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70. equilateral triangle
9. 14; The sum of the interior angle measures of the
71. acute isosceles triangle
polygon is 2 (90°) + 3(180°) + 1440° = 2160°.
72. right scalene triangle
So, the polygon has
73. obtuse scalene triangle
10. a. 720°
74. 55°; obtuse scalene triangle
2160°
+ 2 = 14 sides.
180°
b. 135
7.1 Enrichment and Extension
75. 76°; acute isosceles triangle
1. 61°
2. 130°
3. 58°
4. 50°
76. 24°; right scalene triangle
5. 84°
6. 85°
7. 146°
8. 145°
77. 95°; obtuse scalene triangle
9. regular decagon
78. 33°
79. 52°
80. 76°
Chapter 7
7.1 Start Thinking
1. 540°
2. 720°
3. 900°
2. 70
3. 119
7.1 Warm Up
1. 120
7.1 Cumulative Review Warm Up
2. y = −1
1. x = 1
3. y + 1 =
2
( x − 5)
3
4. y −
11
14
=
( x + 1)
2
5
7.1 Practice A
2. 19-gon
1. 900°
3. interior: 168°, exterior: 12°
4. 84
5. 125
6. m ∠ X = m ∠Y = 75°
7. m ∠ X = m ∠Y = 135°
8. 76
9. 88
11. 144 people
10. 120°
2. 68
d + e + f = 180°, and g + h + i = 180°
because the sum of the interior angles of a triangle
equals 180°. You can add those three equations to
obtain a + b + c + d + e + f + g + h + i = 540°.
m ∠YZV = f + i, m ∠ ZVW = h,
m ∠VWX = g + d + a, m ∠WXY = b, and
m ∠ XYZ = c + e, so
m ∠YZV + m ∠ ZVW + m ∠VWX + m ∠WXY +
m ∠ XYZ = 540°.
7.1 Puzzle Time
THEY DIDN’T WANT TO WAIT FORTY YEARS
FOR A TRAIN
7.2 Start Thinking
yes; Sample answer: The scout could use the Pythagorean
Theorem to determine the distance that should be between
opposite corner posts, the length of the hypotenuse. It
7 inches.
should be approximately 15 feet 7 16
Or, the
scout could make sure that the distances between the
two pairs of opposite corners are the same and not be
concerned about the exact measure. This method uses
the SSS Congruence Theorem (Thm. 5.8).
7.2 Warm Up
REASONS
1. MN ≅ PO ,
1. Given
NO ≅ MP
3. m ∠ X = m ∠Y = 116°
4. m ∠ X = m ∠Y = 130°
5. 56
11. You know that a + b + c = 180°,
1. STATEMENTS
7.1 Practice B
1. 103
10. 12 sides
6. 55
7. interior: 165°, exterior: 15°
2. NP ≅ NP
2. Reflexive Property of
Segment Congruence
(Thm. 2.1)
3. PMN ≅
3. SSS Congruence
Theorem (Thm. 5.8)
NOP
8. 20
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2. STATEMENTS
REASONS
11 − x
= 9 − 7x
5
11 − x = 5(9 − 7 x)
1. Given
1. AB ≅ CD,
AB ⊥ BD,
CD ⊥ BD
2. Reflexive Property of
Segment Congruence
(Thm. 2.1)
2. BD ≅ BD
34 x = 34
4. ∠ ABD ≅ ∠CDB
4. Right Angles
Congruence
Theorem (Thm. 2.3)
x =1
ABD ≅ CDB
5. SAS Congruence
Theorem (Thm. 5.5)
6. Corresponding
parts of congruent
triangles are
congruent.
7.2 Cumulative Review Warm Up
13
;
2
2 x − 8 = 5 + 4 x Write the equation.
−13 = 2 x
−
2.
13
= x
2
Subtraction Property of
Equality
1
(3x + 8) = 2 x − 3
2
3
x + 4 = 2x − 3
2
1
4 = x −3
2
1
7 = x
2
14 = x
Addition Property of
Equality
Subtraction Property of
Equality
Division Property of
Equality
1. x = 14, y = 40
2. a = 10, b = 37
3. u = 62, v = 59
4. s = 9, t = 14
5. ( 2, 1)
6. C ( 2, 0)
9. no; The side lengths of the parallelograms may not
be congruent.
10.
STATEMENTS
REASONS
1. PQRS is a
parallelogram.
