Answers to Exercises 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°. 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. 67.5135ANSWERS TO EXERCISES 61 Answers to Exercises Answers to Exercises CHAPTER 5 • CHAPTER 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°. 5 Number of sides of polygon 7 8 9 10 11 20 55 100 Sum of measures of angles 900° 1080° 1260° 1440° 1620° 3240° 9540° 17640° Number of sides of 5 6 7 8 9 10 12 16 100 equiangular polygon Measure of each angle 108° 120° 1284 7 ° 135° 140° 144° 150° 1571 2 ° 1762 5 ° of equiangular polygon 1. (Lesson 5.1) 2. (Lesson 5.1)
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Answers to Exercises13. 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°.
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.
67.5�
135�
ANSWERS TO EXERCISES 61
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Exe
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CHAPTER 5 • CHAPTER 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°.
5
Number of sides of polygon 7 8 9 10 11 20 55 100
Sum of measures of angles 900° 1080° 1260° 1440° 1620° 3240° 9540° 17640°
Number of sides of 5 6 7 8 9 10 12 16 100equiangular polygon
Measure of each angle 108° 120° 128�47
�° 135° 140° 144° 150° 157�12
�° 176�25
�°of equiangular polygon
1. (Lesson 5.1)
2. (Lesson 5.1)
62 ANSWERS TO EXERCISES
LESSON 5.2
1. 360°
2. 72°; 60°
3. 15
4. 43
5. a � 108°
6. b � 45�13
�°
7. c � 51�37
�°, d � 115�57
�°
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.
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°
16. Yes. �RAC � �DCA by SAS. AD�� � CR� by
CPCTC.
17. Yes. �DAT � �RAT by SSS. �D � �R by
CPCTC.
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1 a � b � 180�
2 c � d � 180�
4 a + b + c + d + e + f =
3 e � f � 180�Subtraction propertyof equality
Addition property of equality
6 b + d + f = �
�?
�?
�?
�?
�? �
�? �
�?
5 a + c + e =
Linear Pair Conjecture
Linear Pair Conjecture
Linear Pair Conjecture
540°
180°
360°
Triangle Sum Conjecture
11. (Lesson 5.2)
LESSON 5.3
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.
Given: Kite BENY with vertex angles �B and �N
Show: Diagonal BN� is the perpendicular bisector
of diagonal YE�.
From the definition of kite, BE�� BY�. From the
Kite Angle Bisector Conjecture, �1 � �2. BX��BX� because they are the same segment. By SAS,
�BXY � �BXE. So by CPCTC, XY�� XE�.
Because �YXB and �EXB form a linear pair, they
are supplementary, so m�YXB � m�EXB �180°. By CPCTC, �YXB � �EXB, or m�YXB �m�EXB, so by substitution, 2m�YXB � 180°, or
m�YXB � 90°. So m�YXB � m�EXB � 90°.
Because XY�� XE� and �YXB and �EXB are right
angles, BN� is the perpendicular bisector of YE�.
11. possible answer: �E � �I
12. possible answer:
The other base is ZI�. �Q and �U are a pair of base
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°.
B O
NE
S H
WI
F
E
BN
W O
S H
ANSWERS TO EXERCISES 63
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9. (Lesson 5.3)
�? and �?
6�1 � and5
�3 ��? Congruence
shortcutDefinition ofangle bisector
4
�?
Given
1 BE � BY
Given
2 EN � YN
Same segment
3�? � �?
�BEN � �? �
?
�?
BN�� BN�
�BYN �2 BN� bisects �B, BN� bisects �N
�4
CPCTCSSS
64 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.
12. Explanations will vary.
80
40 60
60 cm
Cabin
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
EZ�� �38
�, and the slope of YT�� �38
�
18.
There is only one kite, but more than one way to
construct it.
K
R
NF
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�FOA withmidsegment LN
Given
LN � OA3
4
1
Triangle MidsegmentConjecture
�IOA withmidsegment RD
Given
2
5
Two lines parallel to the same line are parallel
�
?�
?
