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10.4 Practice A
1. 20
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
with hypotenuse AB. So, AB is a diameter of the
circle by the Inscribed Right Triangle Theorem
(Thm. 10.12).
10.5 Practice A
1. 202
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. a. 62.3
9. 21
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.
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
10.4 Practice B
1. 90
2. 42
3. 58
4. 48
5. 58
6. 42
7. 96
8. 180
9. x  14, y  38
10.5 Practice B
10. x  72, y  90
measures of the arcs intercepted by the angle and its
vertical angle. So,
1
m4  75  125  100  90.
2
8. 50
11. x  16, y  14
9. 7
10. 70
11. a. 120
12. Sample answer:
STATEMENTS
1.
P
2. AED  BEC
3. CAD  DBC
4. ! AED
! BES
b. 100
REASONS
c. 140
1. Given
12. about 6.8
2. Vertical Angles
Congruence
Theorem (Thm. 2.6)
10.6 Practice A
3. Inscribed Angles of
a Circle Theorem
(Thm. 10.11).
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).
14. yes; Sample answer: mCAB  60 by the
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.
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
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Geometry
Answers
A1
Answers
10.7 Practice A
6.
x
 4   y  1  25
7.
x
 2   y  4  169
1. x 2  y 2  49
x
2.
 5   y  1  25
2
2
2
2
2
2
8. center: 0, 0, radius: 10
3. x 2  y 2  64
4. x 2   y  5
2
 4
5. x 2  y 2  25
x
6.
 3   y  2  841
2
2
7. B
8. A
9. C
10. center: 0, 3, radius: 2
11. Sample answer: The distance from point  3, 3 to
the origin is 3 2, but the radius of the circle is 4,
so the point does not lie on the circle.
12. a. from left to right, top row:
x
x
x
 28   y  44  169,
2
2
 57   y  44  169,
2
2
 86   y  44  169;
2
2
from left to right, bottom row:
x
x
 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
5. x 2  y 2  1
A2
Geometry
Answers
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9. center: 2, 9, radius: 2
10. center: 0,  2, radius: 6
11. center: 1, 0, radius: 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.
b.  2, 4
c. no; The point 4,  5 is about 10.8 miles away
from the epicenter.
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Geometry
Answers
A3