraically you could show that the two methods
are equivalent as follows:
c
a
b d
c
a
bd b bd d
abd bcd
b
d
ad bc
The method of “clearing fractions” results in the
method of “cross multiplying.”
50
ANSWERS TO EXERCISES
15. You get an equation that is always false, so
there is no solution to the equation.
16. Camella is not correct. Because the equation
0 0 is always true, the truth of the equation does
not depend on the value of x. Therefore, x can be
any real number. Camella’s answer, x 0, is only
one of infinitely many solutions.
17.
2x
x
2x
If x equals the measure of the vertex angle, then
the base angles each measure 2x. Applying the
Triangle Sum Conjecture results in the following
equation:
x 2x 2x 180°
5x 180°
x 36°
The measure of the vertex angle is 36° and the
measure of each base angle is 72°.
LESSON 4.3
1. yes
2. no
4
5
9
3. no
5
6
12
Answers to Exercises
4. yes
5. a, b, c
6. c, b, a
7. b, a, c
8. a, c, b
9. a, b, c
10. v, z, y, w, x
11. 6 length 102
12. By the Triangle Inequality Conjecture, the
sum of 11 cm and 25 cm should be greater than
48 cm.
13. b 55°, but 55° 130° 180°, which is
impossible by the Triangle Sum Conjecture.
14. 135°
15. 72°
16. 72°
17. a b c 180° and x c 180°. Subtract c
from both sides of both equations to get x 180 c
and a b 180 c. Substitute a b for 180 c
in the first equation to get x a b.
18. 45°
19. a 52°, b 38°, c 110°, d 35°
20. a 90°, b 68°, c 112°, d 112°, e 68°,
f 56°, g 124°, h 124°
21. By the Triangle Sum Conjecture, the third
angle must measure 36° in the small triangle, but it
measures 32° in the large triangle. These are the
same angle, so they can’t have different measures.
22. ABE
23. FNK
24. cannot be determined
ANSWERS TO EXERCISES
51
13. Cannot be determined. SSA is not a congruence
conjecture.
14. AIN by SSS or SAS
15. Cannot be determined. Parts do not correspond.
16. SAO by SAS
17. Cannot be determined. Parts do not correspond.
18. RAY by SAS
and PR
is (0, 0).
19. The midpoint of SD
Therefore, DRO SPO by SAS.
20. Because the LEV is marking out two triangles
that are congruent by SAS, measuring the length
of the segment leading to the finish will also
approximate the distance across the crater.
21.
22.
LESSON 4.4
1. Answers will vary. Possible answer: If three
sides of one triangle are congruent to three sides of
another triangle, then the triangles are congruent
(all corresponding angles are also congruent).
2. Answers will vary. Possible answer: The picture
statement means that if two sides of one triangle
are congruent to two sides of another triangle, and
the angles between those sides are also congruent,
then the two triangles are congruent.
If you know this:
then you also know this:
Answers to Exercises
3. Answers will vary. Possible answer:
23. a 37°, b 143°, c 37°, d 58°,
e 37°, f 53°, g 48°, h 84°, k 96°,
m 26°, p 69°, r 111°, s 69°; Possible
explanation: The angle with measure h is the vertex
angle of an isosceles triangle with base angles
measuring 48°, so h 2(48) 180°, and h 84°.
The angle with measure s and the angle with
measure p are corresponding angles formed by
parallel lines, so s p 69°.
24. 3 cm third side 19 cm
25. See table below.
13
3
26a. y 6
b. y c. y 4x 2
3
27. (5, 3)
4. SAS
5. SSS
6. cannot be determined
7. SSS
8. SAS
9. SSS (and the Converse of the Isosceles Triangle
Conjecture)
10. yes, ABC ADE by SAS
11. Possible answer: Boards nailed diagonally in
the corners of the gate form triangles in those
corners. Triangles are rigid, so the triangles in the
gate’s corners will increase the stability of those
corners and keep them from changing shape.
12. FLE by SSS
25. (Lesson 4.4)
Side length
1
2
3
4
5
Elbows
4
4
4
4
4
T’s
0
4
8
12
16
Crosses
0
1
4
9
16
…
4n 4
52
ANSWERS TO EXERCISES
n
…
4
20
4
76
361
(n 1)2
LESSON 4.5
1. If two angles and the included side of one triangle
are congruent to the corresponding side and angles
of another triangle, then the triangles are congruent.
2. If two angles and a non-included side of one
triangle are congruent to the corresponding side
and angles of another triangle, then the triangles
are congruent.
If you know this:
then you also know this:
3. Answers will vary. Possible answer:
L
K
M
27.
28a. about 100 km southeast of San Francisco
28b. Yes. No, two towns would narrow it down
to two locations. The third circle narrows it down
to one.
ORE
GO
N
Boise
IDAH
Eureka
Sacramento
San
Francisco
Reno
NEV
O
Elko
ADA
U TA H
LI
FO
RN
Las
Vegas
IA
Los Angeles
Miles
0 50 100
0
100
Kilometers
200
ARIZ
ONA
400
ANSWERS TO EXERCISES
53
Answers to Exercises
22. Construction will show a similar but larger
(or smaller) triangle constructed from a drawn
triangle by duplicating two angles on either end of a
new side that is not congruent to the corresponding
side.
23. Draw a line segment. Construct a perpendicular.
Bisect the right angle. Construct a triangle with
two congruent sides and with a vertex that
measures 135°.
24. 125
25. False. One possible counterexample is a kite.
26. None. One triangle is determined by SAS.
CA
4. ASA
5. cannot be determined
6. SAA
7. cannot be determined
8. ASA
9. cannot be determined
10. FED by SSS
11. WTA by ASA or SAA
12. SAT by SAS
13. PRN by ASA or SAS; SRE by ASA
14. Cannot be determined. Parts do not
correspond.
15. MRA by SAS
16. Cannot be determined.AAA does not guarantee
congruence.
17. WKL by ASA
18. Yes, ABC ADE by SAA or ASA.
slope CD
3 and slope 19. Slope AB
BC 1
3, so AB
BC
, CD
DA
, and
slope DA
DA
. ABC CDA by SAA.
BC
20.
21. The construction is the same as the
construction using ASA once you find the third
angle, which is used here. (Finding the third angle
is not shown.)
13. cannot be determined
14. KEI by ASA
15. UTE by SAS
16.
Answers to Exercises
LESSON 4.6
BD
(same segment), A C
1. Yes. BD
(given), and ABD CBD (given), so DBA CB
by CPCTC.
DBC by SAA. AB
WN
and C W (given), and
2. Yes. CN
RNC ONW (vertical angles), CNR ON
by CPCTC.
WON by ASA. RN
3. Cannot be determined. The congruent parts
lead to the ambiguous case SSA.
IT
4. Yes. S I, G A (given), and TS
(definition of midpoint), so TIA TSG by
IA
by CPCTC.
SAA. SG
FR
and UO
UR
(given), and UF
5. Yes. FO
(same segment), so FOU FRU by SSS.
UF
O R by CPCTC.
MA
and ME
MR
(given), and
6. Yes. MN
M M (same angle), so EMA RMN by
SAS. E R by CPCTC.
EU
and BU
ET
(given), and
7. Yes. BT
UT UT (same segment), so TUB UTE by
SSS. B E by CPCTC.
8. Cannot be determined. HLF LHA by
and HF
are not corresponding sides.
ASA, but HA
9. Cannot be determined. AAA does not guarantee congruence.
10. Yes. The triangles are congruent by SAS.
11. Yes. The triangles are congruent by SAS, and
the angles are congruent by CPCTC.
and DF
to form ABC and DEF.
12. Draw AC
because all were drawn with
AB CB DE FE
DF
for the same reason.
the same radius. AC
ABC DEF by SSS. Therefore, B E by
CPCTC.
17.
18. a 112°, b 68°, c 44°, d 44°, e 136°,
f 68°, g 68°, h 56°, k 68°, l 56°, m 124°;
Possible explanation: f and g are measures of base
angles of an isosceles triangle, so f g. The vertex
angle measure is 44°, so subtract 44° from 180° and
divide by 2 to get f 68°. The angle with measure
m is the exterior angle of a triangle. Add the remote
interior angle measures 56° and 68° to get
m 124°.
19. ASA. The “long segment in the sand” is a
shared side of both triangles
20. (4, 1)
21. See table below.
22. Value C is always decreasing.
23. x 3, y 10
21. (Lesson 4.6)
54
Number of sides
3
4
5
6
7
…
12
…
n
Number of struts needed
to make polygon rigid
0
1
2
3
4
…
9
…
n3
ANSWERS TO EXERCISES
3. See flowchart below.
4. See flowchart below.
LESSON 4.7
1. See flowchart below.
2. See flowchart below.
1. (Lesson 4.7)
1
SE SU
}?
2
ESM USO
4
E U
}?
3
Given
Given
}
}
5
? ?
MS SO
}? CPCTC
ASA Congruence
Conjecture
1 2
}? Vertical Angles Conjecture
2. (Lesson 4.7)
1
3
CI IM
Definition
of midpoint
Given
2
Answers to Exercises
I is midpoint of CM
4
I is midpoint of BL
6
IL IB
of
}? Definition
midpoint
}? Given
}
}
? ?
MB
}? CL
7
}? CIL MIB
CPCTC
by SAS
5 1 2
}? Vertical Angles Conjecture
3. (Lesson 4.7)
1
3
NS is a median
Given
2
Same segment
4
S is a midpoint
Definition
of median
6
WS SE
ESN
WSN ?
}
}
? SSS
Definition
of midpoint
5
7
E
W ?
}
?
}
CPCTC
WN NE
}? Given
4. (Lesson 4.7)
1 NS is an
NS NS
2
angle bisector
}
1 ?
}
2
Definition of ?
angle bisector
Given
3
W E
}
? Given
4
5
WNS ENS
? ?
}
}
? SAA
}
6
WN NE
}
? CPCTC
7 NEW is
isosceles
of
}? Definition
isosceles triangle
NS NS
}? Same segment
ANSWERS TO EXERCISES
55
5. See flowchart below.
BC
, CD
AD
6. Given: ABC with BA
is the angle bisector of ABC.
Show: BD
12. ACK by SSS
13. a 72°, b 36°, c 144°, d 36°, e 144°,
f 18°, g 162°, h 144°, j 36°, k 54°,
m 126°
14. The circumcenter is equidistant from all three
vertices because it is on the perpendicular bisector
of each side. Every point on the perpendicular
bisector of a segment is equidistant from the
endpoints. Similarly, the incenter is equidistant
from all three sides because it is on the angle
bisector of each angle, and every point on an angle
bisector is equidistant from the sides of the angle.
15. ASA. The fishing pole forms the side.
“Perpendicular to the ground” forms one angle.
“Same angle on her line of sight” forms the other
angle.
2
16. 7
17.
A
D
1
2
B
C
BA ⬵ BC
CD ⬵ AD
BD ⬵ BD
Given
Given
Same segment
䉭ABD ⬵ 䉭CBD
SSS
⬔1 ⬵ ⬔2
CPCTC
Answers to Exercises
→
BD bisects ⬔ABC
Definition of angle
bisector
7. The angle bisector does not go to the midpoint
of the opposite side in every triangle, only in an
isosceles triangle.
, because it is across from the smallest angle
8. NE
, which is across
in NAE. It is shorter than AE
from the smallest angle in LAE.
9. The triangles are congruent by SSS, so the two
central angles cannot have different measures.
10. PRN by ASA; SRE by ASA
11. Cannot be determined. Parts do not
correspond.
18.
y
4 Y
X
B
O
B' Y'4
O'
x
X'
5. (Lesson 4.7)
1
SA NE
3
? Given
ᎏ
2
SE NA
3 4
AIA Conjecture
ESN ANS
4 ? 1 2
ᎏ
? AIA Conjecture
ᎏ
? Given
ᎏ
5
SN SN
Same segment
56
ANSWERS TO EXERCISES
6
? ᎏ
?
ᎏ
? ASA
ᎏ
7 ? SA NE
ᎏ
? CPCTC
ᎏ
This proof shows that in a parallelogram,
opposite sides are congruent.
7.
LESSON 4.8
1.
2.
3.
4.
5.
6.
6
90°; 18°
45°
See flowchart below.
See flowchart below.
1
Isosceles 䉭ABC
with AC ⬵ BC
and CD bisects
⬔C
2
AD ⬵ BD
AC ⬵ BC
2
Given
Same segment
D is
midpoint
of AB
⬔ACD ⬵ ⬔DCB
CPCTC
䉭ADC ⬵ 䉭BDC
AD ⬵ BD
Def. of midpoint
7
CD is angle
bisector of ⬔ACB
Def. of angle bisector
8
CD is altitude
of 䉭ABC
Conjecture B
(Exercise 5)
CPCTC
4
䉭ADC ⬵ 䉭BDC
SSS
6
Conjecture A
(Exercise 4)
4
3
CD ⬵ CD
Given
5
Given
3
1
9
CD ⬜ AB
Def. of altitude
8. Yes. First show that the three exterior triangles
are congruent by SAS.
CD is a median
Def. of median
Answers to Exercises
4. (Lesson 4.8)
1
}
2 ?
CD is the bisector of C
Given
1 2
Definition of
angle bisector
3
ABC is isosceles
with AC BC
5
}? SAS
Given
4
ADC BDC
CD CD
Same segment
5. (Lesson 4.8)
1
ABC is isosceles with
AC BC, and CD is
the bisector of C
2
ADC BDC
4
}? CPCTC
Conjecture A
Given
3
1 and 2 form
a linear pair
Definition of
linear pair
1 2
5
6
Congruent
supplementary
angles are 90
1 and 2 are
supplementary
Linear Pair
Conjecture
1 and 2 are
right angles
7
AB
CD
}
Definition of ?
perpendicular
}
is an altitude 8 ?
CD
Definition of
altitude
ANSWERS TO EXERCISES
57
9.
C
A
B
D
Drawing the vertex angle bisector as an auxiliary
segment, we have two triangles. We can show them
to be congruent by SAS, as we did in Exercise 4.
Then, A B, by CPCTC. Therefore, base angles
of an isosceles triangle are congruent.
10. The proof is similar to the one on page 245,
but in reverse, and using the Converse of the
Isosceles Triangle Conjecture.
11.
5
14. y 3x 16
15. 120
16. (4, 6) or (4, 0) or any point at which the
x-coordinate is either 1 or 7 and the y-coordinate
does not equal 3
17. Hugo and Duane can locate the site of the
fireworks by creating a diagram using SSS.
Fireworks
340 m/s • 3 s
= 1.02 km
340 m/s • 5 s
= 1.7 km
Hugo
1.5 km
Duane
18. Cn H2n
H H
H
30⬚
H
Answers to Exercises
H
12. a 128°, b 128°, c 52°, d 76°, e 104°,
f 104°, g 76°, h 52°, j 70°, k 70°, l 40°,
m 110°, n 58°
13. between 16 and 17 minutes
58
ANSWERS TO EXERCISES
H
H
C
C
C
C
C
H
H
C
C
H
H
H
H
H
CHAPTER 4 REVIEW
P
⬔L
A
y
⬔A
⬔P
L
34. Construct P. Mark off the length PB on one
ray. From point B, mark off the two segments that
intersect the other ray of P at distance x.
S2
S1
x
x
P
z
B
ANSWERS TO EXERCISES
59
Answers to Exercises
1. Their rigidity gives strength.
2. The Triangle Sum Conjecture states that the
sum of the measures of the angles in every triangle
is 180°. Possible answers: It applies to all triangles;
many other conjectures rely on it.
3. The angle bisector of the vertex angle is also the
median and the altitude.
4. The distance between A and B is along the
segment connecting them. The distance from A
to C to B can’t be shorter than the distance from A
to B. Therefore, AC CB AB. Points A, B, and C
form a triangle. Therefore, the sum of the lengths
of any two sides is greater than the length of the
third side.
5. SSS, SAS, ASA, or SAA
6. In some cases, two different triangles can
be constructed using the same two sides and
non-included angle.
7. cannot be determined
8. ZAP by SAA
9. OSU by SSS
10. cannot be determined
11. APR by SAS
12. NGI by SAS
13. cannot be determined
14. DCE by SAA or ASA
15. RBO or OBR by SAS
UT
by
16. AMD UMT by SAS, AD
CPCTC
17. cannot be determined
18. cannot be determined
AL
by
19. TRI ALS by SAA, TR
CPCTC
KV
by
20. SVE NIK by SSS, EL
overlapping segments property
21. cannot be determined
22. cannot be determined
23. LAZ IAR by ASA, LRI IZL by
ASA, and LRD IZD by ASA
24. yes. PTS TPO by ASA or SAA
25. ANG is isosceles, so A G. However,
the sum of mA mN mG 188°. The
measures of the three angles of a triangle must sum
to 180°.
26. ROW NOG by ASA, implying that
OG
. However, the two segments shown are
OW
not equal in measure.
27. a g e d b f c. Thus, c is the
longest segment, and a and g are the shortest.
28. x 20°
29. Yes. TRE SAE by SAA, so sides are
congruent by CPCTC.
30. Yes. FRM RFA by SAA. RFM FRA by CPCTC. Because base angles are
congruent, FRD is isosceles.
31. x 48°
32. The legs form two triangles that are congruent
by SAS. Because alternate interior angles are
congruent by CPCTC, the seat must be parallel to
the floor.
33. Construct P and A to be adjacent. The
angle that forms a linear pair with the conjunction
of P and A is L. Construct A. Mark off the
length AL on one ray. Construct L. Extend the
unconnected sides of the angles until they meet.
Label the point of intersection P.
37. Possible method: Construct an equilateral
triangle and bisect one angle to obtain 30°.
Adjacent to that angle, construct a right angle
and bisect it to obtain 45°.
38. d, a b, c, e, f
Answers to Exercises
35. See flowchart below.
36. Given three sides, only one triangle is possible;
therefore, the shelves on the right hold their shape.
The shelves on the left have no triangles and move
freely as a parallelogram.
35. (Chapter 4 Review)
TMI RME
2
}
}
}
? ?
Vertical angles ?
ME
TMI EMR
TM
3 ?
5
M is midpoint
? ?
?
and IR
of TE
? Definition
? SAS
1 ?
of midpoint
? Given
4 ?
?
MR
IM
? Definition of midpoint
}
}
60
} }
}
} }
}
ANSWERS TO EXERCISES
}
}
}
T E
or
R I
6
? ?
} }
?
} CPCTC
RE
TI
7
}? }?
of
}? Converse
AIA Conjecture
Answers to Exercises
CHAPTER 5 • CHAPTER
5
13. 17
14. 15
15. the twelfth century
16. The angles of the trapezoid measure 67.5° and
112.5°; 67.5° is half the value of each angle of a regular
octagon, and 112.5° is half the value of 360° 135°.
CHAPTER 5 • CHAPTER
LESSON 5.1
1. See table below.
2. See table below.
3. 122°
4. 136°
5. 108°; 36°
6. 108°; 106°
7. 105°; 82°
8. 120°; 38°
9. The sum of the interior angle measures of the
quadrilateral is 358°. It should be 360°.
10. The measures of the interior angles shown sum
to 554°. However, the figure is a pentagon, so the
measures of its interior angles should sum to 540°.
11. 18
12. a 116°, b 64°, c 90°, d 82°, e 99°,
f 88°, g 150°, h 56°, j 106°, k 74°,
m 136°, n 118°, p 99°; Possible explanation:
The sum of the angles of a quadrilateral is 360°, so
a b 98° d 360°. Substituting 116° for a
and 64° for b gives d 82°. Using the larger
quadrilateral, e p 64° 98° 360°.
Substituting e for p, the equation simplifies to
2e 198°, so e 99°. The sum of the angles of a
pentagon is 540°, so e p f 138° 116° 540°. Substituting 99° for e and p gives f 88°.
135⬚
67.5⬚
1. (Lesson 5.1)
Number of sides of polygon
Sum of measures of angles
7
8
9
10
11
20
55
100
900° 1080° 1260° 1440° 1620° 3240° 9540° 17640°
2. (Lesson 5.1)
Number of sides of
equiangular polygon
Measure of each angle
of equiangular polygon
5
6
7
8
9
10
12
16
100
108° 120° 12874° 135° 140° 144° 150° 1571° 1762°
2
5
ANSWERS TO EXERCISES
61
Answers to Exercises
17. Answers will vary; see the answer for
Developing Proof on page 259. Using the Triangle
Sum Conjecture, a b j c d k e f l g h i 4(180°), or 720°. The four angles in
the center sum to 360°, so j k l i 360°.
Subtract to get a b c d e f g h 360°.
18. x 120°
19. The segments joining the opposite midpoints
of a quadrilateral always bisect each other.
20. D
21. Counterexample: The base angles of an
isosceles right triangle measure 45°; thus they are
complementary.
LESSON 5.2
360°
72°; 60°
15
43
a 108°
1
6. b 453°
3
5
7. c 517°, d 1157°
8. e 72°, f 45°, g 117°, h 126°
9. a 30°, b 30°, c 106°, d 136°
10. a 162°, b 83°, c 102°, d 39°, e 129°,
f 51°, g 55°, h 97°, k 83°
11. See flowchart below.
12. Yes. The maximum is three. The minimum is
zero. A polygon might have no acute interior
angles.
Answers to Exercises
1.
2.
3.
4.
5.
13. Answers will vary. Possible proof using
the diagram on the left: a b i 180°, c d h 180°, and e f g 180° by the Triangle
Sum Conjecture. a b c d e f g h i 540° by the addition property of equality.
Therefore, the sum of the measures of the angles of
a pentagon is 540°. To use the other diagram,
students must remember to subtract 360° to
account for angle measures k through o.
14. regular polygons: equilateral triangle and
regular dodecagon; angle measures: 60°, 150°,
and 150°
15. regular polygons: square, regular hexagon, and
regular dodecagon; angle measures: 90°, 120°,
and 150°
CR
by
16. Yes. RAC DCA by SAS. AD
CPCTC.
17. Yes. DAT RAT by SSS. D R by
CPCTC.
11. (Lesson 5.2)
1
a b 180
?
Linear Pair Conjecture 2
c d 180
?
Linear Pair Conjecture 3
e f 180
?
Linear Pair Conjecture 62
ANSWERS TO EXERCISES
4
540°
?
a+b+c+d+e+f= Addition property
of equality
180°
5
?
a+c+e=
? Triangle Sum Conjecture
6
360°
?
b+d+f=
Subtraction property
of equality
13. possible answer:
LESSON 5.3
W
1. 64 cm
2. 21°; 146°
3. 52°; 128°
4. 15 cm
5. 72°; 61°
6. 99°; 38 cm
7. w 120°, x 45°, y 30°
8. w 1.6 cm, x 48°, y 42°
9. See flowchart below.
10. Answers may vary. This proof uses the Kite
Angle Bisector Conjecture.
S
2
1
N
X
H
is the other base. S and H are a pair of
OW
base angles. O and W are a pair of base angles.
HO
.
SW
14. Only one kite is possible
F
because three sides determine
N
B
a triangle.
E
15.
S
H
Y
B
O
I
E
Given: Kite BENY with vertex angles B and N
is the perpendicular bisector
Show: Diagonal BN
.
of diagonal YE
BY
. From the
From the definition of kite, BE
Kite Angle Bisector Conjecture, 1 2. BX
because they are the same segment. By SAS,
BX
XE
.
BXY BXE. So by CPCTC, XY
Because YXB and EXB form a linear pair, they
are supplementary, so mYXB mEXB 180°. By CPCTC, YXB EXB, or mYXB mEXB, so by substitution, 2mYXB 180°, or
mYXB 90°. So mYXB mEXB 90°.
XE
and YXB and EXB are right
Because XY
is the perpendicular bisector of YE
.
angles, BN
I
11. possible answer: E I
16. infinitely many, possible construction:
B
E
Z
Q
N
17. 80°, 80°, 100°, 100°
18. Because ABCD is an isosceles trapezoid, A
BH
B. AGF BHE by SAA. Thus, AG
by CPCTC.
19. a 80°, b 20°, c 160°, d 20°, e 80°,
f 80°, g 110°, h 70°, m 110°, n 100°;
Possible explanation: Because d forms a linear pair
with e and its congruent adjacent angle,d 2e 180°.Substituting d 20° gives 2e 160°,so e 80°.
Using theVerticalAngles Conjecture and d 20°, the
unlabeled angle in the small right triangle measures
20°, which means h 70°. Because g and h are a
linear pair, they are supplementary, so g 110°.
