Answers  Chapter10

Answers
ABC is not a right triangle because the side
lengths do not satisfy the Pythagorean Theorem
(Thm. 9.1).
6. no;
Chapter10
10.1 Start Thinking
1.
ABC is a right triangle because the side
lengths satisfy the Pythagorean Theorem
(Thm. 9.1).
7. yes;
y
(3, 4)
4
2
−4
4
2
8. r = 8
x
−2
9. r = 20
11. −
10. 5
−4
(−4, −3)
3
and 4
2
12. Sample answer:
Sample answer: no; It does not pass through the
center.
2.
y
4
A
2
−4
C
−2
4
2
x
−2
B
−4
(3, −4)
Sample answer: radius; It connects the center of the
circle with a point on the circle.
13. a. 40 ft; By the External Tangent Congruence
3. Sample answer: yes; no
8
Theorem (Thm. 10.2), the sidewalks are the
same length.
b. 60 ft
y
4
(5, 0)
−8
4
10.1 Practice B
8 x
1. CE , EF
2. r = 9
3. r = 6
2
10.1 Cumulative Review Warm Up
1. yes; obtuse
2. yes; obtuse
3. yes; right
4. yes; acute
5. yes; obtuse
6. no
2. AB, AD


5. EG
A100 Geometry
Answers
7. no;
3. BD, CH
ABC is not a right triangle because the side
lengths do not satisfy the Pythagorean Theorem
(Thm. 9.1).
8. r = 12
10. 5
10.1 Practice A
4. CH
ABC is a right triangle because the side
lengths satisfy the Pythagorean Theorem
(Thm. 9.1).
6. yes;
10.1 Warm Up
1.  A
3. CF
5. AG, H
4. BD
1. r = 5
2. CF , BD


−8
9. r = 1.4
11. − 23 and 7
12. when the two circles are concentric; There are no
points of intersection and no segment joining the
centers of the circles.
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Answers
13. a. about 19.2 ft
7. no; They are arcs of the same circle, but
 = 120° and mUVT
 = 150°.
m STV
b. AE = BC and DE = CD, so BD = AD by
the SSS Similarity Theorem (Thm. 8.4).
8. a. 45°
10.1 Enrichment and Extension
1. 9
2.
b. 14.4°
412 ≈ 20.3
10.2 Practice B
3. 19.6
1. semicircle; 180°
2. minor arc; 74°
4. It is given that IM and JL are tangent segments.
3. major arc; 286°
4. minor arc; 42°
They intersect at point K. Because tangent segments
from a common point to a circle are congruent,
KI = KL and KM = KJ . By the Addition
Property of Equality, KI + KM = KL + KJ . The
Segment Addition Postulate (Post. 1.2) shows that
IM = KI + KM and JL = KL + KJ . So, by the
Transitive Property of Equality, IM = JL and so
IM ≅ JL by the definition of congruent segments.
10.1 Puzzle Time
.
m
AC = m BD
 and OP
 have the same angle measure,
6. no; NM
but they are arcs of circles that are not congruent.
7. yes; They are arcs of the same circle and
 = 42°.
m
AC = m BD
8. 22.5°
HE WAS SERVING PI
9. a. 135°
10.2 Start Thinking
b. 225°
60 min or 1 h
1. 180°
2. 270°
3. 60°
4. 54°
5. 288°
6. 312°
10. a. 170°
b. 34 sec
10.2 Enrichment and Extension
10.2 Warm Up
1. 18.6 in.
2. 119°
1. 29%
5. yes; They are arcs of the same circle and
3. 35°
3. a. 6 times
b. 60.4%
10.2 Cumulative Review Warm Up
1.
13
7
2. 12
3. 62
2. about 19.1 cm
4.
8
3
4.
10.2 Practice A
1. minor arc; 55°
2. major arc; 245°
3. semicircle; 180°
4. minor arc; 120°
5. a. 32°
a. 72°
b. about 5.9 in.
b. 208°
c. about 29.4 in.
c. 105°
d. about 59.4 in.2
d. 260°
6. yes; They are arcs of congruent circles and
 = mGH
.
m FG
e. A =
1 2  180  °  180  ° 
nd  cos 
  sin 
 