1. Given
2. PQ ≅ SR
2. Parallelogram
Opposite Sides
Theorem
(Thm. 7.3)
3. QT ≅ TS
3. Parallelogram
Diagonals
Theorem
(Thm. 7.6)
4. PT ≅ TR
4. Parallelogram
Diagonals
Theorem
(Thm. 7.6)
Write the equation.
Distributive Property
Subtraction Property
of Equality
5.
Addition Property
of Equality
7. B (1, 1)
8. Two angles are 50°, and two angles are 130°.
Division Property of Equality
x = 14;
Distributive Property
7.2 Practice A
x = −
− 8 = 5 + 2 x Subtraction Property of
Equality
Multiplication Property
of Equality
11 + 34 x = 45
3. Perpendicular lines
form right angles.
6. AD ≅ BC
Write the equation.
11 − x = 45 − 35 x
3. ∠ ABD and ∠CDB
are right angles.
5.
1.
x = 1;
3.
PQT ≅ RST
5. SSS Congruence
Theorem
(Thm. 5.8)
Multiplication Property
of Equality
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A77
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7.2 Practice B
e. sometimes; when the parallelogram is a square
1. x = 11, y = 8
2. u = 66, v = 38
3. a = 7, b = 42
4. c = 15, d = 48
5. (0, 4 )
6. C ( 2, − 2 )
7.
STATEMENTS
REASONS
1. CEHF is a
parallelogram.
1. Given
2. CE ≅ FH
2. Parallelogram
Opposite Sides
Theorem
(Thm. 7.3)
3. CE = FH
3. Definition of
segment
congruence
4. D bisects CE .
G bisects FH .
1
CE
2
1
GH = FH
2
5. CD =
f. never; The angles are supplementary by the
Consecutive Interior Angles Theorem
(Thm. 3.4).
7.2 Enrichment and Extension
1. ( − 9, − 7 ),
2. ( 2, − 2 ),
(5, − 7 ), ( −1, 5)
(− 4, 6), (8, 4 )
3. ( a + 2, b + 3),
(
) (a, 2b − b2 ), (2a2
5.
7. CD ≅ GH
7. Definition of
segment
congruence
8. CF ≅ EH
8. Parallelogram
Opposite Sides
Theorem
(Thm. 7.3)
9. ∠ C ≅ ∠ H
9. Parallelogram
Opposite Angles
Theorem
(Thm. 7.4)
10. CDF ≅ HGE
10. SAS Congruence
Theorem
(Thm. 5.5)
6.
Number of
diagonals (d)
3
0
4
2
5
5
6
9
7
14
n( n − 3)
− 2, b − 3)
)
8. 21 sides
7. 35; 65
2
(a
− a, b2
Number of
sides (n)
5. Definition of
segment bisector
6. Substitution
+ 6, b + 3),
4. a, b2 ,
4. Given
6. CD = GH
(a
9. a. Sample answer:
Player 1
Player 2
Player 6
Player 3
Player 5
Player 4
b. 15 total games
8. a. always; Parallelogram Opposite Sides Theorem
(Thm. 7.3)
b. sometimes; when the parallelogram is a square
c. sometimes; when the parallelogram is a square
d. always; Parallelogram Opposite Angles Theorem
(Thm. 7.4)
c. n +
n( n − 3)
2
=
n( n − 1)
2
7.2 Puzzle Time
ON THE BEACH
7.3 Start Thinking
1. no; Sample answer: AB ≅ BC ≅ CD
A78
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2. no; Sample answer: ∠ E or ∠ H is a right angle.
5.
4
3. no; Sample answer: ∠ X ≅ ∠ Z
1.
4
3
2.
34
4.
45 = 3 5
−4
C
−2
2
A
3. 2
5. −
2
B
7.3 Warm Up
y
3
5
D
−4
6. 5
Because BC = AD = 6, BC ≅ AD. Because
both BC and AD are horizontal line segments,
their slope is 0, and they are parallel. BC and AD
are opposite sides that are both congruent and
parallel. So, ABCD is a parallelogram by the
Opposite Sides Parallel and Congruent Theorem
(Thm. 7.9).
7.3 Cumulative Review Warm Up
1. original: If a triangle is right, then it contains two
acute angles; converse: If a triangle contains two
acute angles, then it is a right triangle; inverse: If a
triangle is not right, then it does not contain two
acute angles; contrapositive: If a triangle does not
contain two acute angles, then it is not a right
triangle; The original and contrapositive are true.