K
E
12. possible answer:
O
I
U
. Q and U are a pair of base
The other base is ZI
angles. Z and I are a pair of base angles.
9. (Lesson 5.3)
1 BE BY
Given
2 EN YN
Given
BN
BN
3
? ᎏ
?
ᎏ
Same segment
BYN
4
?
BEN ᎏ
? Congruence
ᎏ
shortcut
SSS
2
5 1 ? and
ᎏ
? 4
3 ᎏ
? CPCTC
ᎏ
bisects B, BN
bisects N
BN
6
? and ᎏ
?
ᎏ
Definition of
angle bisector
ANSWERS TO EXERCISES
63
Answers to Exercises
T
W
Answers to Exercises
LESSON 5.4
1. three; one
2. 28
3. 60°; 140°
4. 65°
5. 23
6. 129°; 73°; 42 cm
7. 35
8. See flowchart below.
9. Parallelogram. Draw a diagonal of the original
quadrilateral. The diagonal forms two tri-angles.
Each of the two midsegments is parallel to the
diagonal, and thus the midsegments are parallel to
each other. Now draw the other diagonal of the
original quadrilateral. By the same reasoning, the
second pair of midsegments is parallel. Therefore,
the quadrilateral formed by joining the midpoints is
a parallelogram.
10. The length of the edge of the top base
measures 30 m. We know this by the Trapezoid
Midsegment Conjecture.
11. Ladie drives a stake into the ground to create a
triangle for which the trees are the other two
vertices. She finds the midpoint from the stake to
each tree. The distance between these midpoints is
half the distance between the trees.
13. If a quadrilateral is a kite, then exactly one
diagonal bisects a pair of opposite angles. Both the
original and converse statements are true.
14. a 54°, b 72°, c 108°, d 72°, e 162°,
f 18°, g 81°, h 49.5°, i 130.5°, k 49.5°,
m 162°, n 99°; Possible explanation: The third
angle of the triangle containing f and g measures
81°, so using the Vertical Angles Conjecture, the
vertex angle of the triangle containing h also
measures 81°. Subtract 81° from 180° and divide by
2 to get h 49.5°. The other base angle must also
measure 49.5°. By the Corresponding Angles
Conjecture, k 49.5°.
15. (3, 8)
16. (0, 8)
17. coordinates: E(2, 3.5), Z(6, 5); the slope of
83, and the slope of YT
83
EZ
18.
R
F
N
Cabin
K
There is only one kite, but more than one way to
construct it.
12. Explanations will vary.
80
60 cm
40
60
8. (Lesson 5.4)
1 FOA with
3
midsegment LN
Given
midsegment RD
64
ANSWERS TO EXERCISES
}
?
Triangle Midsegment Conjecture
2 IOA with
Given
LN OA
RD
OA
4
}
?
Triangle Midsegment
Conjecture
RD
LN
5
}?
Two lines parallel to the
same line are parallel
10.
LESSON 5.5
1.
2.
3.
4.
5.
6.
7.
34 cm; 27 cm
132°; 48°
16 in.; 14 in.
63 m
80
63°; 78°
Vb
T
Vc
S
L
A
11. (b a, c)
12. possible answer:
8.
R
a
O
D
b
c
d
P
P
9.
R
13. See flowchart below.
14. The parallelogram linkage is used for the
sewing box so that the drawers remain parallel to
each other (and to the ground) so that the contents
cannot fall out.
Vh
15. a 135°, b 90°
16. a 120°, b 108°, c 90°, d 42°, e 69°
Vw
13. (Lesson 5.5)
1 LEAN is a
parallelogram
?
Given ᎏ
2
EA LN
? Definition of
ᎏ
parallelogram
3 AEN LNE
AIA Conjecture
EAL NLA
4
?
ᎏ
AIA Conjecture
LN
AE
5 ?
ᎏ
7
AET LNT
6
ET NT
and LA
bisect
EN
each other
CPCTC
9
?
ᎏ
ASA
LT
AT
8
?
ᎏ
CPCTC
?
ᎏ
Definition of
segment bisector
Opposite sides
congruent
ANSWERS TO EXERCISES
65
Answers to Exercises
b d
c a
17. x 104°, y 98°. The quadrilaterals on the
left and right sides are kites. Nonvertex angles are
congruent. The quadrilateral at the bottom is an
isosceles trapezoid. Base angles are congruent, and
consecutive angles between the bases are
supplementary.
18. a 84°, b 96°
19. No. The congruent angles and side do not
correspond.
20.
Answers to Exercises
21. Parallelogram. Because the triangles are
congruent by SAS, 1 2. So, the segments are
66
ANSWERS TO EXERCISES
parallel. Use a similar argument to show that the
other pair of opposite sides is parallel.
1
2
22. Kite or dart. Radii of the same circle are
congruent. If the circles have equal radii, a rhombus
is formed.
USING YOUR ALGEBRA SKILLS 5
1.
y
(0, 1)
(1, –1)
2.
x
y
(3, 8)
(0, 4)
x
3.
y
4. y x 2
74
6
5. y 1
3 x 13
6. y x 1
7. y 3x 5
2
8
8. y 5x 5
9. y 80 4x
10. y 3x 26
1
11. y 4x 3
6
12. y 5x
13. y x 1
2
43
14. y 9x 9
(2, 9)
(0, 6)
Answers to Exercises
x
ANSWERS TO EXERCISES
67
LESSON 5.6
17.
E
V
L
O
1. Sometimes true; it is true only if the
parallelogram is a rectangle.
false
true
18. Constructions will vary.
2. Always true; by the definition of rectangle, all
the angles are congruent. By the Quadrilateral Sum
Conjecture and division, they all measure 90°, so
any two angles in a rectangle, including consecutive
angles, are supplementary.
3. Always true by the Rectangle Diagonals
Conjecture.
4. Sometimes true; it is true only if the rectangle is
a square.
A
B
B
E
false
Answers to Exercises
5. Always true by the Square Diagonals Conjecture.
6. Sometimes true; it is true only if the rhombus is
equiangular.
false
7. Always true; all squares fit the definition of
rectangle.
8. Always true; all sides of a square are congruent
and form right angles, so the sides become the legs
of the isosceles right triangle and the diagonal is
the hypotenuse.
9. Always true by the Parallelogram Opposite
Angles Conjecture.
10. Sometimes true; it is true only if the
parallelogram is a rectangle. Consecutive angles of
a parallelogram are always supplementary, but are
congruent only if they are right angles.
11. 20
12. 37°
13. 45°, 90°
14. DIAM is not a rhombus because it is not
equilateral and opposite sides are not parallel.
15. BOXY is a rectangle because its adjacent sides
are perpendicular.
16. Yes. TILE is a rhombus, and a rhombus is a
parallelogram.
68
ANSWERS TO EXERCISES
E
E
I
S
20. Converse: If the diagonals of a quadrilateral
are congruent and bisect each other, then the
quadrilateral is a rectangle.
Given: Quadrilateral ABCD with diagonals
BD
. AC
and BD
bisect each other
AC
Show: ABCD is a rectangle
A
8
7
true
K
19. one possible construction:
P
true
A
K
D
1
2
E
6
5
B
3
4
C
Because the diagonals are congruent and bisect
BE
DE
EC
. Using the
each other, AE
Vertical Angles Conjecture, AEB CED and
BEC DEA. So AEB CED and AED
CEB by SAS. Using the Isosceles Triangle
Conjecture and CPCTC, 1 2 5 6,
and 3 4 7 8. Each angle of the
quadrilateral is the sum of two angles, one from
each set, so for example, mDAB m1 m8.
By the addition property of equality, m1 m8 m2 m3 m5 m4 m6 m7. So mDAB mABC mBCD mCDA. So the quadrilateral is equiangular. Using
CD
.
1 5 and the Converse of AIA, AB
Using 3 7 and the Converse of AIA,
AD
. Therefore ABCD is an equiangular
BC
parallelogram, so it is a rectangle.
21. If the diagonals are congruent and bisect each
other, then the room is rectangular (converse of the
Rectangle Diagonals Conjecture).
90°. Because 4 and 5 form a linear pair,
m4 m5 180°. Substitute 90° for m4 and
solve to get m5 90°. By definition of congruent
angles, 5 3, and 5 and 3 are alternate
BC
by the Converse of the
interior angles, so AD
Parallel Lines Conjecture. Similarly, 1 and 5 are
CD
by the
congruent corresponding angles, so AB
Converse of the Parallel Lines Conjecture. Thus,
ABCD is a parallelogram by the definition of
parallelogram. Because it is an equiangular
parallelogram, ABCD is a rectangle.
28. a 54°, b 36°, c 72°, d 108°, e 36°,
f 144°, g 18°, h 48°, j 48°, k 84°
29. possible answers: (1, 0); (0, 1); (1, 2); (2, 3)
8
86
30. y 9x 9 or 8x 9y 86
7
12
31. y 1x
0 5 or 7x 10y 24
32. velocity 1.8 mi/h; angle of path 106.1°
clockwise from the north
Answers to Exercises
22. The platform stays parallel to the floor
because opposite sides of a rectangle are parallel
(a rectangle is a parallelogram).
23. The crosswalks form a parallelogram: The
streets are of different widths, so the crosswalks are
of different lengths. The streets would have to meet
at right angles for the crosswalks to form a rectangle.
The corners would have to be right angles and the
streets would also have to be of the same width for
the crosswalk to form a square.
24. Place one side of the ruler along one side of
the angle. Draw a line with the other side of the
ruler. Repeat with the other side of the angle. Draw
a line from the vertex of the angle to the point
where the two lines meet.
25. Rotate your ruler so that each endpoint of the
segment barely shows on each side of the ruler.
Draw the parallel lines on each side of your ruler.
Now rotate your ruler the other way and repeat the
process to get a rhombus. The original segment is
one diagonal of the rhombus. The other diagonal
will be the perpendicular bisector of the original
segment.
26. See flowchart below.
27. Yes, it is true for rectangles.
Given: 1 2 3 4
Show: ABCD is a rectangle
By the Quadrilateral Sum Conjecture, m1 m2 m3 m4 360°. It is given that all
four angles are congruent, so each angle measures
2 mi/h
60
1.5 mi/h
26. (Lesson 5.6)
1
QU AD
Given
2
QD AU
? Given
3
4
QUD ADU
? SSS
5
1 2
3 4
? CPCTC
DU DU
Same segment
8
QU UA AD DQ
Given
6
7 QUAD is a
QU AD
QD AU
parallelogram
Converse of the
Parallel Lines
Conjecture
Definition of
parallelogram
9 QUAD is a
rhombus
?
Definition of rhombus
ANSWERS TO EXERCISES
69
5. parallelogram
6. sample flowchart proof:
LESSON 5.7
1. See flowchart below. 2. See flowchart below.
3. See flowchart below.
2
5
4. 1 SP OA
SOP APO
SP OA
Given
Given
3
2
IY GO
Opposite sides of
rectangle are congruent
1 2
3
YOG OYI
Definition of rectangle
4
SAS
6
AIA Conjecture
4
1
YO OY
Same segment
YOG OYI
SAS
3 4
CPCTC
PO PO
7
Same segment
5
PA SO
CPCTC
Converse of
AIA Conjecture
8
YG IO
SOAP is a parallelogram
Definition of parallelogram
1. (Lesson 5.7)
2
SO KA
4 3 4
AIA
Answers to Exercises
Definition of
parallelogram
SK
1
3
SOAK is a
parallelogram
Given
5
?
OA ᎏ
1 2
? Definition of
? AIA
ᎏ
ᎏ
parallelogram
6
7
?
SOA ᎏ
AKS
? ASA
ᎏ
? SA
SA ᎏ
? Same segment
ᎏ
2. (Lesson 5.7)
1
Parallelogram BATH
with diagonal BT
Given
4
Parallelogram BATH
with diagonal HA
? Given
ᎏ
2
BAT THB
3
BAT THB
CPCTC
Conjecture proved in
Exercise 1
ATH
THA
5
6
?
?
BAH = ᎏ
HBA ᎏ
?
?
CPCTC ᎏ
ᎏ
Conjecture proved
in Exercise 1
3. (Lesson 5.7)
RT
WA
1
? ᎏ
?
ᎏ
? Given
ᎏ
AT
WR
2 ?
?
ᎏ ᎏ
WRT TAW
4
? Given
ᎏ
WT
WT
3
70
? ᎏ
?
ᎏ
? Same segment
ᎏ
ANSWERS TO EXERCISES
? ᎏ
?
ᎏ
? SSS
ᎏ
2
5 1 ?
ᎏ
WA
RT
6
? CPCTC
ᎏ
3
7 4 ?
ᎏ
? CPCTC
ᎏ
? ?
ᎏ
ᎏ
?
ᎏ
TA
RW
8
by Converse of the Parallel
Lines Conjecture
9 WATR is a
parallelogram
? Definition of
?
? ᎏ
ᎏ
ᎏ
parallelogram
?
Converse
of
the
ᎏ
Parallel Lines Conjecture
7.
1 BEAR is a parallelogram
Given
2
AB
RE
3 BR
EA
Given
4 AR
EB
Parallelogram Opposite
Sides Conjecture
5
Parallelogram Opposite
Sides Conjecture
EBR ARB RAE BEA
SSS
6
EBR ARB RAE BEA
CPCTC
7
BEAR is a rectangle
Definition of rectangle
9
6
12
3
6 inches
Answers to Exercises
is parallel to ZT
, corresponding
8. Because AR
3 and 2 are congruent. Opposite sides of
parallelogram ZART are congruent so AR TZ.
Because the trapezoid is isosceles, AR PT, and
substituting gives ZT PT making PTZ
isosceles and 1 and 2 congruent. By
substitution, 1 and 3 are congruent.
9. See sample flowchart below.
10. If the fabric is pulled along the warp or the weft,
nothing happens. However, if the fabric is pulled
along the bias, it can be stretched because the
rectangles are pulled into parallelograms.
11. 30° angles in 4-pointed star, 30° angles in
6-pointed star; yes
12. He should measure the alternate interior
angles to see whether they’re congruent.If they are,
the edges are parallel.
: y 2x 3; QI
: y 21x 2
13. ES
14. (12, 7)
1
15. 3
16.
9. (Lesson 5.7)
2
TH
GR
Given
1
Isosceles trapezoid
GTHR
3
GT
GT
Same segment
Given
4
5
RGT HTG
SAS
6
TR
GH
CPCTC
RGT HTG
Isosceles Trapezoid
Conjecture
ANSWERS TO EXERCISES
71
distance between the two points is twice the length
of the midsegment.
7. x 10°, y 40°
8. x 60 cm
9. a 116°, c 64° 10. 100
11. x 38 cm
12. y 34 cm, z 51 cm
13. See table below.
14. a 72°, b 108°
15. a 120°, b 60°, c 60°, d 120°, e 60°,
f 30°, g 108°, m 24°, p 84°; Possible
explanation: Because c 60°, the angle that forms
a linear pair with e and its congruent adjacent
angle measures 60°. So 60° 2e 180°, and
e 60°. The triangle containing f has a 60° angle.
The other angle is a right angle because it forms a
linear pair with a right angle. So f 30° by the
Triangle Sum Conjecture. Because g is an interior
angle in an equiangular pentagon, divide 540° by 5
to get g 108°.
16. 15 stones
17. (1, 0)
18. When the swing is motionless, the seat, the bar
at the top, and the chains form a rectangle. When
you swing left to right, the rectangle changes to a
parallelogram. The opposite sides stay equal in
length, so they stay parallel. The seat and the bar
at the top are also parallel to the ground.
19. a = 60°, b = 120°
CHAPTER 5 REVIEW
1. 360° divided by the number of sides
2. Sample answers: Using an interior angle, set the
interior angle measure formula equal to the angle
and solve for n. Using an exterior angle, divide into
360°. Or find the interior angle measure and go
from there.
3. Trace both sides of the ruler as shown at right.
Answers to Exercises
4. Make a rhombus using the double-edged
straightedge, and draw a diagonal connecting the
angle vertex to the opposite vertex.
5. Sample answer: Measure the diagonals with
string to see if they are congruent and bisect each
other.
6. Sample answer: Draw a third point and connect
it with each of the two points to form two sides of a
triangle. Find the midpoints of the two sides and
connect them to construct the midsegment. The
13. (Chapter 5 Review)
Opposite sides
are parallel
Opposite sides
are congruent
Opposite angles
are congruent
Diagonals bisect
each other
Diagonals are
perpendicular
Diagonals are
congruent
Exactly one line
of symmetry
Exactly two lines
of symmetry
72
ANSWERS TO EXERCISES
Kite
Isosceles
trapezoid
Parallelogram
Rhombus
Rectangle
Square
No
No
Yes
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
Yes
No
No
Yes
No
Yes
No
Yes
No
No
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
Yes
Yes
No
20.
24. possible answers:
Resultant
vector
R
x
F
Y
900 km/h
x
F
z
R
Y
D
z
50 km/h
D
Speed: 901.4 km/h. Direction: slightly west of
north. Figure is approximate.
E
21.
R
S
25.
26.
27.
28.
1x
2
1x
2
Q
20 sides
12 cm
See flowchart below.
possible answer:
A
B
4
22. possible answers:
R
R
x
y
C
Given: Parallelogram ABCD
CD
and AD
CB
Show: AB
See flowchart below.
L
23.
N
Answers to Exercises
F
L
2
D
Y
Y
F
1
3
E
x
L
z
P
27. (Chapter 5 Review)
2
Definition of rhombus
?
ᎏ
1 DENI is
a rhombus
? Given
ᎏ
DI
?
DE ᎏ
3
NI
?
NE ᎏ
?
Definition of ᎏ
rhombus
DN
4
?
DN ᎏ
2, 4
DEN DIN
5 ? ?
ᎏ
ᎏ
? SSS
ᎏ
6 1 ?
ᎏ
?
3 ᎏ
? CPCTC
ᎏ
7
DN bisects
IDE and INE
? Definition of
ᎏ
angle bisector
? Same segment
ᎏ
28. (Chapter 5 Review)
2
AD CB
4 1 3
5
1 ABCD is a parallelogram
Given
BD BD
Same segment
3
AB CD
Definition of
parallelogram
8
6 2 4
AIA
AB CD
CPCTC
AIA
Definition of
parallelogram
7 ABD CDB
ASA
9
AD CB
CPCTC
ANSWERS TO EXERCISES
73
Answers to Exercises
CHAPTER 6 • CHAPTER
6
CHAPTER 6 • CHAPTER
LESSON 6.1
1.
3.
5.
6.
50°
30°
76 in.
possible answer:
2. 55°
4. 105°
11. Construct a line and label a point T on it. On
the line, mark off two points, L and M, each on
the same side of T at distances r and t from T,
respectively. Construct circle L with radius r.
Construct circle M with radius t.
r
L
M
T
t
Answers to Exercises
Target
7. Possible answer: The perpendicular to the
tangent passes through the center of the circle. Use
the T-square to find two diameters of the Frisbee.
The intersection of these two lines is the center.
. Construct a line through point T
8. Construct OT
.
perpendicular to OT
r
T
O
, OY
, and OZ
. Construct tangents
9. Construct OX
through points X, Y, and Z.
Z
O
t
Y
X
10. Construct tangent circles M and N. Construct
equilateral MNP. Construct circle P with radius s.
M
s
s
N
P
74
ANSWERS TO EXERCISES
12. Start with the construction from Exercise 11.
, mark off length TK so that TK s
On line LM
and K is on the opposite side of T from L and M.
Construct circle K with radius s.
r
L
M
T
t
K
s
13. Sample answer: If the three points do not lie
on the same semicircle, the tangents form a circumscribed triangle. If two of the three points are
on opposite sides of a diameter, the tangent lines to
those two points are parallel. If the points lie on the
same semicircle, they form a triangle outside the
circle, with one side touching (called an exscribed
triangle).
14. sample answer: internally tangent: wheels on a
roller-coaster car in a loop, one bubble inside
another; externally tangent: touching coins, a
snowman, a computer mouse ball and its roller
balls
15. Constructions will vary.
16. 360° 30° 90° 90° 150° and
150°
%
41.6
360°
17. Angles A and B must be right angles, but this
would make the sum of the angle measures in the
quadrilateral shown greater than 360°.
18a. rhombus
18b. rectangle
18c. kite
18d. parallelogram
19. 78°
20. 9.7 km
21. x 55° 55° 180° and 40° y y 180°, so x y 70°
11
22. 2
1
LESSON 6.2
5 cm
5 cm
15. M(4, 3), N(4, 3), O(4, 3)
16. The center of the circle is the circumcenter of
the triangle. Possible construction:
17. possible construction:
O
18. 13.8 cm
19. They can draw two chords and locate the
intersection of their perpendicular bisectors. The
radius is just over 5 km.
20. See flowchart below.
20. (Lesson 6.2)
1
AB CD
? Given
ᎏ
2
AO CO
All radii of a circle
are congruent
3
AOB COD
4
? ᎏ
?
ᎏ
? SSS Congruence
ᎏ
Conjecture
5
COD
?
AOB ᎏ
? CPCTC
ᎏ
BO DO
? All radii of a circle are congruent
ᎏ
ANSWERS TO EXERCISES
75
Answers to Exercises
1. 165°, definition of measure of an arc
2. 84°, Chord Arcs Conj.
3. 70°, Chord Central Angles Conj.
4. 8 cm, Chord Distance to Center Conj.
68°; mB 34° (Because OBC is
5. mAC
isosceles, mB mC, mB mC 68°,
and therefore mB 34°.)
6. w 115°, x 115°, y 65°; Chord Arcs
Conjecture
7. 20 cm, Perpendicular to a Chord Conj.
8. w 110°, x 48°, y 82°, z 120°; definition
of arc measure
9. x 96°, Chord Arcs Conjecture; y 96°,
Chord Central Angles Conjecture; z 42°,
Isosceles Triangle Conjecture and Triangle Sum
Conjecture.
10. x 66°, y 48°, z 66°; Corresponding
Angles Conjecture, Isosceles Triangle Conjecture,
Linear Pair Conjecture
11. The length of the chord is greater than the
length of the diameter.
12. The perpendicular bisector of the segment
does not pass through the center of the circle.
13. The longer chord is closer to the center; the
longest chord, which is the diameter, passes
through the center.
14. The central angle of the smaller circle is larger,
because the chord is closer to the center.
1
21. y 7 x; (0, 0) is a point on this line.
22a. true
C
A
X
B
Answers to Exercises
D
is the perpendicular
Possible explanation: If AB
, then every point on AB
is
bisector of CD
equidistant from endpoints C and D. Therefore
AD
and BC
BD
. Because CD
is not the
AC
, C is not equidistant
perpendicular bisector of AB
from A and B. Likewise, D is not equidistant from
and BC
are not congruent, and
A and B. So, AC
and BD
are not congruent. Thus ACBD
AD
has exactly two pairs of consecutive congruent
sides, so it is a kite.
22b. false, isosceles trapezoid
22c. false, rectangle
23. 140°, 160°, 60°; 180° minus the measure of
the intercepted arc
76
ANSWERS TO EXERCISES
24. Using the Tangent Conjecture, 2 and 4
are right angles, so m2 m4 180°.
According to the Quadrilateral Sum Conjecture,
m1 m2 m3 m4 360°, so by the
Subtraction Property of Equality, m1 m3 180°, or m1 180° m3. The
measure of a central angle equals the measure of
.
its arc, so by substitution, m1 180° mAB
25. the circumcenter of the triangle formed by
the three light switches
26. 7
27. Station Beta is closer.
Downed aircraft
N
4.6 mi
48
7.2 mi
72
38
Alpha
28. D
Beta
312
8.2 mi
LESSON 6.3
22. a 108°; b 72°; c 36°; d 108°;
e 108°; f 72°; g 108°; h 90°; l 36°;
m 18°; n 54°; p 36°
23. (2, 1); Possible method: Plot the
three points. Construct the midpoint and the
perpendicular bisector of the segments connecting
two different pairs of points. The center is the point
of intersection of the two lines. To check, construct
the circle through the three given points.
24. See flowchart below.
25. Start with an equilateral triangle whose vertices
are the centers of the three congruent circles. Then
locate the incenter/circumcenter/orthocenter/
centroid (all the same point because the triangle is
equilateral) to find the center of the larger circle. To
find the radius, construct a segment from the
incenter of the triangle through the vertex of the
triangle to a point on the circle.
It works on acute and right triangles.