4
  n    n  
5. 73
6. 12
7. 126
10.2 Puzzle Time
THE CRAB APPLE
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Geometry A101
Answers
Answers
10.3 Start Thinking
1. sometimes true; If a chord passes through the center
of the circle, then it is a diameter.
2. always true; By definition, a chord is a segment
whose endpoints are on a circle and a diameter
always satisfies this definition.
3. sometimes true; Because a radius is half the
measure of the diameter, it is possible to draw
infinitely many chords within the circle that have a
measure equal to the radius. However, there are
also infinitely many chords that do not have the
same measure as the radius. For example, all the
diameters do not have the same measure.
4. never true; It is possible for a chord to have the
same measure as a diameter, but it will never be
longer. A diameter is the longest possible chord in a
circle.
10.3 Warm Up
1. 6
2
2. 3
 = 110°
3. m 
AD = mBE
2. 10
4. 100°
5. 7
6. 11
.
8. yes; AB is a perpendicular bisector of QR
9. about 12.8 units
10.3 Cumulative Review Warm Up
10. about 30.4 units
11. Sample answer:
STATEMENTS
REASONS
1. PQ is the diameter
of U .
 ≅ QS

PT
1. Given
2. PT ≅ QS
2. Congruent
Corresponding
Chords Theorem
(Thm. 10.6)
3. UP ≅ UQ ≅ UT
3. Definition of
radius of circle
3. 11
3
≅ US
4. PUT ≅ QUS
1. Given B is the midpoint of EC and DA, you can
conclude that EB ≅ BC and AB ≅ BD. Because
∠ EBA and ∠CBD are vertical angles, you can
conclude that they are also congruent. Then by the
SAS Congruence Theorem (Thm. 5.5), you can
conclude that  AEB ≅  DCB
2. You are given ∠ BDE ≅ ∠ BED and ∠ A ≅ ∠ C .
Then if you conclude DE ≅ DE by the reflexive
property, you have  AED ≅ CDE by the AAS
Congruence Theorem (Thm. 5.11).
2. 160°
4. 65°
4. SSS Congruence
Theorem (Thm. 5.8)
12. Sample answer: You could also use the SAS
 ≅ QS
 , so
Congruence Theorem (Thm. 5.5). PT
m∠PUT ≅ m∠QUS by the Congruent Central
Angles Theorem (Thm. 10.4).
10.3 Enrichment and Extension
1. 60°
2. 19.2°
3. 53.1°
4. 90°
5. 103.5°
6. 180°
7. no; no; Sample answer:
10.3 Practice A
1. 115°
7. 3
3. 11
80° A
P
1.5
Q
5. 4
6. a. yes; AB is a perpendicular bisector of MN .
O
b. no; AB is not perpendicular to MN .
7. 18
8. 6
9. 6 10 ≈ 19 units
10. D
10.3 Practice B
8. 30 units
10.3 Puzzle Time
BECAUSE IT WANTED THE SCHOOL TO HAVE A
LITTLE SPIRIT
1. In a circle, if two chords are congruent, then their
corresponding minor arcs are congruent.
A102 Geometry
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10.4 Start Thinking
10.4 Practice B
 ; m∠A = m∠ B ; Sample answer:
m∠BMC = mBC
1. 90°
2. 42°
3. 58°
4. 48°
Because MB and MA are radii of the same circle,
we can conclude that they are congruent. With this
information, we can conclude that ∠A ≅ ∠B by the
Base Angles Theorem (Thm. 5.6);
m∠BMC = m∠A + m∠B by the Exterior Angle
Theorem (5.2). Because ∠A ≅ ∠B , by substitution
m∠BMC = m∠A + m∠A or m∠BMC = 2m∠A.
 ; then m∠ BC
 = 2 m∠A or