The converse and inverse are false.
6.
y
F
4
2. original: If two lines have the same slope, then they
are parallel; converse: If two lines are parallel, then
they have the same slope; inverse: If two lines do
not have the same slope, then they are not parallel;
contrapositive: If two lines are not parallel, then
they do not have the same slope; All statements
are true.
3. original: If there is ice on the road, then I will not
go shopping; converse: If I do not go shopping,
then there is ice on the road; inverse: If there is not
ice on the road, then I will go shopping;
contrapositive: If I go shopping, then there is not
ice on the road; The original and contrapositive are
true. The converse and inverse are false.
7.3 Practice A
1. Parallelogram Opposite Angles Converse Theorem
4 x
2
E
−4
G
−2
−2
−4
−6
−8
−10
x
−2
−4
H
−6
Because EF = GH = 5 and EH = FG = 13,
EF ≅ GH and EH ≅ FG. Because both pairs of
opposite sides are congruent, quadrilateral EFGH is
a parallelogram by the Parallelogram Opposite
Sides Converse (Thm. 7.7).
(Thm. 7.8)
2. Parallelogram Diagonals Converse Theorem
(Thm. 7.10)
3. 12
4. 4
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Geometry
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A79
Answers
7. STATEMENTS
1. ∠ A ≅ ∠ ABE
REASONS
5.
X
4
y
1. Given
2. AE ≅ BE
2. Base Angles
Theorem (Thm. 5.6)
3. AE ≅ CD
3. Given
4. BE ≅ CD
4. Transitive Property
of Segment
Congruence
Theorem (Thm. 2.1)
5. BC ≅ DE
5. Given
6. BCDE is a
parallelogram.
6. Parallelogram
Opposite Sides
Converse
Theorem (Thm. 7.7)
8. a. Because AB || CD and AB ≅ CD, ABDE is a
parallelogram by the Opposite Sides Parallel and
Congruent Theorem (Thm. 7.9).
−4
−2
2
7.3 Practice B
1. Opposite Sides Parallel and Congruent Theorem
4 x
W
Z
−4
Because WX = YZ = 5, WX ≅ YZ . Because
both WX and YZ are vertical line segments, their
slope is undefined, and they are parallel. XW and
YZ are opposite sides that are both congruent and
parallel. So, WXYZ is a parallelogram by the
Opposite Sides Parallel and Congruent Theorem
(Thm. 7.9).
6.
4
2
b. Because ABDC is a parallelogram, CE || DF .
From the diagram, you can see that CD || EF .
Because the opposite sides are parallel, CDFE is
a parallelogram.
c. no; You are only given that one pair of opposite
sides are parallel, which is not enough
information to prove that it is a parallelogram.
d. m ∠ ACD = 35°, m ∠ DCE = 145°,
m ∠ CEF = 35°, m ∠ EFD = 145°
Y
2
y
B
A
−4
2
−2
−2
D
4
x
C
−4
Because AD = BC = 18 and
AB = CD = 40, AD ≅ BC and AB ≅ CD.
Because both pairs of opposite sides are congruent,
quadrilateral ABCD is a parallelogram by the
Parallelogram Opposite Sides Converse (Thm. 7.7).
(Thm. 7.9)
2. Parallelogram Opposite Sides Converse Theorem
(Thm. 7.7)
3. 11
A80
4. 35
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7. STATEMENTS
REASONS
7.3 Enrichment and Extension
1. ∠ A ≅ ∠ FDE
1. Given
1. yes
2. yes
3. no
4. yes
2. AB || CD
2. Alternate Interior
Angles Converse
(Thm. 3.6)
5. no
6. yes
7. yes
8. yes
3. ∠ AFB ≅ ∠ DFE
3. Vertical Angles
Congruence
Theorem
(Thm. 2.6)
4. F is the midpoint
of AD.
4. Given
5. AF ≅ DF
5. Definition of
midpoint
6.  ABF ≅ DEF
6. ASA Congruence
Theorem
(Thm. 5.10)
7. AB ≅ ED
7. Corresponding
sides of congruent
triangles are
congruent.
8. D is the midpoint
of CE.