20. The camera can be placed anywhere on the
major arc (measuring 268°) of a circle such that the
row of students is a chord intersecting the circle to
form a minor arc measuring 92°. This illustrates
the conjecture that inscribed angles that intercept
the same arc are congruent (Inscribed Angles
Intercepting Arcs Conjecture).
21. two congruent externally tangent circles with
half the diameter of the original circle
26. mA 60°, mB 36°, mC 90°;
60° 36° 90° 180°
24. (Lesson 6.3)
1 OR CD
2
ORC and ORD
are right angles
? Definition of
ᎏ
perpendicular lines
3
6
OCD is isosceles
7 C ? D
ᎏ
? Given
ᎏ
5
OC OD
?
ᎏ
All radii of
a circle are
congruent
? Definition of
ᎏ
isosceles triangle
mORC ⫽ 90
mORD ⫽ 90
? Definition of
ᎏ
right angle
?
ᎏ
7. Isosceles Triangle
Conjecture
4
mORC ⫽ mORD
(ORC ORD)
Substitution
?
ᎏ
property
8 OCR ?
ᎏ
?
ᎏ
ODR
SAA
9 CR ? DR
ᎏ ?
ᎏ
CPCTC
10 OR bisects CD
? Definition of bisect
ᎏ
ANSWERS TO EXERCISES
77
Answers to Exercises
1. 65°
2. 30°
3. 70°
4. 50°
5. 140°, 42° 6. 90°, 100° 7. 50°
8. 148°
9. 44°
10. 142°
11. 120°, 60°
12. 140°, 111° 13. 71°, 41°
14. 180°
15. 75°
16. The two inscribed angles intercept the same
arc, so they should be congruent.
17. BFE DFA (Vertical Angles Conjecture).
BGD FHD (all right angles congruent).
Therefore, B D (Third Angle Conjecture).
, mD 1 mEC
, AC
EC
mB 21 mAC
2
18. Possible answer: Place the corner so that it is
an inscribed angle. Trace the inscribed angle. Use
the side of the paper to construct the hypotenuse
of the right triangle (which is the diameter). Repeat
the process. The place where the two diameters
intersect is the center.
19. possible answer:
Answers to Exercises
LESSON 6.4
mABD by the
1. Proof: mACD 21mAD
Inscribed Angle Conjecture. ACD ABD.
2. Proof: By the Inscribed Angle Conjecture,
mACB 21mAD
B 21(180°) 90°. ACB is a
right angle.
3. Proof: By the Inscribed Angle Conjecture,
and mL 1mYCI
1(360° mC 21mYLI
2
2
1 mYLI ) 180° 2mYLI 180° mC. L
and C are supplementary. (A similar proof can be
used to show that I and Y are supplementary.)
2m2 4. Proof: 1 2 by AIA. mBC
2m1 mAD by the Inscribed Angle Conjecture.
AD
.
BC
5. True. Opposite angles of a parallelogram are
congruent. If it is inscribed in a circle, the opposite
angles are also supplementary. So they are right
angles, and the parallelogram is indeed
equiangular, or a rectangle.
and DB
. Diagonal RG
6. Construct diagonals RG
is the perpendicular bisector of BD by the Kite
Diagonal Bisector Conjecture. Because kite BRDG
and RG
are chords of
is inscribed in the circle, BD
the circle. By the Perpendicular Bisector of a Chord
Conjecture, the perpendicular bisector of BD
passes through the center of the circle. Because RG
is a chord of the circle that passes through the
is a
center, by the definition of diameter, RG
diameter.
, , and AR
. Because they
7. Draw in radii GR
ER , TR
TR
.
are radii of the same circle, GR
ER and AR
By the Parallel Lines Intercepted Arcs Conjecture,
ET
. Their central angles must also be
GA
congruent, so GRA ERT. Thus GRA ET
by CPCTC. GATE is
ERT by SAS, so GA
an isosceles trapezoid.
8. x 65°, y 40°, z 148°; half the measure of
the intercepted arc
9. Because the radii of a circle are congruent,
OB
, so OAB is isosceles. By the Isosceles
OA
Triangle Conjecture, 2 3, so m2 m3.
By the Triangle Sum Conjecture, m1 m2 m3 180°, so by substitution, m1 2(m2) 180°, or m1 180° 2(m2).
78
ANSWERS TO EXERCISES
, so
BC
By the Tangent Conjecture, OB
mOBC 90°. By Angle Addition, m2 m4 mOBC, so m2 m4 90°,
or m2 90° m4. Substitute 90° m4
for m2 in the earlier equation: m1 180° 2(90° m4). Simplifying gives m1 2(m4).
m1, substitution gives mAB
Because mAB
.
2(m4), or m4 21mAB
10a. S; An equilateral triangle is equiangular, but a
rhombus is not equiangular.
10b. A
10c. N; Only one diagonal is the perpendicular
bisector of the other.
10d. A
10e. S; An equilateral triangle has rotational
symmetry and three lines of symmetry; a
parallelogram has rotational symmetry but no line
of symmetry.
11. L
12. J
13. K
14. D
15. E
16. B
17. H
18. G
19. N
20. a b b a 180°, so a b 90°
2
21. 2
1
RS
. OP
OQ
OR
22. It is given that PQ
because all radii in a circle are congruent.
OS
Therefore, OPQ ORS by SSS and 2 1
by CPCTC. Now we move on to the smaller right
triangles inside OPQ and ORS. It is given that
PQ
and OV
RS
. Therefore, OTQ and
OT
OVS are right angles by the definition of
perpendicular lines and OTQ OVS because
all right angles are congruent. Thus, OTQ OV
by CPCTC.
OVS by SAA and OT
LESSON 6.5
1. d 5 cm
2. C 10 cm
12
3. r m
11
4. C 5.5 or 2
18. (Lesson 6.5)
1
2
? MA
MT
ᎏ
Given
? SA
ST
ᎏ
Given
3
MAST is a kite
? Definition of kite
ᎏ
is the perpendicular bisector of AT
MS
4 ?
ᎏ
? Kite Diagonal Bisector Conjecture
ᎏ
ANSWERS TO EXERCISES
79
Answers to Exercises
5. C 12 cm
6. d 46 m
7. C 15.7 cm
8. C 25.1 cm
9. r 7.0 m
10. C 84.8 in.
11. 565 ft
12. C 6 cm
13. 16 in.
14. Trees grow more in years with more rain;
244 yr
15. 1399 tiles
16. g 40°, n 30°, x 70°; y 142°; z 110°;
Conjecture: The measure of the angle formed by
two intersecting chords is one-half the sum of the
measures of the two intercepted arcs. In the
mLG
.
diagrams, mAEN 21mAN
and mLNG 1mLG
17. mAGN 21mAN
2
by the Inscribed Angle Conjecture. mAEN mAGN mLNG by the Triangle Exterior
mLG
Angle Conjecture. So, mAEN 21mAN
by substitution and the distributive property.
18. See flowchart below.
19. b 90°, c 42°, d 70°, e 48°, f 132°,
g 52°
5
150°
20. 360° or 12
21. The base angles of the isosceles triangle have a
measure of 39°. Because the corresponding angles
are congruent, m is parallel to n.
22. 10x 2y
LESSON 6.6
Answers to Exercises
1. 4398 km/h
2. 11 m/s
3. 37,000,000 revolutions
4. 637 revolutions
5. Mama; C 50 in.
6. 168 cm
7. d 7.6 ft. The table will fit, but the chairs may
be a little tight in a 12-by-14 ft room. 12 chairs 192 in., 12 spaces 96 in., C 288 in., d 91.7 in.
7.6 ft.
8. 0.35 ft/s
9. a 37.5°, s 17.5°, x 20°; y 80°; z 61°;
Conjecture: The measure of an angle formed by
two secants that intersect outside a circle is
one-half the difference of the larger arc measure
and the smaller arc measure.
80
ANSWERS TO EXERCISES
and mSAN 1mSN
by
10. mESA 21mEA
2
the Inscribed Angle Conjecture. mSAN mESA mECA by the Triangle Exterior Angle
1mEA
mECA by
Conjecture. So, 21mSN
2
substitution. With a little more algebra, mECA 1
mSN mEA .
2
11. Both triangles are isosceles, so the base angles
in each triangle are congruent. But one of each base
angle is part of a vertical pair. So, a b by the
Vertical Angles Conjecture and transitivity.
12. C
13. 38°
14. 48°
15. 30 cm
400 rev 2 . 2 6 ft 1 min
16. 1 min 1 rev 60 s 1089 ft/s
17. 12 cm third side 60 cm. This is based on
the Triangle Inequality Conjecture.
USING YOUR ALGEBRA SKILLS 6
50
40
Cost ($)
1
1. 2, 3
2. (3, 3)
2
3. 5, 3
4. (7, 2)
5. infinitely many solutions
6. no solution
7. 4. The lines intersect at the solution point, (7, 2).
The circumcenter is 272 , 178 . Only two
perpendicular bisectors are needed because
the third bisector intersects at the same point.
9a. Plan A: y 4x 20; Plan B: y 7x;
x 6 h 40 min, y $46.67
y
9b.
y
y 2x 16
5
30
Plan B
20 (0, 20)
10
2
5
( 7, 2)
x
10
4
6
Hours
x
The point of intersection shows when both plans
cost the same, the answer for 9a.
9c. Plan B; 721 hours, with Plan A
10. (3, 1), (0, 3), (3, 1)
y 12 x 32
–5
–10
y
y 13 x 2
x
6. The lines are parallel (the slopes are the same,
but the y-intercepts are different).
y
There is no solution.
y = –2x + 9
5
x
–5
–5
y = –2x – 1
–10
8. Possible solution using the equations
y 21x 1 and y 32x 134 :
1
y 2x 1
2
14 1
3x 3 2x 1
4x 28 3x 6
22 7x
22
x
7
1 22
y 2 7 1
11a.
y 43x 23
y x 12
11b. (6, 6)
11c. (6, 6); (6, 6)
11d. The diagonals intersect at their midpoints,
which supports the conjecture that the diagonals of
a parallelogram bisect each other.
2
13
12. y 3x 3
13. (4, 7)
3
15. 3, 2
69 6
14. 1,
4 7
16. The circumcenter is the midpoint of the
hypotenuse. For a right triangle, the perpendicular
bisectors of the legs are two of the midsegments
of the triangle. As with any pair of triangle
midsegments, they intersect at the midpoint of
the third side.
18
y 7
ANSWERS TO EXERCISES
81
Answers to Exercises
5. The lines are the same. There are infinitely many
solutions.
4
6 _23 , 46 _23
Plan A
Answers to Exercises
LESSON 6.7
4
1. 3 in.
2. 8 m
3. 14 cm
4. 9 m
5. 6 ft
6. 4 m
7. 27 in.
8. 100 cm
9. 217 m/min
10. 4200 mi
11. Desks are about 17 meters from the center.
About four desks will fit because an arc with
one-half the radius and the same central angle
will be one-half as long as the outer arc.
12. The measure of the central angle is 7.2°
because of the Corresponding Angles Conjecture.
Therefore, 500 376.20 C, so C 25,000 mi.
13. 18°/s. No, the angular velocity is measured in
degrees per second not in distance per second, so it
is the same at every point on the carousel.
14. Outer horse 2.5 m/s, inner horse 1.9 m/s.
One horse has traveled farther in the same amount of
time (tangential velocity), but both horses have
rotated the same number of times (angular velocity).
15. a 50°, b 75°, x 25°; y 45°; z 35°;
Conjecture: The measure of the angle formed by
an intersecting tangent and secant to a circle is
one-half the difference of the larger intercepted arc
measure and the smaller intercepted arc measure.
16. Let z represent the measure of the exterior
.
and tangent PB
angle of PBA formed by AB
1 By the Tangent Chord Conjecture, z 2mAB . By
.
the Inscribed Angle Conjecture, mBAP 21mBC
By the Triangle Exterior Angle Conjecture, z mBPA mBAP, or mBPA z mBAP.
82
ANSWERS TO EXERCISES
1mBC
.
By substitution, mBPA 21mAB
2
Using the distributive property, mBPA 1
(mAB mBC ).
2
17. a 70°, b 110°, c 110°, d 70°, e 20°,
f 20°, g 90°, h 70°, k 20°, m 20°,
n 20°, p 140°, r 80°, s 100°, t 80°,
u 120°
18. possible answer:
19. 170°
8
289
20. y 1x
5 15
21. 45°
22. The sum of the lengths is 8 cm for Case 1,
Case 2, Case 3, and Case 10.
23. 6
24. Yes, as long as the three points are noncollinear; possible answer: connect the points with
segments, then find the point of concurrency of
the perpendicular bisectors (same as circumcenter
construction).
CHAPTER 6 REVIEW
23. Sample answer: Construct a right angle and
and RT
with any
label the vertex R. Mark off RE
lengths. From point E, swing an arc with radius RT.
From point T, swing an arc with radius RE. Label
the intersection of the arcs as C. Construct the
and RC
. Their intersection is the
diagonals ET
center of the circumscribed circle. The circle’s
radius is the distance from the center to a vertex. It
is not possible to construct an inscribed circle in a
rectangle unless it is a square.
E
C
R
T
24. Sample answer: Construct acute angle R. Mark
off equal lengths RM and RH. From points M and
H, swing arcs of lengths equal to RM. Label the
intersection of the arcs as O. Construct RHOM. The
intersection of the diagonals is the center of the inscribed circle. Construct a perpendicular to a side to
find the radius. It is not possible to construct a circumscribed circle unless the rhombus is a square.
H
O
R
M
4
32
25. 4x 3y 32, or y 3x 3
26. (3, 2)
27. d 0.318 m
28. Melanie: 151 m/min or 9 km/h; Melody:
94 m/min or 6 km/h.
2(6357)
2(6378)
29. 1.849 1.852 1.855 360 60
360 60
30.
Possible
location
7700 ft
5280 ft
5500 ft
D
T
Possible
location
ANSWERS TO EXERCISES
83
Answers to Exercises
1. Answers will vary.
2. Draw two nonparallel chords. The intersection
of their perpendicular bisectors is the center of the
circle.
Fold the paper so that two semicircles coincide.
Repeat with two different semicircles. The center is
the intersection of the two folds.
Place the outside or inside corner of the L in the
circle so that it is an inscribed right angle. Trace the
sides of the corner. Draw the hypotenuse of the right
triangle (which is the diameter of the circle). Repeat.
The center is the intersection of the two diameters.
3. The velocity vector is always perpendicular to
the radius at the point of tangency to the object’s
circular path.
4. Sample answer: An arc measure is between 0° and
360°.An arc length is proportional to arc measure
and depends on the radius of the circle.
5. 55°
6. 65°
7. 128°
8. 118°
9. 91°
10. 66° 11. 125.7 cm 12. 42.0 cm
13. 15 cm
14. 14 ft
15. 2 57° 2 35° 180°
16. 84° 56° 56° 158° 360°
1 (180° 108°) 36° 17. mEKL 21 mEL
2
YL
by Converse of the Parallel
mKLY. KE
Lines Conjecture.
360° 56° 152° 152° mMI
.
18. mJI
1 mMI
mMJI. By the
mJMI 21 mJI
2
Converse of the Isosceles Triangle Conjecture,
JIM is isosceles.
2mKEM 140°. mKI
19. mKIM
140° 70° 70° mMI . mIKM 1 1 mMI mKI mIMK. By the Converse of the
2
2
Isosceles Triangle Conjecture, KIM is isosceles.
20. Ertha can trace the incomplete circle on paper.
She can lay the corner of the pad on the circle to
trace an inscribed right angle. Then Ertha should
mark the endpoints of the intercepted arc and use
the pad to construct the hypotenuse of the right
triangle, which is the diameter of the circle.
21. Sample answer: Construct
perpendicular bisectors of two
sides of the triangle. The point
at which they intersect (the
circumcenter) is the center of
the circle. The distance from
the circumcenter to each vertex
is the radius.
22. Sample answer: Construct the incenter (from
the angle bisectors) of the triangle. From the incenter,
which is the center of the circle, construct a perpendicular to a side. The distance from the incenter to
the foot of the perpendicular is the radius.
200
31. ft 63.7 ft
32. 8 m 25.1 m
8
33. The circumference is 346
0 2(45) 12; the
diameter is 12 cm.
34. False. 20° 20° 140° 180°. An angle with
measure 140° is obtuse.
35. true
D
36. false
C
Answers to Exercises
A
53. False. The ratio of the circumference to the
diameter is .
24 cm
24 cm
54. false; 24 24 48 48 96
48 cm
55. true
56. This is a paradox.
57. a 58°, b 61°, c 58°, d 122°, e 58°,
f 64°, g 116°, h 52°, i 64°, k 64°,
l 105°, m 105°, n 105°, p 75°, q 116°,
r 90°, s 58°, t 122°, u 105°, v 75°,
w 61°, x 29°, y 151°
58. TAR YRA by SAS, TAE YR E
by SAA
59. FTO YTO by SAA, SAS, or SSS;
FLO YLO by SAA, SAS, or SSS;
FTL YTL by SSS, SAS, or ASA
60. PTR ART by SSS or SAS;
TPA RAP by SSS, SAS, SAA, or ASA;
TLP RLA by SAA or ASA
61. ASA
84°, length of AC
11.2 35.2 in.
62. mAC
63. x 63°, y 27°, w 126°
64. sample answer:
B
37. true 38. true 39. true 40. true
41. False. (7 2) 180° 900°. It could have seven
sides.
42. False. The sum of the measures of any triangle
is 180°.
43. False. The sum of the measures of one set of
exterior angles for any polygon is 360°. The sum of
the measures of the interior angles of a triangle is
180° and of a quadrilateral is 360°. Neither is greater
than 360°, so these are two counterexamples.
44. False. The consecutive angles between the
bases are supplementary.
45. False. 48° 48° 132° 180°
46. False. Inscribed angles that intercept the same
arc are congruent.
47. False. The measure of an inscribed angle is
half the measure of the arc.
48. true
and BD
bisect each other, but AC
is
49. False. AC
.
not perpendicular to BD
D
C
A
B
48 cm
30
150
65. See table below.
66a. The circle with its contents has 3-fold rotational symmetry, the entire tile does not.
66b. No, it does not have reflectional symmetry.
67.
68. 9.375 cm
50. False. It could be isosceles.
51. False. 100° 100° 100° 60° 360°
CD
52. false; AB
C
A
69.
90
90
70.
D
B
65. (Chapter 6 Review)
84
n
1
2
3
4
5
6
f (n)
5
1
3
7
11
15
ANSWERS TO EXERCISES
...
n
...
20
. . . 9 4n . . .
71
Answers to Exercises
CHAPTER 7 • CHAPTER
7
11.
CHAPTER 7 • CHAPTER
LESSON 7.1
1. Rigid; reflected, but the size and shape do not
change.
2. Nonrigid; the shape changes.
3. Nonrigid; the size changes.
4.
5.
P
P
ᐉ
6.
,
7. possible answer: a boat moving across the water
8. possible answer: a Ferris wheel
9a. Sample answer: Fold the paper so that the
images coincide, and crease.
or
18.
19. P(a, b), Q(a, b), R(a, b)
20. possible construction:
P
9b. Construct a segment that connects
two corresponding points. Construct the
perpendicular bisector of that segment.
10a. Extend the three horizontal segments onto
the other side of the reflection line. Use your
compass to measure lengths of segments and
distances from the reflection line.
10b.
21. 50th figure: 154 (50 shaded, 104 unshaded);
nth figure: 3n 4 (n shaded, 2(n 2) unshaded)
22. 46
23. It is given that 1 2, and 2 3
because of the Vertical Angles Conjecture, so
1 3. Segment DC is congruent to itself.
DCE and DCB are both right angles, so they
are congruent. Therefore, DCB DCE by
CE
by CPCTC.
ASA, and BC
16. (Lesson 7.1)
Number of sides of
regular polygon
3
4
5
6
7
8
...
n
Number of reflectional
symmetries
3
4
5
6
7
8
...
n
Number of rotational
symmetries ( 360°)
3
4
5
6
7
8
...
n
ANSWERS TO EXERCISES
85
Answers to Exercises
12. reflectional symmetry
13. 4-fold rotational and reflectional symmetry
14. reflectional symmetry
15. 7-fold symmetry: possible answers are F or J.
9-fold symmetry: possible answers are E or H.
Basket K has 3-fold rotational symmetry but not
reflectional symmetry.
16. See table below; n, n
17.
LESSON 7.2
1.
y translation
5
x
5
6. Rules that involve x or y changing signs,
or switching places, produce reflections.
If both x and y change signs, the rule produces a
rotation. Rules that produce translations involve a
constant being added to the x and/or y terms.
5, 0 is the translation vector for Exercise 1.
7. (x, y) → (x, y)
8. (x, y) → (x, y)
9.
N
2. reflection
y
Cue ball
8
W
E
8 ball
x
–5
Answers to Exercises
S
10. There are two possible points, one on the N
wall and one on the W wall.
–8
3.
reflection
y
N
5
H
W
–4
x
7
T
y
S
11.
–5
4.
E
H''
reflection
N
5
W
E
T
x
5
H
12. by the Minimal Path Conjecture
Proposed freeway
5.
rotation
y
5
5
Mason
x
Perry
13.
14.
86
H'
S
ANSWERS TO EXERCISES
,
15. possible answer: HIKED
16. one, unless it is equilateral, in which case it
has three
17. two, unless it is a square, in which case it has
four
18.
19. sample construction:
20. sample construction:
21. false; possible counterexample: trapezoid
with two right angles
22. false; possible counterexample: isosceles
trapezoid
Answers to Exercises
ANSWERS TO EXERCISES
87
LESSON 7.3
1. 10, 10
2. A 180° rotation. If the centers of rotation differ,
rotate 180° and add a translation.
3a. 20 cm
3b. 20 cm, but in the opposite direction
4a. 80° counterclockwise
4b. 80° clockwise
5. 180°
6. 3 cm
7. possible answer:
11. Answers may vary. Possible answer: reflection
across the figure’s horizontal axis and 60°
clockwise rotation.
12.
13.
,
14. Sample answer: Draw a figure on an overhead
transparency and then project the image onto a
screen.
15. possible answers: rotational: playing card,
ceiling fan, propeller blade; reflectional: human
body, backpack
16. one: yes; two: no; three: yes
′O
′A
N ′′
A
H
′
A′
H ′′
O
N
′′
′H
O
Answers to Exercises
′N
8. possible answer:
Center of rotation
17. possible answer:
A
9.
18a.
18b.
a b
d e
10. Two reflections across intersecting lines yield
a rotation. The measure of the angle of rotation is
twice the measure of the angle between the lines of
reflection, or twice 90°, or 180°.
88
ANSWERS TO EXERCISES
O
5
12
B
9
0
12
7
14 11
?
? 11
0 13
?
?
20
c
a
f
d
2a
2b
d–e
3c
0
3b
4c
?
? ? ?
?
?
0
d
f
LESSON 7.4
1. Answers will vary. 2. Answers will vary.
3. 33.42
4. 34.6
5. 32.4.3.4
6. 3.4.6.43.42.6
7. 33.4232.4.3.4
8. 3632.4.12
9a. The dual of a square tessellation is a square
tessellation.
9b. The dual of a hexagon tessellation is a triangle
tessellation.
9c. If tessellation A is the dual of tessellation B,
then tessellation B is the dual of tessellation A.
10. The dual is a 34 38 tessellation of isosceles
right triangles.
11.
15. Answers will vary.
1
16. y 2x 4
y
4
–3
5
x
–6
17. possible answer: TOT
18.
12.
Answers to Exercises
N
8-ball
W
E
Cue ball
13. A ring of ten pentagons fits around a decagon,
and another decagon can fit into any two of the
pentagons. But another ring of pentagons around
the second decagon doesn’t leave room for a third
decagon.
14.
S
ANSWERS TO EXERCISES
89
LESSON 7.5
1. Answers will vary.
2. The dual is a 5354 tessellation.
By the Triangle Sum Conjecture, a b c 180°.
Around each point, we have 2(a b c) 2 180° 360°. Therefore, a triangle will fill the
plane edge to edge without gaps or overlaps. Thus, a
triangle can be used to create a monohedral tiling.
6. three ways
7.
8. y 2x 3
3.
y
Answers to Exercises
8
4. Yes. The four angles of the quadrilateral
will be around each point of intersection in the
tessellation.
a
a
c
c
5.
b
b
b
a
c
c
a
b a
90
b a
c
b
c b a
a
c
b
c b a
ANSWERS TO EXERCISES
c
a
c b
5
–2
x
LESSON 7.6
1.