�
?
OA� � RD��
LN� � RD��Triangle Midsegment Conjecture
8. (Lesson 5.4)
LESSON 5.5
1. 34 cm; 27 cm
2. 132°; 48°
3. 16 in.; 14 in.
4. 63 m
5. 80
6. 63°; 78°
7.
8.
9. � Vh
� Vw
D
P
P
O
R
R
T S
AL
10.
11. (b � a, c)
12. possible answer:
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.
15. a � 135°, b � 90°
16. a � 120°, b � 108°, c � 90°, d � 42°, e � 69°
a
a
b
b
c
c
d
d
� Vb
� Vc
ANSWERS TO EXERCISES 65
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ET � NT
EA � LN
2
LEAN is a parallelogram 7
CPCTC
1
AIA Conjecture
3 �AEN � �LNE
�?
�?
4
AIA Conjecture
�? 6
ASA
�?
8
CPCTC
�?
9
Definition of segment bisector
�?
5
Opposite sides congruent
�?
Given �EAL � �NLA �AET � �LNT
AE�� LN�AT�� LT�
EN� and LA� bisect
each other
Definition of
parallelogram
13. (Lesson 5.5)
66 ANSWERS TO EXERCISES
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.
21. Parallelogram. Because the triangles are
congruent by SAS, �1 � �2. So, the segments are
parallel. Use a similar argument to show that the
other pair of opposite sides is parallel.
22. Kite or dart. Radii of the same circle are
congruent. If the circles have equal radii, a rhombus
is formed.
1
2
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USING YOUR ALGEBRA SKILLS 5
1.
2.
3. y
x
(2, 9)
(0, 6)
y
x
(3, 8)
(0, 4)
y
x(0, 1)
(1, –1)
ANSWERS TO EXERCISES 67
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4. y � �x � 2
5. y � ��163� x � �
71
43�
6. y � x � 1
7. y � �3x � 5
8. y � �25
� x � �85
�
9. y � 80 � 4x
10. y � �3x � 26
11. y � ��14
� x � 3
12. y � �65
� x
13. y � x � 1
14. y � � �29
� x � �493�
68 ANSWERS TO EXERCISES
LESSON 5.6
1. Sometimes true; it is true only if the
parallelogram is a rectangle.
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.
5. Always true by the Square Diagonals Conjecture.
6. Sometimes true; it is true only if the rhombus is
equiangular.
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.
false
true
falsetrue
false
true
17.
18. Constructions will vary.
19. one possible construction:
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
AC�� BD�. AC� and BD� bisect each other
Show: ABCD is a rectangle
Because the diagonals are congruent and bisect
each other, AE�� BE�� DE�� EC�. Using the
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, m�DAB � m�1 � m�8.
By the addition property of equality, m�1 �m�8 � m�2 � m�3 � m�5 � m�4 � m�6 �m�7. So m�DAB � m�ABC � m�BCD �m�CDA. So the quadrilateral is equiangular. Using
�1 � �5 and the Converse of AIA, AB� � CD�.
Using �3 � �7 and the Converse of AIA,
BC� � AD��. Therefore ABCD is an equiangular
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).
A
E
B
CD
81 2
3
456
7
I E
P S
A
B E
K A
B
K
E
E V
L O
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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 f lowchart below.
27. Yes, it is true for rectangles.
Given: �1 � �2 � �3 � �4
Show: ABCD is a rectangle
By the Quadrilateral Sum Conjecture, m�1 �m�2 � m�3 � m�4 � 360°. It is given that all
four angles are congruent, so each angle measures
ANSWERS TO EXERCISES 69
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90°. Because �4 and �5 form a linear pair,
m�4 � m�5 � 180°. Substitute 90° for m�4 and
solve to get m�5 � 90°. By definition of congruent
angles, �5 � �3, and �5 and �3 are alternate
interior angles, so AD�� � BC� by the Converse of the
Parallel Lines Conjecture. Similarly, �1 and �5 are
congruent corresponding angles, so AB� � CD� by the