8. Given
9. ED ≅ CD
9. Definition of
midpoint
10. AB ≅ CD
10. Transitive
Property of
Segment
Congruence
(Thm. 2.1)
11. ABCD is a
parallelogram.
11. Opposite Sides
Parallel and
Congruent
Theorem
(Thm. 7.9)
9.
1
11
10. G ( − 4, 1), H (1, 13)
11. yes; It is given that PQRS and QTSU are
parallelograms. Because the diagonals of a
parallelogram bisect each other, PX = RX and
TX = UX . Because PR and TU are diagonals of
PTRU that bisect each other, PTRU is a
parallelogram.
12. You would need to show that one angle is
supplementary to each consecutive angle.
7.3 Puzzle Time
FRIENDSHIP
7.4 Start Thinking
Sample answer: For both the square and the rhombus:
The diagonals bisect each other, the diagonals form
right angles, the diagonals form congruent triangles,
and the opposite angles are congruent. For the square
only: The diagonals are congruent.
7.4 Warm Up
1. 51°
2. 39°
3. 90°
4. 33°
5. 33°
6. 22°
7.4 Cumulative Review Warm Up
1. sometimes; An isosceles triangle could also be an
acute or an obtuse triangle.
2. sometimes; A right triangle could also be an
isosceles triangle.
3. always; An equilateral triangle will always have
8. no; You cannot determine if a quadrilateral is a
parallelogram by only knowing the values of the
angles. You must also know the orientation of the
angles, and whether the congruent angles are
opposite of each other.
9. a. m ∠ FCG = 135°, m ∠ BCF = 45°,
m ∠ D = 135°
b. Parallelogram Opposite Angles Converse
Theorem (Thm. 7.8)
c. 8x
three 60° angles.
4. never; A right triangle always has one side that is
the longest, so it cannot be equilateral.
7.4 Practice A
1. 90°
2. 23°
3. 67°
5. 12
6. 19
7. 22
4. 5
8. rectangle; The sides are perpendicular and not
congruent.
9. rectangle, rhombus, square; The diagonals are
congruent and perpendicular.
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Geometry
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A81
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10. STATEMENTS
REASONS
1. PSUR is a
rectangle.
1. Given
2. m∠U = 90°
m∠ P = 90°
2. Definition of a
rectangle
3. ∠U ≅ ∠ P
3. Transitive Property
of Angle
Congruence
(Thm. 2.2)
4. PS ≅ RU
4. Definition of a
rectangle
5. PQ ≅ TU
5. Given
6. PQS ≅ UTR
6. SAS Congruence
Theorem (Thm. 5.5)
7. QS ≅ RT
7. Corresponding parts
of congruent
triangles are
congruent.
9. STATEMENTS
REASONS
1. ∠ XWY ≅ ∠ XYW
1. Given
2. XW ≅ XY
2. Converse of the
Base Angles
Theorem
(Thm. 5.7)
3. WXYZ is a
parallelogram.
3. Given
4. XY ≅ WZ
4. Definition of a
parallelogram
WX ≅ YZ
5. Transitive
Property of
Segment
Congruence
(Thm. 2.1)
5. WX ≅ XY ≅
11. a. It is a rectangle; By definition, all four angles are
right angles.
b. It is a rhombus; By definition, all four sides are
congruent.
c. It is a square; By definition, all four sides are
congruent and all four angles are right angles.
d. 90°
e. 45°
7.4 Practice B
1. rhombus; It has four congruent sides, but it does not
have four right angles.
YZ ≅ WZ
6. WXYZ is a rhombus.
6. Definition of a
rhombus
10. no, Because a similarity transformation maintains
the shape of an object, the corresponding angles
remain congruent. A rhombus may not have all
right angles, but a square always will.
11. yes; The quadrilateral is a rectangle or square,
which are both parallelograms.
12. no; Because the quadrilateral is not a rectangle, the
other two angles are not 90°. So, the opposite
angles are not congruent and the quadrilateral is
not a parallelogram by the contrapositive of the
Parallelogram Opposite Angles Theorem
(Thm. 7.4).
13. yes; If the rectangle is a square, the side lengths of
2. square, rectangle, rhombus; Any square is also a
rectangle and a rhombus.
3. 90°
4. 37°
5. 53°
6. 16
7. 24
8. ( 2, 5)
the triangle will be congruent. So, the triangles will
be isosceles.