2.
3.
4.
5.
6.
7.
Answers will vary.
Answers will vary.
Answers will vary.
regular hexagons
squares or parallelograms
squares or parallelograms
2
12. y 3x 3; the slope is the opposite sign.
y
5
–10
10
x
13. 3.4.6.4 4.6.12
440 rev 2 28 ft 1 min 1290 ft/s
14. 1 min 60 s
1 rev
8.
9. Answers will vary.
10. Answers will vary.
11.
B
E
A
S
ANSWERS TO EXERCISES
91
Answers to Exercises
15. Possible explanations:
15a. true; The kite diagonal between vertex angles
is the perpendicular bisector of the other diagonal;
in a square, diagonals would bisect each other
15b. False; it could be an isosceles trapezoid.
15c. False; it could be a rectangle.
15d. true; Parallel lines cut off congruent arcs of a
circle, so inscribed angles (the base angles of the
trapezoid) are congruent.
LESSON 7.7
1. equilateral triangles.
2. regular hexagons.
3.
Answers to Exercises
4.
9. true
10. true
11. False; it could be a kite or an isosceles
trapezoid.
12. The path would be 14 of Earth’s circumference,
approximately 6280 miles, which will take
126 hours, or around 5 14 days.
13a. Using the Reflection Line Conjecture, the
line of reflection is the perpendicular bisector of
and BB
. Because these segments are both
AA
perpendicular to the reflection line, they are
is parallel to
parallel to each other. Note that if AB
the reflection line, quadrilateral AABB will be a
rectangle instead of a trapezoid.
13b. Yes; it has reflectional symmetry, so legs and
base angles are congruent.
13c. greatest: near each of the acute vertices;
least: at the intersection of the diagonals (where A,
C, and B become collinear and A, C, and B
become collinear)
14a.
5. Answers will vary.
6. Answers will vary.
7. sample design:
8. False; they must bisect each other in a
parallelogram.
92
ANSWERS TO EXERCISES
14b.
31
5 6
0
4
128
108
28 15
0
? ? 13?
3
5
9
8 7
?
? 6 0 ?
?
9 2
2
?
1 10
29
30
50
LESSON 7.8
1. parallelograms
2. parallelograms
3.
5. Answers will vary.
6. Answers will vary.
7. Circumcenter is (3, 4); orthocenter is (10, 8).
8.
9.
4.
10.
Answers to Exercises
ANSWERS TO EXERCISES
93
USING YOUR ALGEBRA SKILLS 7
1
1. y 6x
Answers to Exercises
2. y 2x 2
3. Centroid is 2, 23; orthocenter is (0, 5).
94
ANSWERS TO EXERCISES
4. Centroid is (4, 0); orthocenter is (3, 0).
4
5. 1, 3
6. (1, 1)
7. (5, 8)
22.
CHAPTER 7 REVIEW
T
H
23. Use a grid of squares. Tessellate by translation.
24. Use a grid of equilateral triangles. Tessellate by
rotation.
25. Use a grid of parallelograms. Tessellate by
glide reflection.
26. Yes. It is a glide reflection for one pair of sides
and midpoint rotation for the other two sides.
Answers to Exercises
1. true
2. true
3. true
4. true
5. true
6. true
7. False; a regular pentagon does not create a
monohedral tessellation and a regular hexagon
does.
8. true
9. true
10. False; two counterexamples are given in
Lesson 7.5.
11. False; any hexagon with one pair of opposite
sides parallel and congruent will create a
monohedral tessellation.
12. This statement can be both true and false.
13. 6-fold rotational symmetry
14. translational symmetry
15. Reflectional; color arrangements will vary, but
the white candle must be in the middle.
16. The two towers are not the reflection (or
even the translation) of each other. Each tower
individually has bilateral symmetry. The center
portion has bilateral symmetry.
17. Answers will vary.
18. Answers will vary.
19. 3632.4.3.4; 2-uniform
20. 4.82; semiregular
1
21. y 2x
27. No.Because the shape is suitable for glide
reflection,the rows of parallelograms should
alternate the direction in which they lean (row 1
leans right,row 2 leans left,row 3 leans right,and
so on).
28.
y
x
ANSWERS TO EXERCISES
95
Answers to Exercises
CHAPTER 8 • CHAPTER
8
CHAPTER 8 • CHAPTER
LESSON 8.1
1. 228 m 2
2. 41.85 cm 2
3. 8 yd
4. 21 cm
5. 91 ft 2
6. 182 m 2
7. 96 in2
8. 210 cm
9. A 42 ft 2
10. sample answer:
6 cm
12 cm
48 cm2
Answers to Exercises
4 cm
8 cm
11.
12.
13.
14.
48 cm2
3 square units
10 square units
712 square units
sample answers:
8 cm
16 cm
64 cm2
23. 500 cm2
24a. smallest: 191.88 cm2; largest: 194.68 cm2
24b. Answers will vary. Sample answer: about
193 cm2.
24c. Answers will vary. The smallest and largest
area values differ at the ones place, so the digits
after the decimal point are insignificant compared
to the effect of the limit of precision in the
measurements.
25a. In one Ohio Star block, the sum of the red
patches is 36 in2, the sum of the blue patches is
72 in2, and the yellow patch is 36 in2.
25b. 42
25c. About 1814 in2 of red fabric, about 3629 in2
of blue fabric, and about 1814 in2 of yellow
fabric. The border requires 5580 in2 (if it does
not need the extra 20%).
26. 100; 36 64. The area of the square on the
longer side is the same as the sum of the areas on
the other two legs.
27. a 76°,b 52°,c 104°,d 52°,e 76°,
f 47°, g 90°, h 43°, k 104°, m 86°.
Explanations will vary.
28. sample construction:
64 cm2
4 cm
8 cm
15. possible answer:
16 cm
16 cm
4 cm
A
30°
M
30°
16 cm
16 cm
16. 23.1 m2
17. 2(4)(3) 2(5.5)(3) 57 m2
18. For a constant perimeter, area is maximized by
a square. 100 m 4 25 m per side; A 625 m 2.
1
19. 530
20. 112
21. 96 square units
22. 32 square units
96
ANSWERS TO EXERCISES
29a.
29c.
29b.
LESSON 8.2
1. 20 cm2
2. 49.5 m2
3. 300 square units
4. 60 cm2
5. 6 cm
6. 9 ft
7. 30 ft
8. 5 cm
9. 16 m
10. 168 cm
11. 12 cm
12. 3.6 ft; 10.8 ft
13. sample answer:
.)
17. 12(To see why, draw altitude PQ
18. more than half, because the top card
completely covers one corner of the bottom card
19a. 86 in. of balsa wood and 960 in2 of Mylar
19b. 56 in. (or less, if he tilts the kite)
20. 3600 shingles (to cover an area of 900 ft2)
21. The isosceles triangle is a right triangle
because the angles on either side of the right
angle are complementary. If you use the trapezoid
area formula, the area of the trapezoid is
1
(a b)(a b). If you add the areas of the
2
three triangles, the area of the trapezoid is
1
c 2 ab.
2
22. A b1 B
h
D
9 cm
12 cm
9 cm
14. sample answer:
4 cm
7 cm
8 cm
7 cm
10 cm
9 cm
15. sample answer:
46 cm
12 cm
45 cm
35 cm
12 cm
56 cm
16. The length of the base of the triangle equals
the sum of the lengths of both bases of the
trapezoid.
C
Given: trapezoid ABCD with height h. area
of ABD 21 hb1; area of BCD 21hb2; area
of trapezoid sum of areas of two triangles
21hb1 b2
23. 1141 square units
24. 7 square units
25. 70 m
26. 144 cm2
27. 828 ft 2; 144 ft
28. 1440 cm2; 220 cm
29a. incenter
29b. orthocenter
29c. centroid
30. a 34°, b 68°, c 68°, d 56°, e 56°,
f 90°, g 34°, h 56°, m 56°, n 90°,
p 34°. Possible explanation: Let O be the center
112° by the Inscribed Angle
of the circle. mBC
by the Central
Conjecture, and d e mBC
Angle Conjecture. OBA is congruent to OCA
is a semicircle, so
by SSS, so d e 56°. DEC
mDE 68°. By the Inscribed Angle Conjecture,
p 34°. Using OEC and the Triangle Sum
Conjecture, n 90°.
31. 32.623.6.3.6
2 cm
3 cm
7 cm
3 cm
9 cm
ANSWERS TO EXERCISES
97
Answers to Exercises
12 cm
b2
LESSON 8.3
Answers to Exercises
1a. 121,952 ft 2
1b. 244 gal of base paint and 488 gal of finishing
paint
2. He should buy at least four rolls of wallpaper.
(The area of each roll is 125 ft2. The total surface
area to be papered is 480 ft2.) If paper cut off at the
corners is wasted, he’ll need 5 rolls.
3. 1552 ft2; 776 ft 2 more surface area
4. 21
5. 336 ft2; $1780
98
ANSWERS TO EXERCISES
6. $760
7. 220 terra cotta tiles, 1107 blue tiles; $1598.15
8. 72 cm 2
9. AB 16.5 cm, BD 15.3 cm
10. 60 cm2 by either method
11. Because AOB is isosceles, mA 20° and
82°.
mAOB 140°. mA
A
B 140° and mCD
mBD
because parallel lines intercept
mAC
360° 140° 82°
69°.
congruent arcs on a circle. 2
12. E
11. a2 2ab b2
USING YOUR ALGEBRA SKILLS 8
1. x 2 6x 5
a
b
a
a2
ab
a
b
ab
b2
b
2. 2x 2 7x
x5
x
x
12. a2 b2
b
a2
ab
ab
b 2
5
x
x2
x1 x
a
13. (x 15)(x 4)
1
2x 7 x
x
15
x
x2
15x
4
4x
60
14. (x 12)(x 2)
x
12
x
x2
12x
2
2x
24
7
15. (x 5)(x 4)
3. 6x 2 19x 10
3x 2
x
x
x
x
5
x2
5x
4x
20
2
x
x
4
2x 5
16. (x 3)2 (x 3)(x 3)
x
x
4. (3)(2x 1)
5. (x 5)(x 3)
x5
3
x
3
3
x2
3x
3x
9
Answers to Exercises
5
x
17. (x 6)(x 6)
5
x
6
x2
6x
6x
36
x
2x 1
x3
x2
x
x
3
x
6
18. (2x 7)(2x 7)
1
6. (2x 3)(x 4)
x
x4
2x 3
x
3
x
4
7. x 2 26x 165
x
15
x2
15x
11x
165
x
11
9. x 2 8x 16
x
4
x
x2
4x
4
4x
16
8. 12x 2 13x 35
3x
7
4x
12x 2
28x
5
15x
35
10. 4x 2 25
2x
5
2x
4x 2
10x
5
10x
25
2x
7
2x
4x 2
14x
7
14x
49
19. x 4 or x 1
20. x 10 or x 3
21. x 3 or x 8
1
22. x 2 or x 4
h
23a and b.
h
h4
1
23c. 2[h (h 4)]h 48
23d. h 8 or h 6. The height cannot be
negative, so the only valid solution is h 6.
The height is 6 feet, one base is 6 feet, and the
other base is 10 feet.
ANSWERS TO EXERCISES
99
LESSON 8.4
2092 cm2
2. 74 cm
256 cm
4. 33 cm2
63 cm
6. 490 cm2
57.6 m
8. 25 ft
42 cm2
10. 58 cm2
1
1
1 1
11. a 2s; A 2asn 2 2s s 4 s 2
12. It is impossible to increase its area, because a
regular pentagon maximizes the area. Any
dragging of the vertices decreases the area.
(Subsequent dragging to space them out more
evenly can increase the area again, but never
beyond that of the regular pentagon.)
13. 996 cm2
14. 497 cm2
15. total surface area 13,680 in2 95 ft2;
cost $8075
16. Area is 20 square units.
Answers to Exercises
1.
3.
5.
7.
9.
17. Area is 36 square units.
y
y – _4 x 12
3
y – _1 x 6
3
(6, 4)
x
18. Conjecture: The three medians of a triangle
divide the triangle into six triangles of equal area.
Argument: Triangles 1 and 2 have equal area
because they have equal bases and the same
height. Because the centroid divides each median
into thirds, you can show that the height of
triangles 1 and 2 is 31 the height of the whole
triangle. Each has an area 61 the area of the whole
triangle. By the same argument, the other small
triangles also have areas 61 the area of the whole
triangle.
y
y _12x 5
1
(2, 6)
y 2x 10
x
100
ANSWERS TO EXERCISES
h
2
19. nw ny 2x
20. 504 cm2
21. 840 cm2
LESSON 8.5
1. 9 in2
2. 49 cm2
3. 0.8 m2
4. 3 cm
5. 3 in.
6. 0.5 m
7. 36 in2
8. 7846 m2
9. 25 48, or about 30.5 square units
10. 100 128, or about 186 square units
11.
16. A r 2 because the 100-gon almost
completely fills the circle.
17. 456 cm2
18. 36 ft2
19. The triangles have equal area when the point
is at the intersection of the two diagonals. There is
no other location at which all four triangles have
equal area.
2 24° 48°
20. x mDE
21. 90° 38° 28° 28° 180°
22.
24 cm
r 18 cm
12 cm
804 m2
11,310 km2
154 m2
4 times
18 cm
Answers to Exercises
12.
13.
14.
15.
6 cm
ANSWERS TO EXERCISES
101
LESSON 8.6
Answers to Exercises
1. 6 cm2
64
2. 3 cm2
3. 192 cm2
4. ( 2) cm2
5. (48 32) cm2
6. 33 cm2
7. 21 cm2
105
8. 2 cm2
9. 6 cm
10. 7 cm
11. 75
12. 100
13. 42
14. $448
15a.
15b.
15c.
102
ANSWERS TO EXERCISES
15d.
16. sample answer:
17a. (144 36) cm2; 78.54%
17b. (144 36) cm2; 78.54%
17c. (144 36) cm2; 78.54%
17d. (144 36) cm2; 78.54%
18. 480 m2
19. AB 17.0 cm, AG 6.6 cm
90
20. True. If 24 360 2r, then r 48 cm.
360
21. True. If n 24, then n 15.
22. False. It could be a rhombus.
23. true; Triangle Inequality Conjecture
LESSON 8.7
1. 150 cm2
2. 4070 cm2
3. 216 cm2
4. 340 cm2
5. 103.7 cm2
6. 1187.5 cm2
7. 1604.4 cm2
8. 1040 cm2
2
9. 414.7 cm
10. 329.1 cm2
11. area of square 4 area of trapezoid 4 area of triangle
12. $1570
13. sample answer:
15. a 75°, b 75°, c 30°, d 60°,
e 150°, f 30°
16. About 23 days. Each sector is about 1.767 km2.
17. a 50°, b 50°, c 80°, d 100°, e 80°,
f 100°, g 80°, h 80°, k 80°, m 20°,
n 80°. Explanations will vary. Sample
explanation: The angle with measure d corresponds
to the angle forming a linear pair with g. Because
d 100°, by the Parallel Lines Conjecture, the
angle adjacent to g measures 100°, and by the
Linear Pair Conjecture, g 80°. The angle with
measure f corresponds to the angle measuring
100°, so f 100°. The angles measuring g and k
are the base angles of an isosceles triangle, so by
the Isosceles Triangle Conjecture, k 80°.
18. 398 square units
Answers to Exercises
14. sample tiling
33.42/32.4.3.4/44
ANSWERS TO EXERCISES
103
CHAPTER 8 REVIEW
1. B (parallelogram)
3. C (trapezoid)
5. F (regular polygon)
7. J (sector)
9. G (cylinder)
11.
2. A (triangle)
4. E (kite)
6. D (circle)
8. I (annulus)
10. H (cone)
12.
20.
23.
26.
29.
32.
32 cm
21. 32 cm
81 cm2
24. 48 cm
153.9 cm2
27. 72 cm2
300 cm2
30. 940 cm2
Area is 112 square units.
22.
25.
28.
31.
15 cm
40°
30.9 cm2
1356 cm2
y
Apothem
D (6, 8)
A (0, 0)
C (20, 8)
B (14, 0)
x
33. Area is 81 square units.
13.
y
R (4, 15)
U (9, 5)
Answers to Exercises
F (0, 0)
14. Sample answer: Construct an altitude from
the vertex of an obtuse angle to the base. Cut off
the right triangle and move it to the opposite side,
forming a rectangle. Because the parallelogram’s
area hasn’t changed, its area equals the area of the
rectangle. Because the area of the rectangle is
given by the formula A bh, the area of the
parallelogram is also given by A bh.
b
O (4, –3)
34. 6 cm
35. 172.5 cm2
36. sample answers:
b
h
h
15. Sample answer: Make a copy of the trapezoid
and put the two copies together to form a
parallelogram with base b1 b2 and height h.
Thus the area of one trapezoid is given by the
formula A 12 b1 b2h.
b2
b1
h
h
b1
b2
16. Sample answer:Cut a circular region into a large
number of wedges and arrange them into a shape
that resembles a rectangle.The base length of this
“rectangle”is r and the height is r, so its area is r 2.
ThustheareaofacircleisgivenbytheformulaAr 2.
r
␲r
18. 5990.4 cm2
17. 800 cm2
19. 60 cm2 or about 188.5 cm2
104
x
ANSWERS TO EXERCISES
37. 1250 m2
38. Circle. For the square, 100 4s, s 25,
A 252 625 ft2. For the circle, 100 2r,
r 15.9, A (15.9)2 794 ft2.
39. A round peg in a square hole is a better fit.
The round peg fills about 78.5% of area of the
square hole, whereas the square peg fills only about
63.7% of the area of the round hole.
40. giant
41. about 14 oz
42. One-eighth of a 12-inch diameter pie;
one-fourth of a 6-inch pie and one-eighth of a
12-inch pie both have the same length of crust,
which is longer than one-sixth of an 8-inch pie.
43a. 96 ft; 40 ft
43b. 3290 ft2
44. $3000
45. $4160
46. It’s a bad deal. 2r1 44 cm. 2r2 22 cm,
which implies 4r2 44 cm. Therefore r1 2r2.
The area of the large bundle is 4r22 cm2. The
combined area of two small bundles is 2r22 cm2.
Thus he is getting half as much for the same price.
47. $2002
48. $384 (16 gal)
Answers to Exercises
CHAPTER 9 • CHAPTER
9
CHAPTER 9 • CHAPTER
LESSON 9.1
b
a
b a
c
18. Sample answer: Yes, ABC XYZ by SSS.
Both triangles are right triangles, so you can
use the Pythagorean Theorem to find that
CB ZY 3 cm.
m
120°
p
q
n
Or use the Exterior Angle Conjecture to get
q n 120°. By AIA, q m. Substituting,
m n 120°.
22. a 122°, b 74°, c 106°, d 16°, e 90°,
f 74°, g 74°, h 74°, n 74°, r 32°,
s 74°, t 74°, u 32°, v 74°. Possible
explanation: By the Tangent Conjecture, the
quadrilateral containing g has two right angles
formed by radii intersecting tangent lines. Using
c 106° and the Quadrilateral Sum Conjecture,
g 90° 90° 106° 360°, so g 74°.
The angle measures e and f and the measure of the
inscribed angle that intercepts the arc with measure
u sum to 180°. Using e 90° and f 74°, the
inscribed angle measures 16°. Using the Inscribed
Angle Conjecture, u 32°.
T1
23 .
T2
ANSWERS TO EXERCISES
105
Answers to Exercises
1. c 19.2 cm
2. a 12 cm
3. b 5.3 cm
4. d 10 cm
5. s 26 cm
6. c 8.5 cm
7. b 24 cm
8. x 3.6 cm
9. x 40 cm
10. s 3.5 cm
11. r 13 cm
12. 127 ft
13. 512 m2
14. 11.3 cm
15. 3, 4, 5
16. 28 m
17. The area of the large square is 4 area of
triangle area of small square.
c 2 4 21ab (b a)2
c 2 2ab b 2 2ab a 2
c 2 a2 b2
19. 54 cm2
20. 3632.4.3.4
21. Mark the unnamed angles as shown in the
figure below. By the Linear Pair Conjecture,
p 120° 180°, so p 60°. By AIA, m q. By
the Triangle Sum Conjecture, q p n 180°.
Substitute m q and p 60° to get
m 60° n 180°. m n 120°.
Answers to Exercises
LESSON 9.2
1. yes
2. yes
3. no
4. no
5. no
6. no
7. no
8. No, the given lengths are not a Pythagorean
triple.
9. y 25 cm
10. y 24 units
11. y 17.3 m
12. 6, 8, 10
13. 60 cm2
14. 14.1 ft
15. 17.9 cm2
16. 102 m
17a. 1442 cm2
17b. 74.8 cm
18. Sample answer: The numbers given satisfy the
Pythagorean Theorem, so the triangle is a right
triangle; but the right angle should be inscribed
in an arc of 180°. Thus the triangle is not a right
triangle.
106
ANSWERS TO EXERCISES
19. Sample answer: (BD)2 62 32 27;
(BC)2 (BD)2 92 108; then (AB)2 (BC)2 (AC)2 (36 108 144), so ABC is a right
triangle by the Converse of the Pythagorean
Theorem.
20. centroid
3
21. 1
0
22. Because mDCF 90°, mDCE (90 x)°.
Because DCE is isosceles, mDEC (90 x)°.
mD 180° 2 (90 x)° 2x. Because D is a
central angle, 2x a. Therefore, x 21a.
23. The path from C to M to T lies on a straight
line and therefore must be shorter than the path
from C to A to T.
P
A
T
M
C
U
24. 790 square units
25.
26. 19
B
USING YOUR ALGEBRA SKILLS 9
1. 6
2. 5
3. 18
2
4. 147
5. 8
6. 2
3
7. 3
2
8. 2
10
9. 5
3
10. 85
11. 4
6
12. 24
13. 12
5
14. 28
15. 6
23
16. Answers will vary. Possible answer: The length
of the hypotenuse of an isosceles right triangle
with legs of length 3 units is 18 units. This is the
same as 2 2 2 , or 3
2.
18. Possible answer: A right triangle with legs of
lengths 1 and 3 units has a hypotenuse of length
units.
10
10
1
3
A right triangle with legs of lengths 6 and 2 units
has a hypotenuse of length 40
10
10
10 .
2
40
2
6
19. Possible answer: A right triangle with legs of
lengths 2 and 3 units has a hypotenuse of length
13 units.
1
2
1
18
3
13
3
3
5
2
12 . Because y is twice as
20. x 3, y long as x, y 2x, so 12 2
3.
2
2
2
21. 2 3 7
7
3
1
2
2
20
2
4
ANSWERS TO EXERCISES
107
Answers to Exercises
17. The hypotenuse represents 5 units. In the
second right triangle, the legs have lengths 4 and 2,
so the hypotenuse has length 20 . The length of
the hypotenuse is twice the length of the hypotenuse
of the smaller triangle, so 20
5.
2
LESSON 9.3
72
2 cm
13 cm
10 cm, 5
3 cm
10
3 cm, 10 cm
34 cm, 17 cm
72 cm
12
3 cm
50 cm, 100 cm
16 cm2
1
1
10. , 2 2
1.
2.
3.
4.
5.
6.
7.
8.
9.
30°
x
3
2x
45°
11. A 30°-60°-90° triangle must have sides whose
lengths are multiples of 1, 2, and 3 . The triangle
shown does not reflect this rule.
12. Possible answer:
Use 12 3 2 22 and 42 48 2 82.
Answers to Exercises
16. c 2 x 2 x 2 Start with the Pythagorean Theorem.
c 2 2x 2
Combine like terms.
c x
2
Take the square root of both sides.
17. 169
3 m2 292.7 m2
18. 390 m2
19.
3
1
2
3
x
2
x
15°
45°
x
20. Construct an isosceles right triangle with legs
of length a, construct a 30°-60°-90° triangle with
legs of lengths a and a 3 , and construct a right
triangle with legs of lengths a 2 and a 3.
1
2
4
3
3
8
a
1
2
2
3
4
13. possible answer:
4
4
14. Possible answer:
Use 22 12 5 2 and 62 32 45 2.
5
a 3
1
45
5
3
4
3
6
15a. CDA, AEC, AEB, BFA, BFC
15b. MDB, MEB, MEC, MFC, MFA
108
a 5
a 2
21.
8 12.5; that is, the sum of the areas of the
semicircles on the two legs is equal to the area of
the semicircle on the hypotenuse.
