48 ANSWERS TO EXERCISES
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Answers to Exercisesangles P and D can be drawn at each endpoint
using the protractor.
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 m�A � m�B � m�C � 180°, and m�D �m�E � m�F � 180°. By the transitive property,
m�A � m�B � m�C � m�D � m�E � m�F.
You also know that m�A � m�D, and m�B �m�E. You can substitute for m�D and m�E in the
longer equation to get m�A � m�B � m�C �m�A � m�B � m�F. Subtracting equal terms
from both sides, you are left with m�C � m�F.
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
21. false
22. false
23. false
24. true
25. eight; 100
85�
55�
40�P D7 cm
Q
CHAPTER 4 • CHAPTER CHAPTER 4 • CHAPTER
LESSON 4.1
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°
10.
11.
12. First construct �E, using the method used in
Exercise 10.
13.
14. From the Triangle Sum Conjecture
m�A � m�S � m�M � 180°. Because �M is a
right angle, m�M � 90°. By substitution,
m�A � m�S � 90° � 180°. By subtraction,
m�A � m�S � 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
�D is 85°. Then PD� can be drawn to be 7 cm, and
E
RA�L
�G
Fold
�M
�R
�A
E
RA
�L
�G
�M �A
�R
4
LESSON 4.2
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.
16.
17. possible answer:
18. perpendicular
19. parallel
20. parallel
21. neither
22. parallelogram
23. 40
24. New: (6, �3), (2, �5), (3, 0). Triangles are
congruent.
25. New: (3, �3), (�3, �1), (�1, �5). Triangles
are congruent.
0 8 16 24 32 40
Fold 2
Fold 1
Fold 4
Fold 3
105� 60�
45�
M N
G
K
H
P
ED
BA
F
C
ANSWERS TO EXERCISES 49
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50 ANSWERS TO EXERCISES
USING YOUR ALGEBRA SKILLS 4
1. false 2. true
3. not a solution 4. solution
5. not a solution 6. x � 7
7. y � �4 8. x � �8
9. x � 4.2 10. n � ��12
�
11. x � 2 12. t � 18
13. n � �25
� 14a. x � �49
�
14b. x � �49
�; 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:
�a�b
� � �dc
�
bd��ab
�� � bd��dc
���ab
bd
� � �b
dcd�
ad � bc
The method of “clearing fractions” results in the
method of “cross multiplying.”
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.
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°.
2x
2x
x
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LESSON 4.3
1. yes
2. no
3. no
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.
5 6
12
4 5
9
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
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52 ANSWERS TO EXERCISES
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:
3. Answers will vary. Possible answer:
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
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
19. The midpoint of SD� and PR� is (0, 0).
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.
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.
26a. y � 6 b. y � �133� c. y � ��
34
�x � 2
27. (�5, �3)
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Side length 1 2 3 4 5 … n … 20
Elbows 4 4 4 4 4 4 4
T’s 0 4 8 12 16 76
Crosses 0 1 4 9 16 361
25. (Lesson 4.4)
4n � 4 (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:
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.
19. Slope AB�� slope CD�� � 3 and slope BC��slope DA�� � ��
13
�, so AB� � BC�, CD�� � DA��, and
BC� � DA��. �ABC � �CDA by SAA.
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.)
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.
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.
0
0 100 400
50 100 200
Sacramento
CA
LI F
OR
NI A
N E VA D A U T A H
A R I Z O N A
I D A H O
O R E G O N
San Francisco
Eureka
Reno
Las Vegas
Elko
Boise
Los AngelesMiles
Kilometers
K
L
M
ANSWERS TO EXERCISES 53
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54 ANSWERS TO EXERCISES
LESSON 4.6
1. Yes. BD�� BD� (same segment), �A � �C
(given), and �ABD � �CBD (given), so �DBA ��DBC by SAA. � AB�� CB� by CPCTC.