7.4 Enrichment and Extension
1. no; Sample answer: If the diagonals of a
parallelogram are congruent, then it would have to
be a rectangle and have a right angle.
2. yes; Sample answer: If there are congruent
diagonals in a parallelogram, it can be a rectangle
or square with two opposite sides 2 centimeters
long.
3. no; Sample answer: In a parallelogram, consecutive
angles must be supplementary, so all angles must
be right. This would make it a rectangle.
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4. 9
5. 10
5 2
7.
7
2
6. 40
2
8. 16°
8. STATEMENTS
1. ABCD is a
parallelogram
1. Given
2. AB  DC
2. Definition of a
parallelogram
3. AE ≅ AD
3. Given
4. ∠ E ≅ ∠ ADE
4. Base Angles
Theorem
(Thm. 5.6)
5. ∠ ADE ≅ ∠C
5. Definition of a
parallelogram
6. ∠ E ≅ ∠C
6. Transitive
Property of
Angle
Congruence
(Thm. 2.2)
7. ABCE is an
isosceles trapezoid.
7. Isosceles
Trapezoid Base
Angles Converse
(Thm. 7.15)
9. Sample answer: Let parallelogram DFGH have
(
(
G − a, −
)
b 2 − a 2 , F (b, 0),
vertices D a,
)
b 2 − a 2 , and H ( − b, 0), respectively.
The slope of both HG and DF is
b2 − a2
, and
a −b
the slope of both HD and GF is
b2 − a2
. The
a +b
products of the slopes of the pairs HG and GF ,
GF and DF , DF and HD , and HD and HG
are all equal to −1, making each pair of consecutive
segments perpendicular and each angle a right
angle. So, parallelogram DFGH is a rectangle.
REASONS
7.4 Puzzle Time
ALL THE ANGLES
9. a. 90°
7.5 Start Thinking
Sample answer: BD is a perpendicular bisector of
AC . AB ≅ BC , AD ≅ CD, ∠ BAD ≅ ∠ BCD, and
m∠ ABC > m∠ ADC.
c. ∠ XWZ
7.5 Practice B
1. Slope of TU = slope of VW and slope of
7.5 Warm Up
1. 120°
2. 60°
3. 90°
4. 45°
5. 135°
6. 109°
1. x = 4, y = 5.5
3. x = 9, y =
2. x = 24, y = 10
370
PQ ≠ slope of RS ; PQ ≅ RS , so PQRS is not an
isosceles trapezoid.
3. 17
7.5 Practice A
2. 70
UV ≠ slope of TW ; TW ≅ UV , so TUVW is an
isosceles trapezoid.
2. Slope of QR = slope of PS and slope of
7.5 Cumulative Review Warm Up
1. 86
b. 22 in.
4. 61
5. rhombus; ABCD is a quadrilateral with four
3. 6
4. 6
5. isosceles trapezoid; WXYZ has exactly one pair of
parallel sides and one pair of congruent base angles.
6. kite; WXYZ has two pairs of consecutive congruent
congruent sides.
6. kite; DEFG is a quadrilateral with two pairs of
consecutive congruent sides, but opposite sides are
not congruent.
sides, but opposite sides are not congruent.
7. The quadrilateral is not a kite. Because the opposite
sides are congruent, the quadrilateral is a rhombus.
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A83
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7.
REASONS
1. VXYZ is a kite.
1. Given
1. x = 5
2. x = −16
3. x = − 5
2. ∠VXY ≅ ∠VZY
2. Kite Opposite
Angles Theorem
(Thm. 7.19)
4. x = 0
5. x = 15
6. x = − 8
3. ∠WXV ≅ ∠UZV
7. x = 16
8. x = −1
9. x = − 2
10. x = 10
3. Congruent
Supplements
Theorem
(Thm. 2.4)
11. x = 8
12. x = 2
4. VX ≅ VZ
4. Definition of a kite
13. x = − 3
14. x = 3
5. WX ≅ UZ
5. Given
15. x = 4
16. x = − 2
17. equilateral triangle
18. rectangle
19. rhombus
20. parallelogram
21. right triangle
22. square
6.
WXV
≅
UZV
6. SAS Congruence
Theorem
(Thm. 5.5)
8. (6, 0)
9. no; A kite is a quadrilateral and by definition it is a
convex polygon.
10. a. The opposite sides are parallel, and all angles are
right angles; A = ac
bc + ac
 bc − ac 
 + ac =
4
2


7.5 Enrichment and Extension
1. about 56.6 in.; about 418.3 in.