73
22. 6 12.2 chih
23. Extend the rays that form the right angle.
m4 m5 180° by the Linear Pair
Conjecture, and it’s given that m5 90°.
m4 90°. m2 m3 m4 m2 m3 90° 180°. m2 m3 90°.
m3 m1 by AIA. m1 m2 90°.
1
2
a 3
Areas: 4.5 cm2, 8 cm2, 12.5 cm2. 4.5 1
2
2a
a
a
3
4
a 2
ANSWERS TO EXERCISES
24. 80°
2
LESSON 9.4
1. No. The space diagonal of the box is about
33.5 in.
2. 10 m
3. 50 km/h
4. 8 ft
5. area: 60 m2; cost: $7200
6. surface area of prism 27
3 180 cm2 2
226.8 cm ; surface area of cylinder 78 cm2 245.0 cm2
36 cm 20.8 cm
7. 3
8. 48.2 ft; 16.6 lb
9a. 160 ft-lb
9b. 40 lb
9c. 20 lb
10. about 4.6 ft
2 2
2 2
11.
2
2
4
2
2
2
2
2
12.
13. 12 units
14. 18
2 cm
and CB
. ABC ADC 15. Draw radii CD
90°. For quadrilateral ABCD, 54° 90° mC 126° but
90° 360°, so mC 126°. BD
126° 226° 360°.
16. 4, 4
3
17. SAA
18. orthocenter
19. 115°
20. PO
2
4
2
2
2
4
4
Answers to Exercises
2
2
ANSWERS TO EXERCISES
109
LESSON 9.5
1. 5 units
4. 354 m
7. rectangle
2. 45 units
5. 52.4 units
3. 34 units
6. isosceles
y
10
A
8
B
6
4
2
D
2
C
4
6
x
8 10
8. parallelogram
E2
4
x
F
H
8
Answers to Exercises
6
G
10
9. kite
y
L
4
2
K
8
x
I 2
2
J
10. square
y
120°
8 P
M
3 1
19. 2, 2
20. k 2 , m 6
6
12
21. x , y 3
3
22. 96 cm
23. The angle of rotation is approximately 77°.
Connect two pairs of corresponding points.
Construct the perpendicular bisector of each
segment. The point where the perpendicular
bisectors meet is the center of rotation.
24. Any long diagonal of a regular hexagon
divides it into two congruent quadrilaterals. Each
(6 2) 180°
angle of a regular hexagon is 120°, and
6
the diagonal divides two of the 120° angles into 60°
angles. Look at the diagonal as a transversal.
The alternate interior angles are congruent, thus
the opposite sides of a regular hexagon are parallel.
y
2
2
11b. Circle A: (x 1)2 ( y 2)2 64;
Circle B: x 2 (y 2)2 36
11c. (x h)2 (y k)2 (r)2
12. (x 2)2 y 2 25
13. Center is (0, 1), r 9.
14. (x 3)2 (y 1)2 18
15a. 14 units
15b. 176 4
11 units
2 (y y )2 (z z )2
15c. (x
x
)
2
1
2
1
2
1
16. 86.5 units
17. 14 units
18. n (n 2) 3 (n 3) (n 1)
60°
60°
120°
4
120°
O
N
4 2
11a.
2
60°
60°
x
4
y
y
(x, y)
(1, –2)
x
(x, y)
Circle A
110
ANSWERS TO EXERCISES
(0, 2)
x
Circle B
120°
LESSON 9.6
18 cm2 56.5 cm2
(8 16) m2 9.1 m2
456 cm2 1433 cm2
120 cm2 377.0 cm2
32 32
3 m2 45.1 m2
(25 48) cm2 30.5 cm2
64
7. 3 16
3 cm2 39.3 cm2
8. (240 18) m2 183.5 m2
9. 102 cm
10. Possible proof:
and AN
Given: Circle C with tangents AM
A
Show: AM
N
1.
2.
3.
4.
5.
6.
14. 12 6
3 cm 22.4 cm
15. 77 cm 8.8 cm
16. 76 cm
17. Inscribed circle: 3 cm2. Circumscribed circle:
12 cm2. The area of the circumscribed circle is four
times as great as the area of the inscribed circle.
18. 135°
19. (x 3)2 (y 3)2 36
20. The diameter is the transversal, and the
chords are parallel by the Converse of the Parallel
Lines Conjecture. The chords are congruent
because they can be shown to be the same distance
from the center. (Draw a perpendicular from each
chord to the center and use AAS and CPCTC.)
Sample construction:
M
C
A
, NC
, and AC
to the diagram as shown.
Add MC
By the Tangent Conjecture, M and N are right
angles, so AMC and ANC are right triangles.
Using the Pythagorean Theorem, (AM)2 (MC)2
(AC)2 and (AN)2 (NC)2 (AC)2. Solving for
AM and AN, AM (AC)2 (MC)2 and
2
2
and NC
are
AN (AC) (NC) . Because MC
radii of the same circle, they have the same lengths.
So, substitute MC for NC in the second equation:
AN (AC)2 (MC)2 AM. Therefore,
AN
.
AM
11. 18 m
12. 324 cm2 1018 cm2
13. 3931 cm2
21a. Because a carpenter’s square has a right angle
and both radii are perpendicular to the tangents, a
square is formed. The radius is 10 in., therefore the
diameter is 20 in.
10 in.
10
in.
10 in.
d 20 in.
10 in.
21b. Possible answer: Measure the circumference
with string and divide by .
5
22. 6
ANSWERS TO EXERCISES
111
Answers to Exercises
N
CHAPTER 9 REVIEW
1.
2.
3.
4.
20 cm
10 cm
obtuse
26 cm
3 1
5. 2, 2
6. 1, 1
2
2
7. 200
3 cm2 346.4 cm2
8. d 12
2 cm 17.0 cm2
9. 246 cm2
10. 72 in2 226.2 in2
11. 24 cm2 75.4 cm2
12. (2 4) cm2 2.28 cm2
13. 222.8 cm2
14. isosceles right
15. No. The closest she can come to camp is 10 km.
16. No. The 15 cm diagonal is the longer diagonal.
17. 1.4 km; 821 min
18. yes
19. 29 ft
20. 45 ft
21. 50 mi
22. 707 m2
23. 6
3 and 18
24. 12 m
25. 42
26. No. If you reflect one of the right triangles into
the center piece, you’ll see that the area of the kite is
almost half again as large as the area of each of the
other triangles.
Answers to Exercises
7
27. 9
28. The quarter-circle gives the maximum area.
Triangle:
s
____
2
45°
s
45°
s
____
2
s
1
A 2 2
s
2
s
4
2
Square:
_1 s
2
_1 s
2
1 1
s2
A 2s 2s 4
Quarter-circle:
s
2s
___
␲
1
s 4 2r
2s
r 1 2s 2 s 2
A 4 s 2 s2
4
Extra
30°
29. 1.6 m
30.
30°
30°
Or students might compare areas by assuming the
short leg of the 30°-60°-90° triangle is 1. The area
3
of each triangle is then 0.87 and the area of
2
the kite is 3 3 1.27.
112
ANSWERS TO EXERCISES
31. 4; 0; 10. The rule is n2 if n is even, but 0 if n is
odd.
32. 4 in./s 12.6 in./s
33. true
34. true
35. False. The hypotenuse is of length x
2.
36. true
2
37. false; AB x
(x 2 (y2 y1) 2
1) 38. False. A glide reflection is a combination of a
translation and a reflection.
39. False. Equilateral triangles, squares, and regular
hexagons can be used to create monohedral
tessellations.
40. true
41. D
42. B
43. A
44. C
45. C
46. D
47. See flowchart below.
48.
N
8-ball
W
B
E
Cue ball
A
S
49. 34 cm2; 22 4
2 27.7 cm
40
50. 3 cm2 41.9 cm2
51. about 55.9 m
52. about 61.5 cm2
53. 48 cm
54. 322 ft2
Answers to Exercises
47. (Chapter 9 Review)
1
ABCD is
a rectangle
Given
2
ABCD is
a parallelogram
Definition of
rectangle
4
D B
Definition of
rectangle
3 DA CB
Definition of
parallelogram
5
DAC BCA
AIA Conjecture
7
ABC CDA
SAA Congruence
Conjecture
6 AC AC
Same segment
ANSWERS TO EXERCISES
113
Answers to Exercises
CHAPTER 10 • CHAPTER
10 CHAPTER 10 • CHAPTER
24.
Answers to Exercises
LESSON 10.1
1. polyhedron; polygonal; triangles
2. PQR, TUS
3. PQUT, QRSU, RPTS
, PT
, RS
4. QU
5. 6 cm
6. GYPTAN
7. point E
, YE
, PE
, TE
, AE
, NE
8. GE
9. 13 cm
10. D
11. L
12. C
13. G
14. B
15. H
16. E
17. A
18. J
19. J
20. M
21. H
22. I
23.
25.
x
2x
26.
x
2x
27. true
28. False. This statement is true only for a right
prism.
29. true
30. true
31. False. It is a sector of a circle.
32. true
33. false; counterexample:
34. true
35. true
114
ANSWERS TO EXERCISES
39. 60
40. 30
41a. yes
36. See table below.
Possible answers include that the number of lateral
faces of an antiprism is always twice the number
for the related prism; that the number of vertices is
the same for each related prism and antiprism; and
that the number of edges for a prism is three times
the number of faces, while for an antiprism, the
number of edges is twice the number of faces.
37. Answer should include the idea that the
painting “disappears” into the view out the window.
Students might also note the effect created by the
cone-shaped tower appearing similar to the road
disappearing into the distance.
38. 8
␣
␣
41b. yes
41c. no
41d. yes
⫻
⫻
Answers to Exercises
36. (Lesson 10.1)
Triangular
prism
Rectangular
prism
Pentagonal
prism
Hexagonal
prism
Lateral
faces
3
4
5
6
...
n
Total
faces
5
6
7
8
...
n2
Edges
9
12
15
18
...
3n
Vertices
6
8
10
12
...
2n
Triangular
antiprism
Rectangular
antiprism
Pentagonal
antiprism
Hexagonal
antiprism
Lateral
faces
6
8
10
12
...
2n
Total
faces
8
10
12
14
...
2n 2
Edges
12
16
20
24
...
4n
Vertices
6
8
10
12
...
2n
n-gonal
prism
n-gonal
antiprism
ANSWERS TO EXERCISES
115
20.
LESSON 10.2
1.
4.
5.
6.
7.
8.
72 cm3
2.
24 cm3
3.
108 cm3
r
r
160 cm3 502.65 cm3
r
36 cm3 113.10 cm3
324 cm3 1017.88 cm3
See table below.
960 in3
9. QT cubic units
21. true
22. false
12 in.
Q
4 in.
T
23.
24 in.
8 in.
Answers to Exercises
10. sample answer:
24. possible solution: prism
12
12
12
8
3
2
12. 3r 3
13. 13x 3
11. 2x 3
14. Margaretta has room for 0.5625 cord. She
should order a half cord.
15. 170 yd3
16. 5100 lb
17. 11,140
18. The volume of the quilt in 1996 was 4000 ft3.
The quilt panels were stacked 2 ft 8 in. high.
19.
Salt crystal
25. approximately 1.89 m
26. 12 6
2
7. (Lesson 10.2)
Information about
base of solid
Height
of solid
Right triangular
prism
Right rectangular
prism
h
H
116
Right trapezoidal
prism
b2
h
b
Right
cylinder
b
H
r
h
H
b
H
b 6, b2 7,
h 8, r 3
H 20
a. V 480 cm3
d. V 960 cm3
g. V 1040 cm3
j. V 180 cm3
b 9, b2 12,
h 12, r 6
H 20
b. V 1080 cm3
e. V 2160 cm3
h. V 2520 cm3
k. V 720 cm3
b 8, b2 19,
h 18, r 8
H 23
c. V 1656 cm3
f. V 3312 cm3
i. V 5589 cm3
l. V 1472 cm3
ANSWERS TO EXERCISES
14. 78,375 grams
15. 48 in3
16. 4 units3
17. 144x 3 cm3
18. 40,200 gal; 44 h 40 min
19. 71 ft 3
20. 403 barrels
21a. 16
3 cm2
21b. 96 cm2
21c. 80
3 cm2
21d. 24
13 120 cm2
A
22.
D
LESSON 10.3
192 cm3
84 cm3 263.9 cm3
150 cm3
60 cm3
84 cm3 263.9 cm3
384 cm3 1206 cm3
m3
7. 3 cm3
2
8. 3 b 3 cm3
9. 324x 3 cm3; 29.6%
10. See table below.
1
11. V 3M 2H ft 3
1.
2.
3.
4.
5.
6.
D'
3 1
2
C
B
4
H
M
D''
Possible answer: From the properties of reflection,
1 3 and 2 4. m1 m2 90°, so
m3 m4 90°, and m1 m2 m3 m4 180°. Therefore D, C, and D are collinear.
23a. Y (a c, d)
23b. Y (a c, b d)
23c. Y (a c e, b d f )
12. sample answer:
27
16
3
48
13. Mount Etna is larger. The volume for Mount
Etna is approximately 2193 km3, and the volume
for Mount Fuji is approximately 169 km3.
10. (Lesson 10.3)
Information about
base of solid
Height
of solid
Triangular
pyramid
Trapezoidal
pyramid
H
H
b
Rectangular
pyramid
H
h
b2
H
r
h
h
b
Cone
b
b 6, b2 7,
h 6, r 3
H 20
a. V 120 cm3
d. V 240 cm3
g. V 260 cm3
j. V 60 cm3
b 9, b2 22,
h 8, r 6
H 20
b. V 240 cm3
e. V 480 cm3
h. V k. V 240 cm3
b 13, b2 29,
h 17, r 8
H 24
c. V 884 cm3
f. V 1768 cm3
i. V 2856 cm3
l. V 512 cm3
2480
cm3
3
ANSWERS TO EXERCISES
117
Answers to Exercises
M
LESSON 10.4
Answers to Exercises
1.
2.
3.
4.
5.
6.
58.5 in3
32
3 cm3 55.43 cm3
15 cm
11 cm
5.0 cm
8.5 in. 2
3
2 11 in. 63.24 in ;
11 in. 2
3
2 8.5 in. 81.85 in
The short, fat cylinder has greater volume.
7. 257 ft 3
8. 4 cm
9. He must refute the statement.
10. 1502 lb
11. 192.4 gal
12. 13 min
13. Answers will vary, but r 2H should equal
about 14.4 in3.
14. approximately 38 in3
15. 100,000 m 3, or about 314,159 m3;
16,528 loads
EC
because the opposite sides of a
16. AB
BD
because
parallelogram are congruent. EC
the diagonals of a rectangle are congruent. So,
BD
because both are congruent to EC
.
AB
Therefore, ABD is isosceles.
118
ANSWERS TO EXERCISES
17. 8.2 cm
18. x 96°
19.
20.
x
D
45°
x
C
x
45°
45°
21a.
21b.
21c.
21d.
21e.
45°
x
A
A
S
N
S
A
x
3x
x
B
LESSON 10.5
1. 675 cm3
2. 36 cm3 113.1 cm3
3. 47 in3
4. 1798.4 g
5. The gold has mass 2728.5 g, and the platinum
has mass 7529.8 g. The solid cone of platinum has
more mass.
6. 1.5 cm
7. 10.5 g/cm3; silver
8. 8000 cm3
9. The volume of the medallion is 160 cm3.Yes, it
is gold, and the Colonel is who he says he is.
10. 679 cm3
11. approximately 193 lb; 22 fish
9
12. 2 3
13. flowchart proof:
SP
QR
Given
R S
MR
SM
Given
QMR PMS
Vertical Angles
AIA
RMQ SMP
ASA
MP
MQ
CPCTC
M is the midpoint of PQ
Definition of midpoint
14. 15 sides
15a. (1, 3)
15b. (x 1)2 (y 3)2 25
16. 58; 3n 2
Answers to Exercises
ANSWERS TO EXERCISES
119
Answers to Exercises
LESSON 10.6
1. 36 cm3 113.1 cm3
2. 6 cm3 0.533 cm3
9
3
3
3. 3
2 cm 0.884 cm
4. 720 cm3 2262 cm3
5. 30 cm3 94.3 cm3
6. 3456 cm3 10,857 cm3
7. 18 m3 56.6 cm3
8. No. The volume of the ice cream is 85.3 cm3,
and the volume of the cone is 64 cm3.
9. only 20 scoops
148
10. 3 m3, or about 155 m3
11. They have the same volume.
12. 9 in.
13. 3 cm
8192
14. 3 cm3 8579 cm3
15. 18 in3, or about 57 in 3
16. No. The unused volume is 16 cm3, and the
volume of the golf ball is 10.6 cm3.
120
ANSWERS TO EXERCISES
17. approximately 15,704 gallons; 53 days
18. lithium
19. 31 ft
20. ABCD is a parallelogram because the slopes of
and AB
are both 0 and the slopes of BC
and
CD
are both 35.
AD
y
C (3, 5) D (9, 5)
x
B (–3, –5)
A (3, –5)
21. They trace two similar shapes, except that the
one traced by C is smaller by a scale factor of 1:2.
22. The line traces an infinite hourglass shape. Or,
it traces the region between the two branches of a
hyperbola.
23. w 110°, x 115°, y 80°
1. V 972 cm3 3054 cm3
S 324 cm2 1018 cm2
2. V 0.972 cm3 3.054 cm3
S 3.24 cm2 10.18 cm2
3. V 1152 cm3 3619 cm 3
S 432 cm2 1357 cm2
4. S 160 cm2 502.7 cm2
256
5. V 3 cm3 268.1 cm3
6. S 144 cm2 452.4 cm2
7. Area of great circle r 2. Total surface area
of hemisphere 3r 2. Total surface area of
hemisphere is three times that of area of great
circle.
8. 2 gal
9. V 21 34(1.8)3 21(1.8) 2 (4.0) 10.368 m3 32.57 m3; S 21 4(1.8) 2 1
2(1.8)(4.0) 13.68 m2 42.98 m 2
2
10a. approximately 3082 ft2
10b. 13 gal
10c. approximately 9568 bushels
11. 153,200,000 km2
12. The total cost is $131.95. He will stay under
budget.
13. 1.13%
14. 150 cm3 471.2 cm3
1
15. 4
19a. (Lesson 10.7)
n
f(n)
1
2
3
4
5
6
...
n
...
200
2
1
4
7
10
13
. . . 3n 5 . . .
595
19b. (Lesson 10.7)
n
1
2
3
4
5
6
...
n
...
200
f(n)
0
1
3
1
2
3
5
2
3
5
7
n1
... n1 . . .
199
201
ANSWERS TO EXERCISES
121
Answers to Exercises
1
16. 2
3
17. 4
18. The ratio gets closer to 1.
19a and 19b. See tables below.
CB
and AD
CD
by the definition of
20. AB
rhombus, and BD BD because it is the same
segment; therefore ABD CBD by SSS. By
CPCTC, 2 3 and 1 4, which shows
bisects both ABC and ADC. Because
that BD
all four sides of the rhombus are congruent, a
similar proof can be used to show that ABC ADC and thus that A and C are both
.
bisected by diagonal AC
CB
by the definition of rhombus and
21a. AB
BE
because it is the same segment. 1 2
BE
by the Rhombus Angles Conjecture. Therefore,
AEB CEB by SAS.
CE
by CPCTC, so BD
bisects AC
.
21b. AE
21c. 3 4 by CPCTC. Also, 3 and 4 form
a linear pair, so they are supplementary. Because
two angles that are congruent and supplementary
are right angles, 3 and 4 are right angles.
21d. Because 3 and 4 are right angles, the
diagonals are perpendicular.You still need to show
bisects BD
. Use a proof similar to that
that AC
given in 21a to show that AEB AED. Then,
DE
, which shows that AC
by CPCTC, BE
.
bisects BD
LESSON 10.7
USING YOUR ALGEBRA SKILLS 10
A
1. h b
P 2h
P
2. b 2 or b 2 h
Answers to Exercises
3. r 3V
H
2 a2
4. b c
2 • SA
5. a l
P
6. y2 mx 2 x1 y1 or y2 mx 2 mx 1 y1
d
7. v t
The original formula gives distance in terms of
velocity and time.
9
8. F 5C 32
The original formula converts degrees Fahrenheit to
degrees Celsius.
T 2
9. L g 2
The original formula gives the period of a
pendulum (time of one complete swing) in terms
of length and acceleration due to gravity.
10a. V F E 2
10b. See table below.
11a. The corresponding radii are approximately
3.63 cm, 3.30 cm, 3.02 cm, and 2.79 cm.
11b. The corresponding heights are approximately
9.32 cm,10.49 cm,11.61 cm,and 12.70 cm.
11c. The corresponding volumes are approximately
129 cm 3, 120 cm 3, 111 cm 3, and 104 cm 3.
11d. Answers will vary. Sample answer: The cone
with slant height 10 cm has the widest radius, so a
scoop of ice cream is least likely to fall off, and that
cone also has the greatest volume.
11e. Answers will vary. Sample answer: The cone
with slant height 13 cm has the greatest height, so
the cone appears bigger even though it has the same
surface area as the other cones; that cone also has
the smallest radius and smallest volume, so it could
hold less ice cream and still appear to be a bigger
cone.
m1 m2 m3 f f
12a. A 5
12b. average of 60: 31; average of 70: 56; average
of 80: 81; average of 90: 106 (impossible)
10b. (Using Your Algebra Skills 10)
Pentahedron
Hexahedron
Octahedron
Decahedron
Dodecahedron
Number of faces
5
6
8
10
12
Number of edges
8
10
12
16
20
Number of vertices
5
6
6
8
10
122
ANSWERS TO EXERCISES
CHAPTER 10 REVIEW
27.
3
3
m3 0.23 m3
8 32 28. 160 cubic units
Answers to Exercises
1. They have the same formula for volume: V BH.
2. They have the same formula for volume: V 31BH.
3. 6240 cm3
4. 1029 cm3 3233 cm3
5. 1200 cm3
6. 32 cm3
7. 100 cm3 314.2 cm3
8. 2250 cm3 7069 cm3
9. H 12.8 cm
10. h 7 cm
11. r 12 cm
12. r 8 cm
13. 960 cm3
14. 9 m
15. 851 cm3
16. four times as great
17a. Vextra large 54 in3
Vjumbo 201.1 in3
Vcolossal 785.4 in3
17b. 14.5 times as great
18. Cylinder B weighs 38 times as much as cylinder A.
19. 2129 kg; 9 loads
Vsphere 34 r 3
20. H2r. (2r)3 0.524. Thus, 52.4% of
Vbox
the box is filled by the ball.
21. approximately 358 yd3
22. No. The unused volume is 98 in3, and the
volume of the meatballs is 32 in3.
23. platinum
24. No. The ball weighs 253 lb.
25. 256 lb
26. approximately 3 in.
ANSWERS TO EXERCISES
123
Answers to Exercises
CHAPTER 11 • CHAPTER
11 CHAPTER 11 • CHAPTER
Answers to Exercises
USING YOUR ALGEBRA SKILLS 11
3 3
1. 8; 5
AC 3 CD 5 BD
8
, , 2. CD 5 BD 8 BC 13
3
3a. 1
9
3b. 1
4. a 6
5. b 16
6. c 39
7. x 5.6
8. y 8
9. x 12
10. z 6
11. d 1
12. y 5
13. 318 mi
14. 2.01
15. 12 ft by 15 ft
124
ANSWERS TO EXERCISES
16. almost 80 years old
17a. true
3 1
17b. false; 6 2, but 3 1 6 2
17c. true
17d. true
3 1
3 1
17e. false; 6 2, but 2 6
17f. true
18a. arithmetic: 10, 25, 40, 65; geometric: 10, 20,
40, 80 or 10, 20, 40, 80
18b. arithmetic: 2, 26, 50, 74; geometric: 2, 10, 50,
250 or 2, 10, 50, 250
18c. arithmetic: 4, 20, 36, 52; geometric: 4, 12, 36,
108 or 4, 12, 36, 108
19a. Add the numbers and divide by 2.
19b. Possible answer: Multiply the numbers and
take the positive square root of the result.
19c. c ab
; This formula holds for all positive
values of a and b.
