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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