2. Yes. CN�� � WN�� and �C � �W (given), and
�RNC � �ONW (vertical angles), �CNR ��WON by ASA. � RN�� � ON�� by CPCTC.
3. Cannot be determined. The congruent parts
lead to the ambiguous case SSA.
4. Yes. �S � �I, �G � �A (given), and TS�� IT�(definition of midpoint), so �TIA � �TSG by
SAA. � SG�� IA� by CPCTC.
5. Yes. FO�� FR� and UO�� � UR�� (given), and UF�� UF� (same segment), so �FOU � �FRU by SSS.
� �O � �R by CPCTC.
6. Yes. MN��� MA�� and ME�� � MR�� (given), and
�M � �M (same angle), so �EMA � �RMN by
SAS. � �E � �R by CPCTC.
7. Yes. BT�� EU� and BU�� ET� (given), and
UT�� UT� (same segment), so �TUB � �UTE by
SSS. � �B � �E by CPCTC.
8. Cannot be determined. �HLF � �LHA by
ASA, but HA�� and HF� are not corresponding sides.
9. Cannot be determined. AAA does not guaran-
tee congruence.
10. Yes. The triangles are congruent by SAS.
11. Yes. The triangles are congruent by SAS, and
the angles are congruent by CPCTC.
12. Draw AC� and DF� to form �ABC and �DEF.
AB�� CB�� DE�� FE� because all were drawn with
the same radius. AC�� DF� for the same reason.
�ABC � �DEF by SSS. Therefore, �B � �E by
CPCTC.
13. cannot be determined
14. KEI by ASA
15. UTE by SAS
16.
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
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Number of sides 3 4 5 6 7 … 12 … n
Number of struts needed 0 1 2 3 4 … 9 … n � 3to make polygon rigid
21. (Lesson 4.6)
LESSON 4.7
1. See flowchart below.
2. See flowchart below.
3. See flowchart below.
4. See flowchart below.
ANSWERS TO EXERCISES 55
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1 SE � SU
�
?
2�E � �U
�
?
4 � �
? � � �
?
ASA CongruenceConjecture
5
3�1 � �2
�
?
MS � SO
�
?
Given
Given CPCTC
Vertical Angles Conjecture
�ESM � �USO
1 I is midpoint of CM
Given
CI � IM
3
Definition of midpoint
2 I is midpoint of BL
�
?�
?
CPCTC
7
5 �1 � �2
�
?
6
�
?
� �
? � � �
?IL � IB 4
�
?
3 NS � NS
WS � SE
WN � NE
4
1 NS is a median
S is a midpoint2 6 7
5
�
?
�
?�
?
Given Same segment
Definitionof median
Definitionof midpoint
�W � � �
?�WSN � � �
?
2
Definition of �?
NS � NS
3�W � �E
�1 � � �
?
�
?
1 NS is anangle bisector
Given
5 � �
? � � �
? 6
4
�
?
�
?�
?
WN � NE
�
?
7 �NEW isisosceles
Given
Vertical Angles Conjecture
Definition of
midpoint�CIL � �MIB
by SAS
CL�� MB��
Given
�ESN
SSS CPCTC
�E
Given CPCTCSAA
�WNS � �ENS
Definition of
isosceles triangle
Same segment
�2
angle bisector
1. (Lesson 4.7)
2. (Lesson 4.7)
3. (Lesson 4.7)
4. (Lesson 4.7)
56 ANSWERS TO EXERCISES
5. See flowchart below.
6. Given: �ABC with BA�� BC�, CD�� � AD��
Show: BD� is the angle bisector of �ABC.
7. The angle bisector does not go to the midpoint
of the opposite side in every triangle, only in an
isosceles triangle.
8. NE�, because it is across from the smallest angle
in �NAE. It is shorter than AE�, which is across
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.
Given Given
BA � BC CD � AD BD � BD
Same segment
�ABD � �CBD
BD bisects �ABC
SSS
�1 � �2
CPCTC
→
Definition of anglebisector
12B
A
D
C
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.