19d. c 2 • 50 100
10;
c 4 • 36 144
12
3 9
15. 1; 1
LESSON 11.1
y
1. A
2. B
3. possible answer:
Y' (6, 15)
Y (2, 5)
R' (6, 6)
R (2, 2)
O' (18, 6)
O (6, 2)
x
16. Yes, they are similar.
3. possible answer:
y
6
(0, 6)
(5, 6)
4
(7, 3)
2
(1, 1)
5
6. Figure A is similar to Figure C. Possible answer:
If
A
B
and
B
C
, then
A
.
C
7. AL 6; RA 10; RG 4; KN 6
8. No; the corresponding angles are congruent,
but the corresponding sides are not proportional.
9. NY 21; YC 42; CM 27; MB 30
10. Yes; the corresponding angles are congruent,
and the corresponding sides are proportional.
11. x 6 cm, y 3.5 cm
2
12. z 103 cm
13. Yes, the corresponding angles are congruent.
Yes, the corresponding sides are proportional.Yes,
AED ABC.
9
9
14. m 2 cm 4.5 cm, n 4 cm 2.25 cm
17. Possible answer: Assuming an arm is about
three times as long as a face, each arm would be
about 260 ft.
18. Possible answer: Not all isosceles triangles are
similar because two isosceles triangles can have
different angle measures. A counterexample is
shown at right. Not all right triangles are similar
because they can have different side ratios, as in a
triangle with side lengths 3, 4, and 5 and a triangle
with side lengths 5, 12, and 13. All isosceles right
triangles are similar because they have angle
measures 45°, 45°, and 90°, and the side lengths have
the ratio 1:1: 2.
d
21. c
1825
22. Possible answers: Jade might get 4475 of the
2650
profits, or $2,773.18, and Omar might get 4475 of
the profits, or $4,026.82. Or they take out their
investments and they divide the remaining $2,325:
Jade, $1825 $1,162.50 $2,987.50; and Omar,
$3,812.50.
23a.
23b.
19. 36
20. bc
60°
60°
24. approximately 92 gallons
ANSWERS TO EXERCISES
125
Answers to Exercises
5. possible answer:
x
LESSON 11.2
Answers to Exercises
1.
2.
3.
4.
6 cm
40 cm; 40 cm
28 cm
54 cm; 42 cm
37 35
5. No, 3
0 28 .
6. Yes, MOY NOT by SAS.
7. Yes, PHY YHT because YH 12 and
20
16
12
(SSS).Yes, PTY is a right triangle
15
12
9
because 202 152 25 2.
8. TMR THM MHR by AA.
x 15.1 cm, y 52.9 cm, h 28.2 cm
9. Yes,QTA TUR and QAT ARU.
QTA QUR by AA; 632 cm
10. 24 cm; 40 cm
11. Yes, THU GDU and HTU DGU;
52 cm; 42 cm
12. SUN TAN by AA; 13 cm; 20 cm
13. Yes, RGO FRG and GOF RFO.
GOS RFS by AA; 28 cm; 120 cm
14. 20 cm; 21 cm
15. x 50, y 9
126
ANSWERS TO EXERCISES
2 1
16. r R R
2 12
2 1
17. She should order approximately 919 lb every
three months. Explanations will vary.
18. 448
19. The corresponding angles are congruent; the
ratio of the lengths of corresponding sides is 13; the
dilated image is similar to the original.
20. Yes, ABCD ABCD. The ratio of the
perimeters is 12. The ratio of the areas is 14.
y
D' C'
A'
B'
x
21. 118 square units
22. The statue was about 40 ft, or 12 m, tall. To
estimate, you need to approximate the height of a
person (or some part of a person) in the picture,
measure a part of the statue in the picture, calculate
the approximate height of that statue piece, and
assume that the statue has the same proportions
as the average person.
LESSON 11.3
1. 16 m
2. 4 ft 3 in.
3. 30 ft
4. 10.92 m
5. 5.46 m
6. Thales used similar right triangles. The height of
the pyramid and 240 m are the lengths of the legs of
one triangle; 6.2 m and 10 m are the lengths of the
corresponding legs of the other triangle; 148.8 m.
7. 90 m; R and O are both right angles and
P is the same angle in both triangles, so
PRE POC by AA.
8. 300 cm
6
13. GHF FHK GFK by AA; h 181
3,
9
4
x 71
3 , y 44 13 .
14. sample answer:
1
A 2(8.2)(1.7) 6.97 cm2
1
A 2(3)(4.6) 6.9 cm2
8.2 cm
3 cm
1.7 cm
4.6 cm
45 cm
3 cm
30 cm
2 cm
d
A
1
Answers to Exercises
2
15. 53
16.
B
2
20 cm
4
9.
3
D
t
h
y
x
Possible answer: Walk to the point where the guy
wire touches your head. Measure your height, h;
the distance from you to the end of the guy wire, x;
and the distance from the point on the ground
directly below the top of the tower to the end of the
guy wire, y. Solve a proportion to find the height of
the tower, t: ht xy. Finally, use the Pythagorean
Theorem to find the length of the guy
wire: t 2 y 2 .
10. The triangles are similar by AA (because the
ruler is parallel to the wall), so Kristin can use the
length of string to the ruler, the length of string to
the wall, and the length of the ruler to calculate the
height of the wall; 144 in., or 12 ft.
2
11. MUN MSA by AA; x 313.
12. BDC AEC by AA; y 63.
C
Given: Parallelogram ABCD
CD
and AD
BC
Show: AB
ABCD is a parallelogram
Given
BC
AD
CD
AB
Definition of
parallelogram
Definition of
parallelogram
BD
BD
2 4
1 3
Same segment
AIA Conjecture
AIA Conjecture
ABD CDB
ASA
BC
AD
CD
AB
CPCTC
CPCTC
17a. 4.6.12
17b. 3.12.12 or 3.12 2
ANSWERS TO EXERCISES
127
1 + 5
18. The golden ratio is , or approximately
2
1.618. Here is one possible construction:
C
D
A
X
B
mB 90°
1
Construct BC 2AB
5
AC 2 AB
1
5
AB
AX AD 2 AB 2
5
1
AB
AX 2
Answers to Exercises
AB
1 + 5
2
AX 5 1 2
Therefore, X is the golden cut.
128
ANSWERS TO EXERCISES
19a. Answers will vary.
19b. Possible answer: The shape is an irregular
curve.
19c. Answers will vary. Possible answers: As the
circular track becomes smaller, the curve becomes
more circular; as the track becomes larger, the
curve becomes more pointed near the fixed point.
As the rod becomes shorter, the curve becomes
more pointed near the fixed point; as the rod
becomes longer, the curve becomes more like an
oval. As the fixed point moves closer to the traced
endpoint, the curve becomes more pointed near
the fixed point; as the fixed point moves closer to
the circular track, the curve begins to look like a
crescent moon.
a
c
17. b 1 d 1
a b c
d
b b d d
LESSON 11.4
1.
2.
3.
4.
5.
6.
7.
8.
18 cm
12 cm
21 cm
15 cm
2.0 cm
126 cm2; 504 cm2
16 cm
60 cm
4
9. 49 cm
p b
10. q; q
ab cd
b
d
18. The ratio will be the same as the ratio for the
original rectangle. The ratio is 21 if it can be divided
like this:
It might be any ratio if divided like this:
L
O
V
M
A
19. Yes, by the SSS or the SAS Similarity
Conjecture.
20a. 2a b
20b. 2a b
20c. all values of a and b
20d. no values of a and b
21. AB 3 cm, BC 7.5 cm
Answers to Exercises
11. 6
3 cm
12. 6 cm
5
13
1
2
13. x 33 cm, y 3 cm, z 83 cm
k 7
3
14. B (3, 5), R 14, 7 ; h 4
2
15. 1
H
16.
E
T
Consider similar triangles LVE and MTH with
and HA
. To
corresponding angle bisectors EO
show that the corresponding angle bisectors are
proportional to corresponding sides, for example
EO
EL
, show by AA that LOE MAH,
HA
HM
EL
and then you can show that HEOA HM . You
know that L M. Use algebra to show that
LEO MHA.
ANSWERS TO EXERCISES
129
LESSON 11.5
1. 18 cm2
4. 27 cm2
2. 18
1
5. 4
9
3. 5; 10
6. 1: 3
9
12
,
16d. x 9, y 16, 1
(9)(16) 2 16 h 144
12
17.
m2
m 2
7. or n
2
n
8. Possible answer: Assuming the ad is sold by
area, Annie should charge $6000.
9
9. 5000 tiles
10. 1
11.
7.1 m 14.2 m
14.2 m 6.5 m
10 m
42 m
42 m
18.
1
2
2
4
3
6
4
8
5
10
6
12
y
5
5
x
12b. A(x) 2x 2
y
x
Area
in cm2
70
1
2
60
2
8
50
3
18
4
32
5
50
6
72
40
30
20
10
5
x
12c. The equation for a(x) is linear, so the graph is
a line. The equation for A(x) is quadratic, so the
graph is a parabola.
13. Possible proof: The area of the first rectangle
is bh. The area of the dilated rectangle is rh rb, or
r 2bh. The ratio of the area of the dilated rectangle
r2bh
2
to the area of the original rectangle is bh , or r .
16 20
15. 8
14. 3, 3
16a. (i) 40°; 50°; 40°
(ii) 60°; 30°; 60°
(iii) 22°; 68°; 22°
Conjecture: similar; right triangle
s
h
h
b
16b. (i) h
(ii) x
(iii) m or a
p h
Conjecture: q
h
16c. h pq
130
ANSWERS TO EXERCISES
H
E
L
10
Area in cm2
x
Area
in cm2
Area in cm2
Answers to Exercises
12a. a(x) 2x
O
V
M
A
T
Consider similar triangles LVE and MTH with
and HA
. Show
corresponding altitudes EO
EO
EL
that LOE MAH, then HA HM . You know
and that L M. Because EO
HA are altitudes,
LOE and MAH are both right angles, and
LOE MAH.
EL
So, LOE MAH by AA. Thus HEOA HM ,
which shows that the corresponding altitudes
are proportional to corresponding sides.
19. x 92°; Explanations will vary but should
reference properties of linear pairs, isosceles
triangles, and alternate interior angles formed
by parallel lines.
20. 105.5 cm2
21. 60 ft2
22. true
23. Two pairs of angles are congruent, so the
triangles are similar by the AA Similarity
Conjecture. However, the two sets of corresponding
105
sides are not proportional 8600 135 , so the
triangles are not similar.
24.
Top
Front
Right side
Top: 6 square units; front: 4 square units; right side:
4 square units. The sum of the areas is half the
surface area. The volume of the original solid is
8 cubic units. The volume of the enlarged solid is
512 cubic units. The ratio of volumes is 614 .
LESSON 11.6
1. 1715 cm3
x
1
Surface area in cm2
2
3
4
5
22
88
198
352
550
Volume in cm 3
6
48
162
384
750
18. yes, because 182 242 302 (Converse of the
Pythagorean Theorem)
19a. Possible answer: Fold a pair of corresponding
vertices (any vertex in the original figure and the
corresponding vertex in the image) together and
crease; repeat for another pair of corresponding
vertices; the intersection of the two creases is the
center of rotation.
19b. Possible answer: Draw a segment (a chord)
between a pair of corresponding vertices and
construct the perpendicular bisector; repeat for
another pair of corresponding vertices; the
intersection of the two perpendicular bisectors
is the center of rotation.
20e. Label the third vertex C. Construct segment
D
2x
2
CD, which bisects C. AD
B 3x 3 , or 2:3
y
800 Volume
600
Surface area
400
200
5
x
ANSWERS TO EXERCISES
131
Answers to Exercises
2. 16 cm, 4 cm; 768 cm3 2412.7 cm3;
64
12 cm3 37.7 cm3; 1
3 125
3
3. 5; 2;
7 1500 cm
H 3 64
;
4. 1944 ft3 6107.3 ft3; 2
4 27 32 ft
125
5. 2
7
2
6. 2:5
7. 3
8. $1,953.13
9. 2432 lb
10. Possible answer: No, a 4-foot chicken, similar
to a 14-inch 7-pound chicken, would weigh
3
approximately 282 pounds 41843 7x. It is unlikely
that the legs of the giant chicken would be able to
support its weight.
11. Possible answer: Assuming the body types of
the African goliath frog and the Brazilian gold frog
are similar, the gold frog would weigh about
0.0001 kg, or 0.1 g.
12. surface area ratio 116 , volume ratio 614 . The
dolphin has the greater surface area to volume
ratio.
13. The ratio of the volumes is 217 . The ratio of the
surface areas is 91.
14.
S(x) 22x 2 and V(x) 6x 3.
Possible answer: The surface area equation is
quadratic, so the graph is a parabola, and the
volume equation is cubic, so the graph is a cubic
graph.
15. Possible proof: The volume of the first
rectangular prism is lwh. The volume of the second
rectangular prism is rl rw rh, or r 3lwh. The ratio
r3lwh
3
of the volumes is lwh , or r .
3
16. 9,120 m or approximately 28,651 m3
s
s s2 s
17. 4 or 4
2
2
LESSON 11.7
1
2. 333 cm
3. 45 cm
4. 21 cm
5. 28 cm
6. no
7. José’s method is correct. Possible explanation:
Alex’s first ratio compares only part of a side of the
larger triangle to the entire corresponding side of the
smaller triangle, while the second ratio compares
entire corresponding sides of the triangles.
8. yes
9. 6 cm; 4.5 cm
10. 13.3 cm; 21.6 cm
11. yes; no; no
12. yes; yes; yes
13. 3
2 cm; 6
2 cm
14.
Answers to Exercises
1. 5 cm
E
15.
F
I
J
So two pairs of corresponding sides of XYZ and
XAB are proportional. X X, so
XYZ XAB by the SAS Similarity Conjecture.
Because XYZ XAB, XAB XYZ.
by the Converse of the Parallel
YZ
Hence, AB
Lines Conjecture.
20. Set the screw so that the shorter lengths of the
styluses are three-fourths as long as the longer
lengths.
21. x 4.6 cm, y 3.4 cm
22. x 45 ft, y 40 ft, z 35 ft
343
23. 729
24. She is incorrect. She can make only nine 8 cm
diameter spheres.
25. 6x 2; 24x 2; 54x 2
1
26. 3r
27a. Possible construction method: Use the
triangle-and-circle construction from Lesson 11.3,
.
Exercise 18, to locate the golden cut, X, of AB
Then use perpendicular lines and circles to create a
rectangle with length AB and width AX.
27b. Possible construction method: Construct
golden rectangle ABCD following the method from
by constructing
27a. For square AEFD, locate EF
circle A and circle D each with radius AD. Repeat
the process of cutting off squares as often as
desired. For the golden spiral from point D to point
E, construct circle F with radius EF; select point D,
point E, and circle F and choose Arc On Circle from
the Construct menu.
28. possible answer:
B
16. Extended Parallel/Proportionality Conjecture
17. You should connect the two 75-marks. By the
Extended Parallel/Proportionality Conjecture,
drawing a segment between the 75-marks will
5
form a similar segment that has length 170
0 , or 75%,
of the original.
18. 2064 cm3 6484 cm3
19. possible proof:
a b
c
d
ad cb
ad ab cb ab
a(d b) b(c a)
a(d b) b(c a)
ab
ab
db ca
b
a
132
ANSWERS TO EXERCISES
P
Q
A
C
S
R
D
Given: Circumscribed quadrilateral ABCD, with
points of tangency P, Q, R, and S
Show: AB DC AD BC
Use segment addition to show that each sum of
lengths of opposite sides is composed of four lengths:
AB DC (AP BP) (DR CR) and AD BC (AS DS) (BQ CQ). Using the Tangent
Segments Conjecture, AP AS, BP BQ,
CR CQ, and DR DS, and the four lengths in
each sum are equivalent. Here are the algebraic
steps to show that the whole sums are equivalent.
29.
AB DC (AP BP) (DR CR) Segment
addition.
(AS BQ) (DS CQ) Substitute AS
for AP, BQ
for BP, DS for
DR, and CQ
for CR.
(AS DS) (BQ CQ) Regroup the
measurements
by common
points of
tangency.
Answers to Exercises
AB DC AD BC
Use segment
addition to
rewrite the
right side as
the other sum
of opposite
sides.
ANSWERS TO EXERCISES
133
CHAPTER 11 REVIEW
1. x 24
2. x 66
3. x 6
4. x 17
5. 6 cm; 4.5 cm; 7.5 cm; 3 cm
1
1
6. 46 cm; 72 cm
7. 13 ft 2 in.
8. It would still be a 20° angle.
9.
Answers to Exercises
K
P
L
10. Yes. If two triangles are congruent, then
corresponding angles are congruent and
corresponding sides are proportional with ratio
1
, so the triangles are similar.
1
11. 15 m
12. 4 gal; 8 times
13. Possible answer: You would measure the
height and weight of the real clothespin and the
height of the sculpture.
Wsculpture
Hsculpture 3
Wclothespin
Hclothespin
If you don’t know the height of the sculpture, you
could estimate it from this photo by setting up a
ratio, for example,
Hperson
Hsculpture
Hperson’s photo Hsculpture’s photo
134
ANSWERS TO EXERCISES
14. 9:49
5 125
15. 4; 6
4
16. $266.67
17. 640 cm3
18. 32; 24; 40; 126
19. 841 coconuts
1
20a. 1 to 4 to 2, or 4 to to 2
20b. 3 to 2 to 1
20c. Answers will vary.
21. Possible answer: If food is proportional to
1
body volume, then 8000 of the usual amount of food
is required. If clothing is proportional to surface
1
of the usual amount of clothing is
area, then 400
required. It would take 20 times longer to walk a
given distance.
22. The ice cubes would melt faster because they
have greater surface area.
Answers to Exercises
CHAPTER 12 • CHAPTER
12 CHAPTER 12 • CHAPTER
LESSON 12.1
1.
2.
3.
4.
5.
6.
0.6018
0.8746
0.1405
11.57
30.86
62.08
s
s
r
7. sin A t; cos A t; tan A r
4
3
4
8. sin 5; cos 5; tan 3
7
7
24
;
;
9. sin A 2;
5 cos A 25 tan A 24
10. 30°
11. 53°
30°
24°
a 35 cm
b 15 cm
c 105 yd
d 40°
e 50 cm
f 33°
g 18 in.
approximately 237 m
x 121 ft
6.375
2.2
16-inch pizza
box of ice cream
8
3 cm 13.9 cm
V 288 ft3 905 ft 3, S 144 ft 2 452 ft 3
ANSWERS TO EXERCISES
135
Answers to Exercises
24
24
7
;
sin B 2;
5 cos B 25 tan B 7
12.
13.
14.
15.
16.
17.
18.
19.
20.
21.
22.
23.
24.
25.
26.
27.
28.
Answers to Exercises
LESSON 12.2
1. 9 cm
2. 64°
3. 7 cm
4. 24 m
5. 22 in.
6. 49°
7. 127°
8. 30°
9. 64 cm
10. approximately 655 m
11. approximately 101 m
12. approximately 65 m
13. approximately 188 m
14. approximately 1621 m
15. approximately 1570 m
16a. approximately 974 m
16b. approximately 1007 m
16c. Yes, the height of the balloon would be 1 m
less because you don’t have to account for the
tripod. The distance to a point under the balloon
would not change.
136
ANSWERS TO EXERCISES
17. 45°
18. 60°
19. 60°
20a. 15.04
20b. 2.05
20c. 5.2304
20d. 2.644
1
21. x 3.5, y 97 9.14
22. The block has a volume of 90 cm3 but it
displaces only 62.8 cm3 of water. So not all of the
block is under water, which means it floats.
23. CZ AX BY
24a. decreases, approaching 0
24b. decreases, approaching 0
24c. remains 90°
24d. increases, approaching (AO)2
24e. decreases, approaching 1
24f. decreases, approaching 1
25. R(8, 7) and E(3, 9), or R(4, 3) and
E(1, 1)
LESSON 12.3
1. 329 cm2
2. 4 cm2
3. 11,839 cm2
4. 407 cm2
5. 35 cm
6. 17 cm
7. 30 cm
8. 56°
9. 45°
10. 66°
11a. approximately 2200 m
11b. approximately 1600 m
11c. approximately 1400 m
12. The other two walls were 300 ft and
approximately 413 ft. The area was approximately
45,000 ft2.
13. approximately 48 m
14. 2366 cm; approximately 41°
15. approximately 5°
93 3
3
16. 8 cm , or approximately 6 cm
CD
, A D by the AIA
17. Because AB
Conjecture. Because D and B intercept the
same arc, D B. Therefore A B by the
transitive property. So ABE is isosceles by the
Converse of the Isosceles Triangle Conjecture.
18. Draw line through the two points. Then fold
the paper so that the two points coincide. Draw
another line along the fold. These two lines contain
the diagonals of the square. Now fold the paper so
that the two lines coincide. Mark the vertices on
the other line.
19. AC 60 cm, AE 93.75 cm, AF 117 cm
20. Box 1; about 1.2 in. longer
Answers to Exercises
ANSWERS TO EXERCISES
137
Answers to Exercises
LESSON 12.4
1. 32 cm
2. 47 cm
3. 341 cm
4. 74°
5. 64°
6. 85°
7. approximately 43 cm
8. approximately 30°
9. approximately 116° and 64°
10. approximately 6 min
11. approximately 87.8 ft
12. approximately 139 m
13. approximately 143 m
14. The ladder at approximately a 76° angle is not
safe. The ladder should be at least 6.5 ft and no
more than 14.3 ft from the base of the wall.
15.
c
A
a
b
a
sin A c
b
cos A c
a
tan A b
a
sin A
a c a
c
tan A
cos A
c b b
b
c
16a. PR x 3
16b. Since x2 x 3 2 (2x)2, TPR is a right
triangle by the Converse of the Pythagorean
Theorem. Therefore mTPR 90°.
17a. increases
17b. decreases
17c. increases then decreases
18a. The base perimeters are equal.
18b. The cone has the greater volume.
18c. The cone has the greater surface area.
19a. a translation right 10 units;
(x, y) → (x 10, y)
19b. a rotation 180° about the origin;
(x, y) → (x, y)
138
ANSWERS TO EXERCISES
20. possible answer:
x
y
x
2y
21. possible answer:
x
120°
y
120°
120°
x
x
60°
60°
120°
y
y
60°
60°
2y
2x
22. Answers will vary from simple (circles and
segments) to quite complex. See below for a more
detailed answer.
Students might begin with the simplest case where
all factors are equal. With circles of the same radius
and with endpoints of the segment traveling in the
same direction, at the same speed, and starting
from the same relative position, the midpoint will
trace a circle equal in size to those constructed.
Changing the size of both circles will change the
size of the circle traced, as will changing the
starting positions of the endpoints. The maximum
radius of the traced circle is limited by the radius of
the constructed circles; the smallest observable
trace is a single point.
With all other factors equal and the endpoints of
the segment traveling in opposite directions
around the circles, the midpoint will trace a
segment. Changing the relative speeds of the
endpoints, or the relative radii of the circles, will
produce more complex polar curves, which
students might describe as flowers, radii, or
“spirograph-like.” Changing the distance between
the centers has no effect on the shape of the trace,
only its position.
LESSON 12.5
1. approximately 12.7 m
2. approximately 142 mi/h
3. approximately 51 km
4. approximately 5.9 m
5a. approximately 9.1 km
5b. approximately 255°
6. approximately 240 m
7. approximately 42 ft
8. They must dig at an angle of approximately
71.6° and dig for approximately 25.3 m.
9. approximately 168 km/h
6 cm
10.
For simplicity, let 326n0 . Now use the tangent
ratio to find the length of the apothem.
s
2
tan a
s
a
2 tan Now use the area formula for a regular polygon.
1
A 2aP
1
s
A 2 2 tan ns
2
360
ns
A
4 tan where 2n
13. approximately 12.8 m at an angle of
approximately 48°
14. 1536
3 cm3 2660 cm3
15. The area increases by a factor of 9.
y
(0, 2)
10a. apothem 9.23 cm; area 277 cm2
10b. leg 9.7 cm; area of triangle 27.7 cm2;
area of decagon 277 cm2
10c. They are the same.
11. approximately 108 cm3
12. Sample answer: Any regular polygon can be
divided into n congruent isosceles triangles, each
360°
with a vertex angle of n .
360°
___
n
The altitude from the vertex angle of an isosceles
triangle bisects the angle and the opposite side.
This altitude is also an apothem of the regular
polygon.