2. a = 9
3. AD = 7.08 in., AB = CD = 5.08 in.,
BC = 10.16 in.
(x
5. ( a + c, b + c)
+ y, y + z )
7. any point of the form ( a, a ), where a is a real
number, a > 3.5, and a ≠ 7
7.5 Puzzle Time
INCORRECTLY
b. 11 units
24. a. P = 2(3 x + 4) + 2( 2 x + 7)
c. length = 16 units, width = 15 units
c. 2
4. ( a + 3, 3b)
23. a. x = 9
b. 4
bc − ac
b. A =
4
6.
Cumulative Review
STATEMENTS
25. dilation
26. rotation
27. reflection
28. translation
29. rotation
30. dilation
31. A′( −1, 2)
32. B ′(11, 3)
33. C ′(6, −11)
34. A′( 4, 0)
35. B ′( −16, 2)
36. C ( 2, 5)
37. D (7, 5)
38. 139°
39. 105°
40. 115°
41. 143°
42. x = 3, y = 9
43. x = 5, y = 2
44. a. 8
b. 48°
c. 42°
45. a. 4
b. 17°
c. 73°
46. ∠ DGF
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47. ∠ EFG
48. ∠GDF
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49. ∠ FED
50. ∠ FDE
52. ∠ EDG
53. ∠ DFG
51. ∠ DFE
3. 3
7. 12
54. 11.4; Perpendicular Bisector Theorem (Thm. 6.1)
Theorem (Thm. 6.2)
57. 22; Perpendicular Bisector Theorem (Thm. 6.1)
b. 36 units
8.1 Practice B
Chapter 8
8.1 Start Thinking
Sample answer: The three diagrams are the same
image, but stretched or shrunk into different sizes or
forms; The first resizing is not “similar” to the original
in a geometric sense. The proportions of the map were
not maintained. The second resizing is “similar” to the
original in a geometric sense. It appears to be a dilation
of the original in a geometric sense. It appears to be a
dilation of the original by a factor less than one and
maintains proportionally with the original.
3
; ∠W ≅ ∠ S , ∠ X ≅ ∠ T , ∠ Y ≅ ∠ U ,
2
WX
XY
YZ
ZW
=
=
=
∠ Z ≅ ∠V ,
ST
TU
UV
VS
3. 6
7. a.
4. 9
5. 7 in.
6. 11 ft
7
4
c. 108°
3
2
2. x = 20
d. about 74.2 units
e. about 219.73 square units
9
4. x = −
5
3. x = ± 3
5. x =
2.
3
; ∠ A ≅ ∠ H , ∠ B ≅ ∠ I , ∠C ≅ ∠ J ,
4
AB
BC
CA
=
=
HI
IJ
JH
b. 7.5
8.1 Warm Up
1. x =
9. 9
12. 336 ft 2
11. 3
1.
c. 72 units
8. 60; 540
5.5),  ADC ≅ BDC and  XWZ ≅ YWZ .
Because corresponding parts of congruent triangles
are congruent, BC = 13 and YZ = 39.
Theorem (Thm. 6.2)
56. 36; Converse of the Perpendicular Bisector
6. 67°
5. 3
10. 13, 39; By the SAS Congruence Theorem (Thm.
55. 1.9; Converse of the Perpendicular Bisector
58. a. 3
4. 22.5
64
7
3
2
6. x = − , x = 4
8.1 Cumulative Review Warm Up
1. 120°
2. 60°
3. 60°
4. 60°
5. 75°
6. 45°
f. yes; Because corresponding angles of similar
triangles are congruent, ∠ ABC ≅ ∠ D. By the
corresponding Angles Converse Theorem
(Thm. 3.5), BC || DE.
8.1 Enrichment and Extension
1. Sample answer:
8.1 Practice A
1. 3; ∠ L ≅ ∠ Q , ∠ M ≅ ∠ R , ∠ N ≅ ∠ S ,
LM
MN
NL
=
=
QR
RS
SQ
2.
2. Sample answer:
2
; ∠ A ≅ ∠ E, ∠ B ≅ ∠ F , ∠ C ≅ ∠ G,
5
AB
BC
CD
DA
=
=
=
∠D ≅ ∠H,
EF
FG
GH
HE
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Geometry
Answers
A85