Because m∠BMC = mBC
5. 58°
6. 42°
7. 96°
8. 180°
1
 = m∠A.
m∠ BC
2
10.4 Warm Up
1. m∠ C = 100°, m∠ D = 132°
2. m∠X = 97°, m∠Y = 50°, m∠Z = 33°
3. m∠P = 115°, m∠Q = 115°, m∠R = 65°,
m∠S = 65°
10.4 Cumulative Review Warm Up
1. 18 square units
10. x = 72, y = 90
11. x = 16, y = 14
12. Sample answer:
STATEMENTS
REASONS
1. P
1. Given
2. ∠AED ≅ ∠BEC
2. Vertical Angles
Congruence
Theorem (Thm. 2.6)
3. ∠CAD ≅ ∠DBC
3. Inscribed Angles of
a Circle Theorem
(Thm. 10.11).
4.  AED  BES
4. AAA Similarity
Theorem (Thm. 8.3)
13. yes; Sample answer: ∠ADB and ∠BCA intercept
the same arc, so the angles are congruent by the
Inscribed Angles of a Circle Theorem
(Thm. 10.11).
2. about 39.3 square units
3. about 15.9 square units
14. yes; Sample answer: m∠CAB = 60° by the
10.4 Practice A
1. 21°
9. x = 14, y = 38
2. 144°
3. 58°
4. B; Sample answer: ∠ RQS and ∠ RPS are inscribed
angles that intercept the same arc, so the angles are
congruent by the Inscribed Angles of a Circle
Theorem (Thm. 10.11).
5. x = 110, y = 67
6. x = 99, y = 90
7. x = 39, y = 29
8. Opposite angles should be supplementary, not
congruent; m∠B = 95°
9. a. 62.3°
b. 83.1°
c. acute, scalene; Sample answer: Because
m∠ A = 34.6°, m∠ B = 62.3°, and
m∠C = 83.1°,  ABC has three acute angles
and no congruent sides.
Measure of an Inscribed Angle Theorem
(Thm. 10.10) and m∠ACB = 90° by the Triangle
Sum Theorem (Thm. 5.1).  ABC is a right triangle
with hypotenuse AB. So, AB is a diameter of the
circle by the Inscribed Right Triangle Theorem
(Thm. 10.12).
10.4 Enrichment and Extension
1. m∠1 = m∠ 4 = 45°, m∠ 2 = 20°, m∠3 = 70°
2. 27.70°
3. m∠1 = 60°, m∠2 = 60°, m∠3 = 120°,
m∠4 = 30°
4. m∠1 = 40°, m∠ 2 = 25°, m∠3 = 40°
5. 24°
6. 48°
7. 45°, 135°, 75°, 105°
10.4 Puzzle Time
IT USED ITS HEAD
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Geometry A103
Answers
Answers
10.5 Start Thinking
10.5 Warm Up
Sample answer:
Two chords intersect at the center of the circle.
A
B
The circle is divided into four arcs, and opposite arcs
are congruent.
Two chords intersect within the circle, but not at the
center.
A
D
A
B
D
C
M
C
B
E
E
F
The circle is divided into four arcs. In the diagram of
C, none of the arcs have the same measure. In the
 and the chords are
diagram of  M , m 
AD = mBE
congruent.
Two chords intersect at a point on the circle.
C
M
Q
The circle is divided into three arcs. Of the three arcs,
none may be congruent, two may be congruent, or all
three may be congruent.
Two chords do not intersect.
Y
W
A
2. 2
3. 4
Z
2. 102°
3. 56°
5. 42
6. 35
7. 26
4. 133
8. Sample answer: This finds the supplement of the
angle labeled x°. The measure of the angle should
be one-half the sum of the measures of the arcs
intercepted by the angle and its vertical angle;
1
m∠x = (66° + 66°), so m∠x = 66°.
2
9. 21°
10.5 Practice B
1. 60°
2. 30°
3. 60°
4. 60°
5. 30°
6. 60°
7. D; The measure of ∠4 is one-half the sum of the
measures of the arcs intercepted by the angle and its