16. �27
�
17.
18. y
x
X
B O
y
4
X'O'
Y'B' 4
Y
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3
SN � SN
4
�3 � �41
�?
�?
�? 6 � �
? � � �? 7
5
�?
�?
�?
SA � NE
2
�?
SE � NA
AIA Conjecture
Same segment
Given
Given AIA Conjecture
1 � �2 SA � NE�
ASA
�ESN � �ANS
This proof shows that in a parallelogram,
opposite sides are congruent.
CPCTC
5. (Lesson 4.7)
LESSON 4.8
1. 6
2. 90°; 18°
3. 45°
4. See flowchart below.
5. See flowchart below.
6. 1 Isosceles �ABC
with AC � BC
and CD bisects
�C
Given
2 �ADC � �BDC
Conjecture A(Exercise 4)
3 AD � BD
CPCTC
4 CD is a median
Def. of median
7.
8. Yes. First show that the three exterior triangles
are congruent by SAS.
5 �ADC � �BDC
1 AC � BC
Given
4 AD � BD
Def. of midpoint
Def. of angle bisector
9 CD � AB
Def. of altitude
SSS
6 �ACD � �DCB
CPCTC
2 CD � CD
Same segment
3 D is
midpoint
of AB
Given
7 CD is angle
bisector of �ACB
8 CD is altitude
of �ABC
Conjecture B (Exercise 5)
ANSWERS TO EXERCISES 57
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2
35 �ADC � �BDC
4
�
?
�ABC is isosceleswith AC � BC
1 CD is the bisector of �C
Given Definition of angle bisector
�
? Given
Same segment
CD � CD
2 �ADC � �BDC
1 �ABC is isosceles with AC � BC, and CD is the bisector of �C
Given Conjecture A
Congruent supplementary angles are 90�
4 �1 � �2
�
?
8 �
?
Definition of altitude
7
�
?
3 �1 and �2 form a linear pair
Definition of linear pair
5 �1 and �2 are supplementary
6 �1 and �2 are right angles
Linear Pair Conjecture
CD � AB ���� ����
SAS
�1 � �2
CPCTC
Definition of
perpendicular
CD� is an altitude
4. (Lesson 4.8)
5. (Lesson 4.8)
58 ANSWERS TO EXERCISES
9.
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.
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
30�
C
A D B
14. y � ��53
� x � 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.
18. CnH2n
CC C
CC
C C
H
HH
H
HH
H
HH
HHHH
H
Fireworks
Duane
Hugo
1.5 km
340 m/s • 5 s= 1.7 km
340 m/s • 3 s= 1.02 km
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CHAPTER 4 REVIEW
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
16. �AMD � �UMT by SAS, AD�� � UT� by
CPCTC
17. cannot be determined
18. cannot be determined
19. �TRI � �ALS by SAA, TR�� AL� by
CPCTC
20. �SVE � �NIK by SSS, EL�� KV�� by
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 m�A � m�N � m�G � 188°. The
measures of the three angles of a triangle must sum
to 180°.
26. �ROW � �NOG by ASA, implying that
OW��� OG��. However, the two segments shown are
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.
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.
z
xx
P B
S2
S1
A Ly
P
�P�A
�L
ANSWERS TO EXERCISES 59
An
swe
rs to E
xercise
s
60 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.
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
An
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o E
xerc
ise
s
2
3 5 � �
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4
6 7� �
? � � �
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1�
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? �
? �
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�
? � �
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�
? � �
? M is midpoint
of TE� and IR�
Given
�TMI � �RME
Vertical angles
TM�� � ME��
Definition
of midpoint
�TMI � �EMR
SAS
TI� � RE�
Converse of
AIA Conjecture
�T � �Eor
�R � �I
CPCTC
IM�� MR��
Definition of midpoint
35. (Chapter 4 Review)