–6
_1
A 2
(1, 0)
_1
A 2
1 2 1x
6
369
–6 (0, –6)
5 5 15
16. 4 12
17a. decreases then increases
17b. does not change
18a. a reflection across the x-axis;
(x, y) → (x, y)
18b. a reflection across the y-axis; (x, y) → (x, y)
19. Answers will vary.
360°
___ 360°
___
2n 2n
a
_s
2
_s
2
ANSWERS TO EXERCISES
139
Answers to Exercises
(–3, 0)
USING YOUR ALGEBRA SKILLS 12
4.
y
2
1.
y
x
–5
6
–2
4
–4
f(x) x
2
2
–6 –4 –2
4
6
–6
p(x) 2(x 3) 5
x
–2
The graph of y 2x 1 is the same as the
graph of p(x). The slope of a line is equal to
the vertical stretch of the linear function. The
y-intercept is equal to ah k.
5. a vertical stretch by a factor of 2 and a vertical
translation down 5 units
–4
–6
y
y
10
6
8
4
6
2
4
–2
2
–2
g(x) x 2
x
2
–2
h(x) x 3
x
2
y
2
–3
3
–4
–2
–6
–4
f(x) 2x3 5
–6
Answers to Exercises
2.
y
–8
4
f(x) x
x
2
–5
5
g(x) x 5
x
10
–2
h(x) x 5
–4
6. a horizontal translation right 5 units and a
vertical translation up 2 units
y
6
g(x) (x 5)3 2
4
2
All three graphs are V-shaped, consisting of two
parts that have slopes of 1 and 1. Graph g is a
translation of graph f right 5 units. Graph h is a
translation of graph f down 5 units.
y
3.
2
–4 –2
–2
2
8
x
7. a vertical stretch by a factor of 21 (a vertical
shrink), a horizontal translation right 2 units,
and a vertical translation down 5 units
x
2 4
f(x) x2
y
1
–4 –2
–4
6
–2
2
4
6
8
x
–2
–6
–6
y
h(x) 12 x 2 5
2
–4 –2
2
4
x
–2
–4
g(x) x
–6
y
2
–4 –2
2
–2
–4
4
x
h(x) x 3
A vertical stretch by a factor of 1 reflects the
graph across the x-axis.
140
ANSWERS TO EXERCISES
8. f(x) |x 3| 1; a horizontal translation
right 3 units and a vertical translation down 1 unit
9. f(x) 3x2; a vertical stretch by a factor of 3
10. f(x) (x 3)2; a vertical stretch by a factor
of 1 (a reflection across the x-axis) and a horizontal translation right 3 units
11. f(x) 2|x| 3; A vertical stretch by a factor
of 2 and a vertical translation up 3 units
12. f(x) 3(x 1)2 3; A vertical stretch by a factor of 3, a horizontal translation left 1 unit, and a
vertical translation down 3 units
13. f(x) 21(x 3)2 2; A vertical stretch by a factor of 21 (a vertical shrink), a horizontal translation
right 3 units, and a vertical translation up 2 units
14. Both graphs have the same “hills and valleys”
shape, but one is a horizontal translation of the
other.
17. Possible answer: f(x) 2 sin(x 90°) 1;
a vertical stretch by a factor of 2, a horizontal
translation right 90°, and a vertical translation
down 1 unit
y
p(x) ⫽ sin(x)
2
x
360
–360
–2
q(x) ⫽ cos(x)
15. a horizontal translation; a 1, h 90°, k 0
16. The dashed function in each graph is the parent
function f(x) sin(x).
y
p(x) ⫽ 2sin(x)
y
2
2
–360
360
x
360
–360
x
–2
Answers to Exercises
q (x) ⫽ _12 sin(x)
y
r(x) = –sin(x)
2
360
–360
x
–2
Varying a in the equation of the sine function
causes the same types of vertical stretches as with
other functions.
ANSWERS TO EXERCISES
141
Answers to Exercises
CHAPTER 12 REVIEW
1. 0.8387
2. 0.9877
3. 28.6363
a c a
4. b; b; c
8 15 8
; 5. 1;
7 17 15
s
6. s; t; t
7. 33°
8. 86°
9. 71°
10. 1823 cm2
11. 15,116 cm3
12. Yes, the plan meets the act’s requirements. The
angle of ascent is approximately 4.3°.
13. approximately 52 km
14. approximately 7.3°
15. approximately 22 ft
16. approximately 6568 m
17. approximately 2973 km/h
18. 393 cm2
19. 30 cm
20. 78°
21. 105 cm
22. 51°
23. 759 cm2
24. approximately 25 cm
25. 72 cm2
26. approximately 15.7 cm
27. approximately 33.5 cm2
28. approximately 10.1 km/h at an approximate
bearing of 24.5°
29. False; an octahedron is a polyhedron with
eight faces.
30. false
33.
34.
35.
36.
37.
38.
39.
40.
41.
2
False; the ratio of their areas is mn2 .
true
length of leg opposite T
false; tangent of T length of leg adjacent to T
true
true
true
true
False; the slope of line 2 is m1.
false
42. B
43. C
44. A
45. D
46. B
47. B
48. A
49. B
50. C
51. D
52. C
53. A
100
54. 3 cm3
55. 28 cm3
56. 30.5 cm3
57. 33
58. (x, y) → (x 1, y 3)
59. w 48 cm, x 24 cm, y 28.5 cm
60. approximately 18 cm
61. (x 5)2 (y 1)2 9
62. Sample answer: Each interior angle in a regular
pentagon is 108°. Three angles would have a sum of
324°, or 36° short of 360, which would leave a gap.
Four angles would have a sum exceeding 360° and
hence create an overlap.
63. approximately 99.5 m
64. 30 ft
65. 4 cm
B
66.
A
D
C
66a. mABC 2 mABD
is the perpendicular
66b. Possible answers: BD
. It is the angle bisector of ABC, it
bisector of AC
is a median, and it divides ABC into two
congruent right triangles.
31. true
32. true
142
ANSWERS TO EXERCISES
Answers to Exercises
CHAPTER 13 • CHAPTER
13 CHAPTER 13 • CHAPTER
LESSON 13.1
1. A postulate is a statement accepted as true
without proof. A theorem is deduced from other
theorems or postulates.
2. Subtraction: Equals minus equals are equal.
Multiplication: Equals times equals are equal.
Division: Equals divided by nonzero equals are equal.
3. Reflexive: Any figure is congruent to itself.
C
ABC ABC
B
A
Transitive: If Figure A is congruent to Figure B and
Figure B is congruent to Figure C, then Figure A is
congruent to Figure C.
P
Q
R
S
X
, then
Y
P
Q
X
Y
.
RS
and RS
XY
, then PQ
XY
.
If PQ
Symmetric: If Figure A is congruent to Figure B,
then Figure B is congruent to Figure A.
X
Y
Z
L
then
M
L
If
,
M
X
N
N
.
Y
Z
If XYZ LMN, then LMN XYZ.
4. reflexive property of equality; reflexive
property of congruence
5. transitive property of congruence
19. (Lesson 13.1)
and BO
are radii
AO
1
}?
2
Given
AO BO
AOB is isosceles
3 ?
}
}
Definition of ?
isosceles
Definition of circle
20. (Lesson 13.1)
1
}
1 ? 2
}? Given
2
}? m n
Corresponding
Angles Postulate
3
}
3 ? 4
}? Corresponding
Angles Postulate
ANSWERS TO EXERCISES
143
Answers to Exercises
If
6. subtraction property of equality
7. division property of equality
8. Distributive; Subtraction; Addition; Division
9. Given; Addition property of equality;
Multiplication property of equality; Commutative
property of addition.
10. true, definition of midpoint
11. true, Midpoint Postulate
12. true, definition of angle bisector
13. true, Angle Bisector Postulate
14. false, Line Intersection Postulate
15. false, Line Postulate
16. true, Angle Addition Postulate
17. true, Segment Addition Postulate
18.
• That all men are created equal.
• That they are endowed by their creator with
certain inalienable rights, that among these are life,
liberty, and the pursuit of happiness.
• That to secure these rights, governments are
instituted among men, deriving their just powers
from the consent of the governed.
• That whenever any form of government becomes
destructive to these ends, it is the right of the
people to alter or to abolish it, and to institute new
government, laying its foundation on such
principles and organizing its powers in such
form as to them shall seem most likely to effect
their safety and happiness.
19. See flowchart below.
20. See flowchart below.
28. 30 ft
29. FG 2
6 and DG 2
3 because ABGF
and BCDG are parallelograms. Triangle FGD is
right (mFGD 90°) by the Converse of the
Pythagorean Theorem because 2
6 2 2
3 2
62. But mFGB 128° and mDGB 140° by
conjectures regarding angles in a parallelogram.
So, mFGD 92° because the sum of the angles
around G is 360°. So, mFGD is both 90° and 92°.
30. x 54°, y 126°, a 7.3 m
31a. 4
31b. 1 4
31c. 2 1
Answers to Exercises
21. See flowchart below.
22. See flowchart below.
23. (2n 1) (2m 1) 2n 2m 2 2(n m 1)
24. (2n 1)(2m 1) 4nm 2n 2m 1 4nm 2n 2m 2 1 2(2nm n m 1) 1
25. Let n be any integer. Then the next two
consecutive integers are n 1 and n 2. The sum
of these three integers is (n) (n 1) (n 2) n n 1 n 2. Combining like terms: 3n 3 3(n 1), which is divisible by 3.
26. 299 m
27. sphere, cylinder, cone; cylinder, cone, sphere;
cone, cylinder, sphere
21. (Lesson 13.1)
1
AC BD
}? Given
2
3
D C
BAD
BC
}? AD
}? Given
4
}
5
ABC ?
Congruence
}? SSS
Postulate
}?
CPCTC
AB BA
Reflexive property
of congruence
22. (Lesson 13.1)
2 ? AB
CB
}
Given
BCD
CBD
1 Construct angle
BD
bisector Angle Bisector
Postulate
3
?
ABD }
5
?
BAD }
? Reflexive property
of congruence
}
ANSWERS TO EXERCISES
6
?
}
? SAS Congruence ? CPCTC
Definition
}
}
? angle bisector Postulate
of }
4 BD BD
144
A C
LESSON 13.2
3
10.
1
1. Linear Pair Postulate
2. Parallel Postulate, Angle Addition Postulate,
Linear Pair Postulate, Corresponding Angles
Postulate
3. Parallel Postulate
4. Perpendicular Postulate
5.
3
4
Use the VA Theorem and the transitive property to
get 3 4. Therefore the lines are parallel by the
Converse of the AIA Theorem.
3
11.
2
Use the definition of supplementary angles and the
substitution property to get m1 m1 180°.
Then use the division property and the definition
of a right angle.
6.
1 2
3 4
1
2
Use the definition of a right angle and the
transitive property to get m1 m2. Then use
the definition of congruence to get 1 2.
3
8.
1
1
2
2
3
Use the VA Theorem and the transitive property to
get 1 3. Therefore the lines are parallel by the
CA Postulate.
3
9.
1
3
1
2
1
2
Use the Linear Pair Postulate and the definition of
supplementary angles to get m1 m3 180°.
Then use the CA Postulate and the substitution
property to get m1 m2 180°.
3
12.
1
31
2
2
Use the Linear Pair Postulate and the definition of
supplementary angles to get m1 m3 m1 m2. Then use the subtraction property
and the Converse of the AIA Theorem to get
1 2.
13.
1
A
1
B 2
2
3
3
Construct a transversal across lines 1 and 2; it
will intersect line 3 by the Parallel Postulate. Use
the Interior Supplements Theorem and the
definition of supplementary angles to get m1 m2 180°. Use the CA Postulate and the
substitution property to get m1 m3 180°.
Therefore 1 3 by the Converse of the Interior
Supplements Theorem.
2
Use the VA Theorem and the CA Postulate to get
1 3 and 2 3. Then use the transitive
property to get 1 2.
ANSWERS TO EXERCISES
145
Answers to Exercises
Use the definition of supplementary angles and the
transitive property to get m1 m2 m3 m4. Then use the substitution property and the
subtraction property to get m1 m4.
7.
2
2
3
1
21
1
3
14.
1
2
1
2
Use the definition of perpendicular lines and the
transitive property to get m1 m2. Therefore
lines 1 and 2 are parallel by the Converse of the
AIA Theorem.
15. 2
Answers to Exercises
1
3
By the Triangle Sum Theorem, m1 m2 m3 180°. By the definition of a right triangle,
1 is a right angle. By the definition of right
angle, m1 90°. Using the subtraction property,
m2 m3 90°, so by the definition of
complementary angles, 2 and 3 are
complementary.
16.
Linear Pair
Post.
VA Thm.
CA Post.
Converse of
AIA Thm.
Converse of
AEA Thm.
146
ANSWERS TO EXERCISES
17. 1066 cm3
18.
Area (ft2)
Bottom
6
2 sides
9
Back and front
6
2 rooftops
812
2 gable ends
2
Total
3112
Plywood (4 by 8)
32
The area is less than that of one sheet of plywood.
However, it is impossible to cut the correct size
pieces from one piece. This answer assumes that
the bottom of the dog house is included. If the
bottom is not included, and the gables can be cut
separately from the rectangular part of the front
and back, then the pieces can be cut from a single
sheet of plywood.
19. A(6, 10), B(2, 6), C(0, 0); mapping
rule: (x, y) → (2x 8, 2y 2)
Isosceles Triangle Theorem (CB CA), the reflexive
property, and the SSS Congruence Postulate to get
ACP BCP. Therefore, ACP BCP by
CPCTC.
6. n C
m
LESSON 13.3
1. Case 1: P is collinear with A and B. Use the Line
Intersection Postulate to show that P and E are the
same point and use the definitions of bisector and
midpoint to get AP BP.
Case 2: P is not collinear with A
and B. Use the SAS Congruence Postulate to get
BP
.
AEP BEP. Then use CPCTC to get AP
P
A
Use the Line Intersection Postulate and the
Perpendicular Bisector Theorem to get AP BP
and BP CP. Then use the transitive property and
the Converse of the Perpendicular Bisector
Theorem to prove that point P is on line n.
C
7.
C
P
A
B
B
E
m
D
n
Q
A
Use the Line Intersection Postulate and the Angle
Bisector Theorem to prove that Q is equally distant
and AC
and from AB
and BC
. Then use the
from AB
transitive property and the Converse of the Angle
Bisector Theorem to prove that point Q is on line n.
B
8.
P
A
B
E
2
Case 2: P is not collinear with A and B. Draw
. Use the SSS Congruence
midpoint E and PE
Postulate to get AEP BEP. Then use CPCTC
and the Congruent and Supplementary Theorem
to prove that AEP and BEP are both right
is the perpendicular bisector
angles. Therefore PE
by the definitions of midpoint and
of AB
perpendicular.
C
3. Use the reflexive property and
the SSS Congruence Postulate to get
ABC BAC. Therefore,
A B by CPCTC.
1
5.
B
P
C
A
Draw line BA.
Use the Isosceles Triangle Theorem to get
PAB PBA. Use the Angle Addition
Postulate and the subtraction property to get
BAC ABC. Then use the Converse of the
4
C
Use the Linear Pair Postulate and the definition of
supplementary angles to get m3 m4 180°.
Then use the Triangle Sum Theorem and the
transitive property to get m1 m2 m3 m3 m4. Therefore, m1 m2 m4 by the subtraction property.
D
9.
3 4
A
B
C
A
3
A
A
4. Use the reflexive property and
the ASA Congruence Postulate to
get ABC BAC. Then use
CPCTC and the definition of
isosceles triangle.
B
C
1
2
B
B
Use the Triangle Sum Theorem and the addition
property to get mA m1 m3 mC m4 m2 360°. Then use the Angle Addition
Postulate and the substitution property to get
mA mABC mC mCDA 360°.
C
10. Use the definitions of median
and midpoint to get BM 21BC and
N
M
AN 21AC. Then use the multiplication property and the substitution
B
A
BM
. By the
property to get AN
reflexive property, the Isosceles Triangle Theorem,
and the SAS Congruence Postulate, ABN AM
by CPCTC.
BAM. Therefore, BN
ANSWERS TO EXERCISES
147
Answers to Exercises
2. Case 1: P is collinear with A and B. Use the
definitions of congruence and midpoint to show
. Then use the
that P is the midpoint of AB
definition of perpendicular bisector.
11.
median by CPCTC and the definitions of midpoint
and median.
C
Q
P
C
B
A
Use the Angle Addition Postulate and the
definition of angle bisector to get
mPAB 21 mCAB and
mQBA 21 mCBA. Then use the Isosceles
Triangle Theorem, the multiplication property, and
the substitution property to get PAB QBA.
By the reflexive property and the ASA Congruence
BQ
Postulate, ABP BAQ. Therefore, AP
by CPCTC.
C
12.
T
S
B
Answers to Exercises
A
Use the Isosceles Triangle Theorem, the Right
Angles Are Congruent Theorem, and the SAA
Theorem to get ABT BAS. Therefore,
BT
by CPCTC.
AS
C
13.
A
B
D
median → angle bisector
Use the definitions of median, midpoint, and
isosceles triangle, the reflexive property, and the
SSS Congruence Postulate to prove that
ADC BDC. Then use CPCTC and
the definition of angle bisector.
A
D
B
angle bisector → altitude
Use the definitions of angle bisector and isosceles
triangle, the reflexive property, and the SAS
Congruence Postulate to get ADC BDC.
Then use CPCTC, the Linear Pair Postulate, and
the Congruent and Supplementary Theorem to
prove that ADC and BDC are both right
is the altitude by the
angles. Therefore CD
definitions of perpendicular and altitude.
14. x 6, y 3
15. 21.9 m
16. first image: (0, 3); second image: (6, 1)
17. BC FC makes ABCF a rhombus, so its
diagonals are perpendicular. mFGC 90°, so
GE
, so by
mCFG mFCG 90°. FD
subtraction and transitivity m2 mFCG.
1 FCG by AIA, so 1 2.
3
18a. 3
18b. 18c. 4
3
4
6
19. One possible sequence:
1. Fold A onto B and crease. Label as 1. Label the
midpoint of the arc M.
2. Fold line 1 onto itself so that M is on the crease.
Label as 2.
, and is the desired
M is the midpoint of AB
2
tangent.
1
M
C
2
A
D
B
altitude → median
Use the Right Angles Are Congruent Theorem, the
Isosceles Triangle Theorem, and the SAA Theorem
is the
to get ADC BDC. Therefore, CD
148
ANSWERS TO EXERCISES
A
B
20. mBAC 13°
21a. B
21b. B
1
1
1
22a. x 2(a c), y 2(a b), z 2(b c)
1
1
1
22b. w 2(a b), x 2(b e), y 2(e d),
1
z 2(d a)
LESSON 13.4
1.
D
B
Use x to represent the measures of one pair of
congruent angles and y for the other pair. Use the
Quadrilateral Sum Theorem and the division
property to get x y 180°. Therefore, the
opposite sides are parallel by the Converse of the
Interior Supplements Theorem.
C
2. D
3
2
4
B
A
8
2
C
A
B
5
3
4
B
A
Use the definition of rhombus, the reflexive
property, and the SSS Congruence Postulate to get
ABC ADC. Then use CPCTC and the
bisects
definition of angle bisector to prove that AC
DAB and BCD. Repeat the steps above using
.
diagonal DB
C
4. D
Use the Converse of the Opposite Angles Theorem
to prove that ABCD is a parallelogram. Then use
the definition of rectangle.
C
7. D
A
B
Use the definition of rectangle to prove that ABCD
is a parallelogram and DAB CBA. Then use
the Parallelogram Opposite Sides Theorem, the
reflexive property, and the SAS Congruence Postulate
to get DAB CBA. Finish with CPCTC.
C
8. D
A
B
Use the Parallelogram Opposite Sides Theorem,
the reflexive property, and the SSS Congruence
Postulate to get DAB CBA. Repeat the above
steps to get ADC CBA and DAB BCD.
Then use CPCTC and the transitive property to get
DAB ABC BCD ADC. Finish with
the Four Congruent Angles Rectangle Theorem.
D
C
9.
A
B
A
BC
Use the definition of parallelogram to get AD
and AB DC . Then use the Interior Supplements
Theorem.
C
5. D 2
3
1
A
E
B
CB
.
Use the Parallel Postulate to construct DE
Then use the Parallelogram Opposite Sides
Theorem and the transitive property to prove that
AED is isosceles. Therefore, A B by the
Isosceles Triangle Theorem, the CA Postulate, and
substitution.
D
C
10.
4
B
Use the reflexive property and the SSS Congruence
Postulate to get ABD CDB. Then use
CPCTC and the Converse of the AIA Theorem to
CD
and AD
CB
. Therefore,ABCD is a
get AB
rhombus by the definitions of parallelogram and
rhombus.
A
B
Use the Isosceles Trapezoid Theorem, the reflexive
property, and the SAS Congruence Postulate to get
BD
by CPCTC.
DAB CBA. Then AC
ANSWERS TO EXERCISES
149
Answers to Exercises
Use the AIA Theorem, the reflexive property, and
the SAS Congruence Postulate to get ADC CBA. Then use CPCTC and the Converse of the
DC
.
AIA Theorem to get AB
3. D 7 6 C
1
D
C
A
1
6.
11.
D
2
A
4
1
14.
C
3
Parallelogram
Diagonal Lemma
B
Use the Parallelogram Opposite Angles Theorem,
the multiplication property, and the definition of
angle bisector to get 1 3. Then use the
Converse of the Isosceles Triangle Theorem,
the definition of isosceles triangle, and the
Parallelogram Opposite Sides Theorem to get
BC
DC
AD
.
AB
12.
W
Z
Q
Answers to Exercises
X
P
ASA Congruence
Postulate
Y
Use the Converse of the Angle Bisector Theorem
is the angle bisector of Y. In
to prove that WY
like manner, WY is the angle bisector of W.
Therefore, WXYZ is a rhombus by the Converse
of the Rhombus Angles Theorem.
13. Linear Pair Postulate
CA Postulate
Interior Supplements Theorem
Parallelogram Consec. Angle Theorem
Opposite Sides
Theorem
Line Postulate
Converse of the
IT Theorem
2 diagonals
2 bisectors of sides
isosceles
trapezoid
150
ANSWERS TO EXERCISES
Converse of the Angle
Bisector Theorem
15a. A
15b. S
15c. S
15d. N
16. 2386 ft2
17. See table below.
18. V1 V 2 has length 12.8 and bearing 72.6°.
19a. B
19b. A
20a. 19°
20b. 52°
20c. 52°
20d. 232°
20e. 19°
Lines of symmetry
Rotational symmetry
none
2-fold
trapezoid
none
none
kite
1 diagonal
none
parallelogram
IT Theorem
Double-Edged
Straightedge Theorem
17. (Lesson 13.4)
Name
SSS Congruence
Postulate
Opposite Angles
Theorem
Converse of the Rhombus
Angles Theorem
Angle Addition
Postulate
square
4-fold
rectangle
2 bisectors of sides
2-fold
rhombus
2 diagonals
2-fold
1 bisector of sides
none
LESSON 13.5
B
C
D
O
E
A
10. a 75°, b 47°, c 58°
11. 42 ft3 132 ft 3
3
12a. 4 1
2
12b. 1 3 3
13a. A
13b. N
13c. S
13d. S
13e. A
ANSWERS TO EXERCISES
Answers to Exercises
1. D: Paris is in France; Tucson is in the U.S.;
London is in England. Bamako must be the capital
of Mali.
2. C: The “Sir” in part A shows that Halley was
English; Julius Caesar was an emperor, not a
scientist; Madonna is a singer. Galileo Galilei must
be the answer.
3. No, the proof is claiming only that if two
particular angles are not congruent, then the two
particular sides opposite them are not congruent. It
still might be the case that a different pair of angles
are congruent and that therefore a different pair of
sides are congruent.
4. Yes, this statement is the contrapositive of the
conjecture proved in Example B, so they are
logically equivalent.
5. 1. Assume the opposite of the conclusion;
2. Triangle Sum Theorem; 3. Substitution property
of equality; 4. 0°; Subtraction property of equality
6. Assume ZOID is equiangular. Use the definition
of equiangular and the Four Congruent Angles
Rectangle Theorem to prove that ZOID is a
rectangle. Therefore ZOID is a parallelogram, which
creates a contradiction.
is the altitude to AB
. Use the
7. Assume CD
definitions of altitude, median, and midpoint, the
Right Angles Are Congruent Theorem, and the
SAS Congruence Postulate to get ADC BDC.
BC
, which creates a contradiction.