vertical angle. So,
1
m∠4 = (75° + 125°) = 100° ≠ 90°.
2
8. 50
9. 7
10. 70
11. a. 120°
b. 100°
Y
10.5 Enrichment and Extension
W
1. a. 164°
Y
C
Z
D
Y
1. 202°
12. about 6.8°
B
Z
W
X
X
X
W
10.5 Cumulative Review Warm Up
c. 140°
X
Z
W
3. 84°
10.5 Practice A
E
X
2. 74°
1. 1
C
D
1. 120°
b. 196°
c. 48°
d. 32°
e. 64°
F
Y
f. 80°
Z
The circle is divided into four arcs. Of the four arcs,
you may have none that are congruent, or two, three, or
all four congruent.
A104 Geometry
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10.6 Cumulative Review Warm Up
2. a. 60°
b. 60°
6
1.
c. 2.25
d. 1.125
2.
−13 + 5 53
3. 3 10
2
10.6 Practice A
3 ) ≈ 1.95
(
f. ( 2.25 + 2.25 3 ) ≈ 6.1
e. 1.125
3. Sample answer: Draw chords RU and ST . It is
 ≅ ST
. Because congruent arcs have
given that RU
congruent chords, RU ≅ ST . It is given that
 ≅ TU
 . ∠RUS , ∠URT , ∠TSU , and ∠STR are
RS
 or TU
.
all inscribed angles that intercept either RS
So, all four angles have the same measure and are
congruent. By the SAS Congruence Theorem
(Thm. 5.5), QRU and QST are congruent
triangles. Also, the base angles are all the same,
so they are isosceles triangles. So RQ, UQ, SQ,
and TQ are congruent because corresponding parts
of congruent triangles are congruent. Congruent
segments have equal lengths, so Q is equidistant
from points R, U, S and T that lie on the circle. So,
Q is the center of the circle.
10.5 Puzzle Time
1. 15
2. 2
3. 12
4. 5
5. 6
6. 7
7. 15
8. 12
9. 4
10. 4
11. 4
12. 7
13. about 14.2 ft
10.6 Practice B
1. 10
2. 8
3. 4
4. 4
5. 8
6. 15
7. 9
8. 5
9. 30
10. about 20.1 in.
11. about 139.8 in.
10.6 Enrichment and Extension
1. AC = 16.5, BD = 16.8
2. 40.5
3. a. 60°
b. Sample answer: ∠ACB ≅ ∠FCE by the
Vertical Angles Congruence Theorem
(Thm. 2.6). Because m∠CAB = 60° and
m∠EFD = 60°, then ∠CAB ≅ ∠EFD.
Using the AA Similarity Theorem (Thm. 8.3),
 ABC   FEC .
A PARALLEL
10.6 Start Thinking
1. Sample answer: PX 1 = 7.7, PY1 = 11.6
y
x + 10
x + 10
=
;y =
3
6
2
2. Sample answer: PX 2 = 6.8, PY 2 = 13.2
c. Sample answer:
3. Sample answer: PX 3 = 6.7, PY3 = 13.3
d. Sample answer: y 2 = x ( x + 16)
e. x = 2, y = 6
4. Sample answer: PX 4 = 8.3, PY4 = 10.8
f. 2
Each pair of segments has approximately the same
period.
10.6 Warm Up
3. x =
1. x = 16
2. x = 1
4. x = − 2, 5
5. x = 6 ± 2 11
6. x = − 7, 1
3
,1
2
30; Sample answer: Because
 ABC   FEC and
CF
12
=
= 2, then
AC
6
CE
2
= . Let CE = 2 x and CB = x. So,
CB
1
2 x 2 = 60 by the Segments of Chords Theorem
(Thm. 10.18), which implies x = 30, and
CE = 2 30.
4. OT 2 = OP • OQ and OT 2 = OR • OS by the
Segments of Secants and Tangents Theorem
(Thm. 10.20). So, OP • OQ = OR • OS .
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Geometry A105
Answers
Answers
10. center: (0, 3), radius: 2
10.6 Puzzle Time
BECAUSE THE ELLIPSES ARE TOO ECCENTRIC
FOR THE CIRCLES
6
2
x 2 + y 2 = 4;
y
−2
2
−4
2
4 x
12. a. from left to right, top row:
(x
(x
(x
10.7 Warm Up
1. PQ = 4, midpoint = (0, 8)
7

53, midpoint =  , − 6 
2

2. PQ =
3. PQ = 10
10.7 Cumulative Review Warm Up
1. 63°
2. 42°
4. 117°
10.7 Practice A
1. x 2 + y 2 = 49
− 5) + ( y − 1) = 25
2
= 4
7. B
2
8. A
− 57) + ( y − 44) = 169,
2
2
− 86) + ( y − 44) = 169;
2
2
− 42.5) + ( y − 31) = 169,
2
2
− 71.5) + ( y − 31) = 169
2
2
b. Sample answer: Subtract 3 from the radius to
obtain 100 on the right side of each equation.
10.7 Practice B
1. x 2 + y 2 = 9
2.
(x
− 3) + ( y − 2) = 4
3.
(x
− 4) + ( y + 7) = 16
4.
(x
+ 3) + y 2 = 25
2
2
2
2
2
6.
(x
− 4) + ( y + 1) = 25
7.
(x
− 2) + ( y − 4) = 169
2
2
2
2
8. center: (0, 0), radius: 10
− 3) + ( y + 2) = 841
2
2
5. x 2 + y 2 = 1
5. x 2 + y 2 = 25
(x
(x
(x
2
3. x 2 + y 2 = 64
4. x 2 + ( y + 5)
2
3. 138°
5. 180°
2
− 28) + ( y − 44) = 169,
from left to right, bottom row:
2, midpoint = ( − 5, 4)
117
3