Therefore AC
8. Assume ZO ID. Use the Opposite Sides
Parallel and Congruent Theorem to prove that
ZOID is a parallelogram, which creates a
contradiction.
and perpendicu9. Given: Circle O with chord AB
lar bisector CD
passes through O
Show: CD
does not pass through O. Use the Line
Assume CD
and OA
and the
Postulate to construct OB
. Then use
Perpendicular Postulate to construct OE
the Isosceles Triangle Theorem, the Right Angles
Are Congruent Theorem, and the SAA Theorem to
get OEA OEB. From CPCTC and the
definition of midpoint, prove that E is the midpoint
, which creates a contradiction.
of AB
151
LESSON 13.6
1.
A
X
Use the Cyclic Quadrilateral Theorem, the
Opposite Angles Theorem, and the Congruent and
Supplementary Theorem to get R, C, E, and
T are right angles.
P
5.
1
S
B
2
W
Case 1 The same: Use the Inscribed Angle Theorem
and the transitive property to get A B.
A
T
Z
Y
B
X
W
Answers to Exercises
O
Case 2 Congruent: Use the multiplication
1 mWX
. Then follow
property to get 12 mYZ
2
the steps in Case 1.
2. D
, OT
, and
Use the Line Postulate to construct OS
. Then use the Tangent Theorem, the Converse
OP
of the Angle Bisector Theorem, and the SAA
PT
by
Theorem to get OSP OTP. PS
CPCTC.
6.
A
D
3
1
2
E
C
C
B
B
A
Use the Inscribed Angle Theorem, the addition
property, and the distributive property to get
mDAB
. Then use
mA mC 12 mBCD
the definition of degrees in a circle, the substitution
property, and the definition of supplementary
angles to get A and C are supplementary.
Repeat the steps using mB and mD to get B
and D are supplementary.
3.
B
. Then use
Use the Line Postulate to construct AD
the Inscribed Angle Theorem, the addition
property, and the distributive property to get
mBD
. Therefore,
m2 m3 12 mAC
mBD
by the Triangle Exterior
m1 12 mAC
Angle Theorem and the transitive property.
P
7.
x
A
b
B
C
O
a
A
D
D
C
. Then use
Use the Line Postulate to construct AD
the AIA Theorem, the Inscribed Angle Theorem,
BD
.
and substitution to get AC
4.
T
C
R
E
152
ANSWERS TO EXERCISES
Intersecting Secants Theorem: The measure of an
angle formed by two secants intersecting outside a
circle is half the difference of the measure of the
larger intercepted arc and the measure of the
smaller intercepted arc. Use the Triangle Exterior
Angle Theorem and the subtraction property to get
x b a. Then use the Inscribed Angle Theorem,
the substitution property, and the distributive
mAC
.
property to get x 12 mBD
8.
D
C
A
B
E
Use the Inscribed Angle Theorem to get
1 mBDC 2 mBEC
. By the definition of
mBDC
180°, so by the
semicircle, mBEC
division property, mBDC 90°. By the
definition of right angle, BDC is a right angle.
Because only definitions, properties, and the
Inscribed Angle Theorem are needed to prove this
conjecture, it is a corollary of the Inscribed Angle
Theorem.
9.
Angle Addition
Postulate
Line
Postulate
ASA Congruence
Postulate
SAA
Theorem
IT
Theorem
SAS Congruence
Postulate
Right Angles Are
Congruent Theorem
Converse of the Angle
Bisector Theorem
Midpoint
Postulate
Tangent
Theorem
h
c
a
b
P
16a. 27°
16b. 47°
16c. 67°
16d. 133°
16e. cannot be determined
16f. cannot be determined
17.
Perpendicular
Postulate
Tangent Segments
Theorem
x
A
O
10.
Linear Pair
Postulate
CA
Postulate
Parallel
Postulate
Angle Addition
Postulate
Answers to Exercises
Converse of the
IT Theorem
SSS Congruence
Postulate
11. A(4.0, 2.9), P(1.1, 2.8)
12. 3; 1; 9
, AB
, AC
, CD
, AD
13. BC
14. 32.5
15. As long as point P is inside the triangle,
a b c h.
Proof: Let x be the length of a side. The areas of the
three small triangles are 21 xa, 21 xb, and 21 xc. The area
of the large triangle is 21 xh. So, 21 xa 21 xb 21 xc 1
1
xh. Divide both sides by x. So, a b c h.
2
2
M
P
Arc Addition
Postulate
B
SSS
Congruence
Postulate
VA
Theorem
IT
Theorem
AIA
Theorem
Triangle Sum
Theorem
ASA Congruence
Postulate
Exterior Angle
Sum Theorem
Inscribed Angle
Theorem
Cyclic Quadrilateral
Theorem
Parallelogram
Diagonal Lemma
Opposite Angles
Theorem
Congruent and
Supplementary Theorem
Right Angles
Are Congruent
Theorem
Steps:
.
1. Construct OP
. Label midpoint M.
2. Bisect OP
3. Construct circle with center M and radius PM.
4. Label the intersection of the two circles A and B.
and PB
. PA
and PB
are the
5. Construct PA
required tangents.
OAP and OBP are right angles because they are
inscribed in semicircles.
Parallelogram Inscribed
in a Circle Theorem
ANSWERS TO EXERCISES
153
6.
LESSON 13.7
1.
T
C
V
1
A
B
H
A
G
L
E
Use the Right Angles Are Congruent Theorem,
CASTC, and the AA Similarity Postulate to get
BHT GEV. Therefore, GBVT TVHE by CSSTP. See
the Solutions Manual for the family tree.
A
G
2.
2
B
D
Use the Right Angles Are Congruent Theorem, the
reflexive property, and the AA Similarity Postulate
to get ADC ACB and ACB CDB.
Therefore, ADC ACB CDB by the
transitive property of similarity.
C
7.
h
S
L
M
Answers to Exercises
B
Y
Use the definitions of median and midpoint, the
Segment Addition Postulate, the substitution
property, and the multiplication property to get
BY 12BI and SL 12SM. Then use CASTC,
CSSTP, and the SAS Similarity Theorem to get
BYG SLA. Therefore, BSAG GAYL by CSSTP. See
the Solutions Manual for the family tree.
3.
F
C
2 4
13
D
A
Q
E
B
P
Use the definition of angle bisector, the Angle
Addition Postulate, and the substitution property to
get mACB 2m1 and mDFE 2m2. Then
use CASTC and the AA Similarity Postulate to get
APC DQF. Therefore, DACF CFQP by CSSTP.
C
4.
D 1
A
E
2
B
Use the CA Postulate and the AA Similarity
Postulate to get CDE CAB. Then use CSSTP
CD DA
and the Segment Addition Postulate to get CD
CE EB
DA
EB
. Therefore, by algebra.
CE
CD
CE
C
5.
D 1
A
2
E
B
A
Use the addition property to get DCD
1 CEBE 1.
Then use algebra and the Segment Addition
Postulate to get CCAD CCEB . Therefore ABC DEC by the SAS Similarity Theorem, 1 2
by the CA Postulate.
AB
by CASTC, and DE
154
ANSWERS TO EXERCISES
A
I
x
B
y
D
Use the Three Similar Right Triangles Theorem to
get ADC CDB. Then use CSSTP to get
x
h
.
h
y
8.
b
a
d
c–d
c
Draw the altitude to the hypotenuse, then use the
ratios given by the Three Similar Right Triangles
a
c
2
2
Theorem. In particular, c d a yields a c cd.
Now look at the other small triangle and use bd bc
to get b 2 cd.
C
E
9.
a
B
b
a
A
c
b
D
x
F
Begin by constructing a second triangle, right
triangle DEF (with E a right angle), with legs of
lengths a and b and hypotenuse of length x. The
plan is to show that x c, so that the triangles
are congruent. Then show that C and E are
congruent. Once you show that C is a right angle,
then ABC is a right triangle.
L
10. H
Y
P
E
G
Use the Pythagorean Theorem to write expressions
for the lengths of the unknown legs. Show that the
expressions are equivalent. The triangles are
congruent by SSS or SAS.
11.
CA
Postulate
Segment Addition
Postulate
AA Similarity
Postulate
Parallel/Proportionality
Theorem
12.
Segment Duplication
Postulate
Parallel
Postulate
CA
Postulate
AA Similarity
Postulate
SAS
Congruence
Postulate
SSS Similarity
Theorem
13.
Perpendicular
Postulate
AA Similarity
Postulate
Right Angles
Are Congruent
Theorem
Three Similar
Right Triangles
Theorem
Pythagorean
Theorem
14. approximately 1.5 cm or 6.5 cm
15a. C
15b. A
15c. A
15d. A
15e. C
16a. The vectors are diagonals of your quadrilateral.
16b. A 180° rotation about the midpoint of the
common side; the entire tessellation maps onto
itself.
20a. CDG CFG by SAA; GEA GEB
by SAS; DGA FGB by the HL Theorem
CF
and DA
FB
by CPCTC; CD 20b. CD
DA CF FB (addition property of equality).
CB
, and ABC is isosceles.
Therefore, CA
20c. The figure is inaccurate.
20d. The angle bisector does not intersect the
perpendicular bisector inside the triangle as
shown, except in the special case of an isosceles
triangle, when they coincide.
21. 173 cm, 346 cm, 20 stones. Draw the trapezoid
and extend the legs until they meet to form a
triangle. Use parallel proportionality to find the
rise. The span is twice the rise. Use inverse tangent
to find the central angle measure. Divide into 180°
to find the number of voussoirs.
ANSWERS TO EXERCISES
155
Answers to Exercises
SAS Similarity
Theorem
SSS
Congruence
Postulate
17a. a 180° rotation about the midpoint of any
side
17b. possible answer: a vector running from each
vertex of the quadrilateral to the opposite vertex
(or any multiple of that vector)
18a. 33°
18b. 66°
18c. 57°
18d. 62°
18e. PSQ and PTQ are 90° by the Tangent
Theorem and so are supplementary. So, SPT
and SQT must also be supplementary by the
Quadrilateral Sum Theorem. Therefore, opposite
angles are supplementary, so it’s cyclic.
is a tangent, mPSQ 90°
18f. Because PS
(Tangent Theorem). Because mPSQ 90°, SQ
must be a tangent (Converse of the Tangent
Theorem).
3
19a. 1
1
2
2
3
19b. 6 4 1
USING YOUR ALGEBRA SKILLS 13
y
1. B(a, 0)
2. C(a b, c)
P (c, d )
2 b 2 , D b,
2 b2
3. C a b, a
a
4. possible answer:
y
Answers to Exercises
R (0, 0)
E (a, 0)
T (0, 0)
A (a – c, d )
R (a, 0)
x
00 0
slope of TR
a0 a 0
C (a, b)
T (0, b)
6. possible answer:
d0
d
slope of RA
a c a c
dd
0
slope of AP
a 2c a 2c 0
x
d0 d
slope of PT
c0 c
00 0
slope of RE
a0 a 0
b0 b
slope of CE
aa 0
(undefined)
bb 0
slope of TC
a0 a 0
b0 b
slope of TR
00 0
(undefined)
and AP
have the same slope and are parallel by
TR
the parallel slope property. So TRAP has only one
pair of parallel sides and is a trapezoid by definition.
Opposite sides have the same slope and are
therefore parallel by the parallel slope property.
Two sides are horizontal and two sides are vertical,
so the angles are all congruent right angles. RECT
is an equiangular parallelogram and is a rectangle
by definition.
5. possible answer:
PT (c 0
)2 (d
0)2 c 2 d 2
RA (a c
a)2
(d 0)2 (c)2 d2
2
2
c d
The nonparallel sides of the trapezoid have the
same length. So trapezoid TRAP is isosceles by
definition.
y
7.
U (0, a 3)
E
(–a, 0)
Q
(a, 0)
x
y
I (b, c)
Z b, c
2 2
(
T (0, 0)
Y a+b, c
2
2
)
(
X a,0
2
(
)
R (a, 0)
)
x
.
Let X be the midpoint of TR
a0 00
a
X 2, 2 2, 0
.
Let Y be the midpoint of RI
ab 0c
ab c
Y 2, 2 2, 2
.
Let Z be the midpoint of TI
b0 c0
b c
Z 2, 2 2, 2
, RI
, and TI
,
X, Y, and Z are the midpoints of TR
respectively, by the coordinate midpoint property.
, YZ
, and ZX
are midsegments by definition.
So XY
156
ANSWERS TO EXERCISES
Show: EQU is equilateral
EQ 2a
EU (a
0)2 a
(0 3 )2
2 3a 2
a
2
4a
2a
2 0a
UQ (a
0)
2
3
2 3a 2
a
4a 2
2a
EQ EU UQ
EQU is equilateral
8. Task 1: Given: A rectangle with both diagonals
Show: The diagonals are congruent
y
Task 2:
C (a, b)
T (0, b)
R (0, 0)
E (a, 0)
Task 2:
x
Task 3: Given: Rectangle RECT with diagonals RC
and TE
TE
Show: RC
Task 4: To show that two segments are congruent,
you use the distance formula to show that they
have the same length.
Task 5: RC (a 0
)2 (b
0)2 a2 b2
TE (a 0
)2 (0
b)2 a2 b2
I (c, d)
(
T (0, 0)
)
Y a+c, d
2
2
(
R (a, 0)
)
x
Task 3: Given: TRI and midsegment YZ
1
TR
and YZ TR
Show: YZ
2
Task 4: To show that two segments are parallel, use
the parallel slope property. The segments are
horizontal, so to compare lengths, subtract their
x-coordinates.
Task 5:
00 0
Slope of TR
a0 a 0
d d
0 0
2 2 Slope of YZ ac c
a
2
2
2
The slopes are the same, so the segments are parallel
by the parallel slope property.
TR a 0 a
ac c a 1
1
YZ 2 2 2 2a 2TR
y
P (b, c)
M
T (0, 0)
A (d, c)
N
R (a, 0)
x
One possible set of the coordinates for TRAP is
shown in the figure. By the coordinate midpoint
property, the coordinates of M are 2b, 2c and of N
ad c
are 2 , 2 .
Task 3: Given: Trapezoid TRAP with midsegment
MN
TR
Show: MN
Task 4: To show that the midsegment and bases
are parallel, you need to find their slopes.
Task 5:
slope of MN
c c
0
2
2 0
adb
ad b
2
2
2
00 0
slope of TR
a0 a 0
cc
0
is The slope of PA
d b d b 0. The slopes are
equal, therefore the lines are parallel.
11. Task 1: Given: A quadrilateral in which only
one diagonal is the perpendicular bisector of the
other
Show: The quadrilateral is a kite
y
Task 2:
I (0, b)
T (–a, 0)
K (a, 0)
M(0, 0)
x
E (0, c)
Task 3: Given: Quadrilateral KITE with diagonal
, which is the perpendicular bisector of diagonal
IE
TK
ANSWERS TO EXERCISES
157
Answers to Exercises
TE
because both segments have the same
So RC
length. Therefore the diagonals of a rectangle are
congruent.
9. Task 1: Given: A triangle with one midsegment
Show: The midsegment is parallel to and half the
length of the third side
y
Task 2:
Z c,d
2 2
So the midsegment is half the length of the third
side. Therefore the midsegment of a triangle is
parallel to the third side and half the length of the
third side.
10. Task 1: Given: A trapezoid
Show: The midsegment is parallel to the bases
Show: KITE is a kite
Task 4: To show that a quadrilateral is a kite, you
use the distance formula to show that only two
pairs of adjacent sides have the same length.
Task 5:
2 b2
KI (0
a
)2 (b
0)2 a
IT (0 (
a))2 (b 0)2 a2 b2
TE (0 (
a))2 (c 0)2 a2 c2
Answers to Exercises
EK (a
0
)2 (0
c)2 a 2 c 2
and IT
have the same length and
Adjacent sides KI
adjacent sides TE and EK have the same length,
and because ⏐b⏐ ⏐c⏐ the pairs are not equal in
length to each other. Therefore, KITE is a kite by
definition. Therefore, if only one diagonal of a
quadrilateral is the perpendicular bisector of the
other diagonal, then the quadrilateral is a kite.
12. Task 1: Given: A quadrilateral with midpoints
connected to form a second quadrilateral
Show: The second quadrilateral is a parallelogram
Task 2: y
D (d, e)
(
d, e
2 2
G
)
L b+d, c+e
2
2
A (b, c)
(
)
R a+b, c
2
2
(
Q (0, 0) P a , 0
2
( )
U (a, 0)
)
x
Task 3: Given: Quadrilateral QUAD with
midpoints P, R, L, and G
Show: PRLG is a parallelogram
Task 4: To show that a quadrilateral is a
parallelogram, we need to show that opposite
sides have the same slope.
c
c
0
2
2
c
Task 5: slope of PR
ab a
b
b
2
2
2
e
ce c
2
2
2
e
slope of RL
bd ab
da
da
2
2
2
e ce
c
2
2
2
c
slope of LG
b
d bd
b
2
2
2
e
e
0
2
2
e
slope of GP
d a
da
d
a
2 2
2
158
ANSWERS TO EXERCISES
and LG
have the same slope, and
Opposite sides PR
and GP
have the same slope. So
opposite sides RL
they are parallel by the parallel slope property, and
PRLG is a parallelogram by definition. Therefore
the figure formed by connecting the midpoints of
the sides of a quadrilateral is a parallelogram.
13. Task 1: Given: An isosceles triangle with the
midpoint of the base connected to the midpoint of
each leg, to form a quadrilateral
Show: The quadrilateral is a rhombus
Task 2: y
D a2 , h2
(
)
A (a, h)
F 32a, h2
(
)
x
B (0, 0) E (a, 0) C (2a, 0)
Task 3: Given: Isosceles triangle ABC with
midpoint of base, E, and midpoints of legs, D and
F, connected to form quadrilateral ADEF.
Show: ADEF is a rhombus
Task 4: You need to show that all the sides of
ADEF have the same length.
Task 5: By the distance formula,
AD 2
a
a 2
h
h 2
2
a
2
2
h
2
2
a 2 h 2
2
AF 3a
a
2
2
h
2 h
2
a
2
2
h
2
a 2 h 2
2
DE a
a 2
2
h
0 2
2
a
2
2
h
2
2
a 2 h 2
2
EF 3a
a
2
2
h
2 0
2
a
2
2
h
2
a 2 h 2
2
AD AF DE EF by the transitive property
of equality.
Therefore, ADEF is a rhombus by the definition of
a rhombus.
2
2
CHAPTER 13 REVIEW
140⬚
40⬚
20⬚
40⬚
140⬚
20⬚
20⬚
20⬚
20. False; x 360° 2b so x 90° only if
b 135°.
x
a
b
a
b
x ⫽ 540⬚ ⫺ (180⬚ ⫹ 2b)
x ⫽ 360⬚ ⫺ 2b
21. True (except in the special case of an isosceles
right triangle, in which the segment is not defined
because the feet coincide).
C
EAB DBA by the Isosceles
Triangle Theorem. AEB E
D
BDA by the definition of
X
altitude and by the Right Angles A
B
Are Congruent Theorem.
AB
by the reflexive property of congruence,
AB
BD
by
so AEB BDA by SAA. AE
CPCTC. By the definition of congruence, AC BC
and AE BD, so EC DC by the subtraction
property of equality and the Segment Addition
Postulate. By the division property of equality,
DC
EC
divides the sides of ABC
, so ED
BD
AE
AB
by the
proportionally. Therefore ED
Converse of the Parallel/Proportionality Theorem.
22. true
and
Given: Rhombus ROME with diagonals RM
EO intersecting at B
E
4 B
3
2
R
M
1
O
EO
Show: RM
Because diagonals bisect the angles in a rhombus,
the diagonals are angle bisectors.
Statement
Reason
1. 1 2
1. Rhombus Angles
Theorem
2. RO RE
2. Definition of rhombus
3. RO RE
3. Definition of
congruence
RB
4. RB
4. Reflexive property of
congruence
5. ROB REB
5. SAS Congruence
Postulate
6. 3 4
6. CPCTC
7. 3 and 4 are
7. Definition of linear
a linear pair
pair
8. 3 and 4 are
8. Linear Pair Postulate
supplementary
9. 3 and 4 are right 9. Congruent and
angles
Supplementary
Theorem
EO
10. RM
10. Definition of
perpendicular
ANSWERS TO EXERCISES
159
Answers to Exercises
1. False. The quadrilateral could be an isosceles
trapezoid.
2. true
3. False. The figure could be an isosceles trapezoid
or a kite.
4. true
5. False. The angles are supplementary but not
necessarily congruent.
6. False. See Lesson 13.5, Example B.
7. true
8. perpendicular
9. congruent
10. the center of the circle
11. four congruent triangles that are similar to the
original triangle
12. an auxiliary theorem proven specifically to
help prove other theorems
13. If a segment joins the midpoints of the
diagonals of a trapezoid, then it is parallel to the
bases.
14. Angle Bisector Postulate
15. Perpendicular Postulate
16. Assume the opposite of what you want to
prove, then use valid reasoning to derive a contradiction.
17a. Smoking is not glamorous.
17b. If smoking were glamorous, then this smoker
would look glamorous. This smoker does not look
glamorous, therefore smoking is not glamorous.
18. False. The parallelogram is a rhombus.
19. False. Possible counterexample:
Given: Isosceles ABC with
and BE
; AC
BC
; ED
altitudes AD
AB
Show: ED
23. true
D
E
C
2
1
3
Answers to Exercises
A
F
B
BAD DCB by the Opposite Angles Theorem.
m1 21 mBAD 21 mDCB m2 by
DC
by the
the definition of angle bisector. AB
definition of parallelogram. 2 and 3 are
supplementary by the Interior Supplements
Theorem.
1 and 3 are supplementary by the
substitution property.
FC
by the Converse of the Interior
AE
Supplements Theorem.
24. Use the Inscribed Angle Theorem, the addition
property, and the distributive property to get
mP mE mN mT mA 1
mTN mAT mPA mEP mNE . Because
2
there are 360° in a circle, mP mE mN mT mA 180°.
25. T
OP
by CPCTC. Use
SAA Theorem; thus DY
the Triangle Midsegment Theorem and the
substitution property to get TR 21 (ZO DY).
Also, use the Triangle Midsegment Theorem to
ZO
.
get TR
28a. The quadrilateral formed when the
midpoints of the sides of a rectangle are connected
is a rhombus.
G
C
28b. D
H
F
A
Use the Right Angles Are Congruent Theorem and
the SAS Congruence Postulate to get AEH BEF DGH CGF. Then use CPCTC to
prove that EFGH is a rhombus.
29a. The quadrilateral formed when the
midpoints of the sides of a rhombus are connected
is a rectangle.
I
29b.
P
O
H
Assume mH 45° and mT 45°. Use the
Triangle Sum Theorem, the substitution
property, and the subtraction property to get
mH mT 90°, which creates a
contradiction.Therefore mH 45° or mT 45°.
Y
26.
M
S
T
R
Use the definition of midpoint, the Segment
Addition Postulate, the substitution property, and
MY
1
YS
1
the division property to get TY 2 and YR 2 .
Then use the reflexive property and the SAS
Similarity Theorem to get MSY TRY.
Therefore, MS 21TR by CSSTP and the
TR
by
multiplication property and MS
CASTC and the CA Postulate.
D
Y
27.
1
3
T
R
42
Z
O
P
and DR
. Then
Use the Line Postulate to extend ZO
use the Line Intersection Postulate to label P as the
and DR
. DYR POR by the
intersection of ZO
160
M
Y
L
R
ANSWERS TO EXERCISES
B
E
X
J
N
K
By the Triangle Midsegment Theorem both PM
and ON are parallel to LJ , and both PO and MN
are parallel to IK . Because LJ and IK are
perpendicular, we can use corresponding angles
on parallel lines to prove that the lines that are
adjacent sides of the quadrilateral are also
perpendicular.
30a. The quadrilateral formed when the midpoints
of the sides of a kite are connected is a rectangle.
I
30b. By the Triangle Midsegment
P
M
Theorem both PM and ON are
L
J
, and both PO
and
parallel to LJ
are parallel to IK
. Because
MN
N
LJ and IK are perpendicular, we can O
use the CA Postulate to prove that
K
the lines that are adjacent sides of
the quadrilateral are also perpendicular.
C
31. Use the Line Postulate
and
to construct chords DB
. Then use the Inscribed
AC
B
Angles Intercepting Arcs
P
O
A
Theorem and the AA
Similarity Postulate to get
APC DPB. Therefore,
D
by CSSTP and the multiplication
property, AP PB DP PC.