, midpoint =  −1, − 
2
4


4. PQ =
6.
4 x
the origin is 3 2, but the radius of the circle is 4,
so the point does not lie on the circle.
−4
(x
2
11. Sample answer: The distance from point ( − 3, 3) to
−2
−2
2.
x2 + (y − 3)2 = 4
4
10.7 Start Thinking
4
y
x2 + y2 = 100
9. C
y
8
8
A106 Geometry
Answers
x
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9. center: ( 2, 9), radius: 2
10.7 Enrichment and Extension
1. center: ( − 0.5, 2), radius: 7
(x − 2)2 + (y − 9)2 = 4
y
12
y
8
4
4
−4
−4
8 x
4
+ (y +
2)2
−8
4
12 x
−12
10. center: (0, − 2), radius: 6
x2
2
2. (3, − 2), ( x − 3) + ( y + 2)
2
2
= 26
y
= 36
3. x 2 + ( y + 9.5)
−4
4
8 x
4.
−4
(x
2
= 56.25
− 12) + ( y − 19) = 56.25
2
2
5. a. h = −14 and h = 10
b. 4
11. center: (1, 0), radius: 2
y
(x
(x
+ 2) + ( y + 4) = 16 and
6. a.
(x
+ 5) + y 2 + ( z − 4) = 121
b.
(x
− 10) + ( y + 6) + ( z − 2) = 169
c.
(x
+ 1) + ( y − 2) + ( z + 4) = 59
c.
(x − 1)2 + y2 = 4
2
−2
2
4 x
−2
12. Sample answer: The statement is true. The distance
from point ( − 3, 4) to the origin is 5, and the radius
of the circle is 5, so the point lies on the circle.
13. Sample answer: The statement is false. The distance
(
)
from point 2,
3 to the origin is
7, but the
radius of the circle is 3, so the point does not lie on
the circle.
14. a.
(x + 6)2 + (y − 4)2 = 16
(x − 2)2 + (y − 1)2 = 25
8
y
2
2
+ 2) + ( y + 4) = 484
2
2
2
2
2
2
2
2
2
2
10.7 Puzzle Time
COINCIDE
Cumulative Review
1. x 2 − 10 x + 21
2. j 2 + 4 j + 3
3. c 2 + 4c − 96
4. m 2 − 12 m + 20
5. y 2 + 21 y + 110
6. s 2 + 11s + 30
7. 5q 2 − 17 q − 12
8. 12 p 2 − 40 p − 7
9. − 2 f 2 − 17 f + 84
10. 54b 2 − 36b + 6
11. −15 g 2 + 46 g − 24
12. 21k 2 + 35k − 56
13. x = − 8 and x = 9
14. x = 10 and x = 12
15. x = 6 and x = 12
16. x = − 7 and x = 6
17. x = 1 and x = 7
18. x = − 4 and x = 8
19. x = −10 and x = 11
20. x = 4 and x = 9
C
A
−4
8 x
B
−4
(x + 2)2 + (y + 2)2 = 36
b. ( − 2, 4)
c. no; The point ( 4, − 5) is about 10.8 miles away
from the epicenter.
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Geometry A107
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Answers
21. x = − 9 and x = −1
22. x = −1 and x = 8
23. x = − 4 and x = 2
24. x = − 8 and x = − 5
25. x = − 6 and x = −1
26. x = − 2 and x = 8
27. x = − 8 and x = 7
28. x = − 4 and x = 7
29. x = 43
30. x = 102
31. x = 112
32. x = 18
33. x = 26
34. x = 84
35. x = 133
36. x = 26
37. x = 19
38. x = 1275
39. 34
40. 25
41. 82
42. 34
43. 8
44. 12
45. 9
46. 6
11.1 Start Thinking
10π ≈ 31.4 cm
1. 5π ≈ 15.7 cm
3.
2.
5π
≈ 7.9 cm
2
25π
≈ 13.1 cm
6
2. 90°, 15π
3. 135°, 16π
11.1 Cumulative Review Warm Up
1. 130°
2. 54
164 = 2 41
3.
b. 6
11.1 Practice A
c. 11
d. 17
49. ( − 3, − 3)
 15 7 
, 
 2 2
11 

50.  − 2, 
2

 13 7 
,− 
2
 2
1

51.  −1, − 
2

1. 21 m
2. about 169.6 ft
3. about 47.1 in.
4. 12.4 cm
5. Divide the circumference of the tree by π to find
the diameter of the tree. Because the diameter
is 50 ÷ π ≈ 15.9 inches, which is less than
18 inches, the tree is not suitable for tapping.
 11

, − 6
2

52.  −
53.  −
 5 9
55.  − , 
 2 2
 5 1
56.  − , − 
 2 2
58. 2
29
59.
233
60. 2 145
365
62.
109
63.
54. 
57.
745
569
64. 8
6. about 6.28 cm
7. about 47.1 in.
8. about 7.33 ft
9. about 36.57 mm
π
10. about 86.85 in.
11.
12. 225°
13. 1257 ft
3
11.1 Practice B
65. Sample answer: ∠ABC , ∠ABD, and ∠CBD
66. Sample answer: ∠GFJ , ∠GFH , and ∠JFH
67. 107°
218
73. 8
Chapter 11
1. 120°, 8π
48. a. 2 x + 11
70.
b. 17.6 in.
11.1 Warm Up
47. 18
61.
76. a. 7.6 in.
2
68. 113°
69. 2 130
71. 3
72.
74. 2 102
75. a. 18.4 in.
b. 44.4 in.
A108 Geometry
Answers
65
466
1.
36
π
m
2. 10.8π ft
3. about 44.0 cm
4. 160°
5. 200°
6. about 19.54 m
7. about 24.43 m
8. 280°
9. about 34.21 m
11. about 58.03 ft
10. about 114°
12. about 20.53 cm
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