Download Practice B

Name———————————————————————— Date —————————————
Practice B
Lesson
10.6
For use with the lesson “Find Segment Lengths in Circles”
Find the value of x.
1.
23
23
2.
15
3
6
3.
x
x14
6
x
12
4
x
Find AB and DE.
4.
5.
E
A
x 1 10
D x 1 13
12
x15 6
B
x
A
D
x
x11
6. x 2 10 D
A
E
B
x
x26
x 1 12
B
E
Find the value of x.
7.
x
8.
3
3
10
2
5
x
9.
4
6
4
x
5
Lesson 10.6
10.
R
x 13
S 8 T
11.
10
U
x 23
21
V
V
T
R
3x
2x 1 8
U
27
12.
S
S
x
x
V
R
15
T
20
U
Find the value of x.
13.
10
14.
15.
x
5
x
15
4x
7
9
9
Find PQ.
16.
P
17.
P R
15
S
R 12
18.
N
24
36
48
S
70
P
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
Find RT and TV.
R
10-78
Geometry
Chapter Resource Book
CS10_CC_G_MECR710778_C10L06PB.indd 78
4/27/11 4:25:51 PM
Name———————————————————————— Date —————————————
Practice B Lesson
continued
For use with the lesson “Find Segment Lengths in Circles”
10.6
Find the value of x.
19.
20.
4
3
42
2x 2 1
20
23.
x15
2x
21
28
24.
5
3x
2x 2 4
x
x12
21.
x
22.
25.
2x
3x 1 1
x11
8
10
2x
3x 2 3
2x
x16
x
26.
x12
2x 2 8
x11
2x 2 5
x
27.
x12
x22
x
28. Winch A large industrial winch is enclosed as shown. 15 in.
8 in.
Lesson 10.6
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
There are 15 inches of the cable hanging free off of the
winch’s spool and the distance from the end of the cable
to the spool is 8 inches. What is the diameter of the spool?
29. Storm Drain The diagram shows a cross-section of a large
storm drain pipe with a small amount of standing water.
The distance across the surface of the water is 48 inches and
the water is 4.25 inches deep at its deepest point. To the nearest
inch, what is the diameter of the storm drain pipe?
4.25 in.
48 in.
30. Basketball The Xs show the positions of two basketball teammates relative to the circular “key” on a basketball
court. The player outside the key passes the ball to the
player on the key. To the nearest tenth of a foot, how
long is the pass?
5 ft
6 ft
12 ft
Geometry
Chapter Resource Book
CS10_CC_G_MECR710778_C10L06PB.indd 79
10-79
4/27/11 4:25:52 PM
Lesson 10.5 Apply Other Angle
Relationships in Circles,
continued
18. Sample answer: Because
1
m∠ D 5 ​ }2 ​(m​AG ​2 m​CE ​), you can use the given
C  C 
C 5 2m​CE ​
C to substitute:
information m​AG ​
1
C 2 m​CE ​
C)  5 ​ 12 ​m​CE ​
C . Because in
m∠ D 5 ​ 2 ​(2m​CE ​
1
C 2 m​BF ​
C) , you
the small circle m∠ D 5 ​ 2 ​ (m​BHF ​
1
C 2 m​BF ​
C)  5
can write m∠ D 5 ​ 2 ​(3608 2 m​BF ​
C
C . Therefore, ​ 12 ​mCE ​
C 
1808 2 ​BF ​
​  5 1808 2 m​BF ​
which gives m​C
CE ​ 1 2m​C
BF ​ 5 3608.
} 
} 
} 
} 
} 
1. 3, 9; 15 2. 4, x; 8 3. 4, 6; 12 4. x, 18; 22.5
5. x, 12; 16 6. x, 15; 25 7. 16; 8 8. 18; 6
}
35
9. 10; 2​Ï 15 ​  10. 12 11. 4 12. 12.5 13. }
​ 3  ​
14. 12 15. 7.4 16. 9 17. 9 18. 5.7 19. 7.7
20. 3 21. 28.1 22. 2 23. 1.9 24. 5.6
25. a. 50 cm b. 80 cm, 80 cm c. 128 cm
d. 178 cm e. 89 cm
Practice Level B
1. 15 2. 2 3. 4 4. AB 5 16, DE 5 17
5. AB 5 21, DE 5 23 6. AB 5 32, DE 5 24
7. 5 8. 6 9. 7 10. RT 5 20, TV 5 16
Study Guide
1. 2268 2. 828 3. 1808 4. 1098 5. 408
Problem Solving Workshop:
Mixed Problem Solving
Copyright © Houghton Mifflin Harcourt Publishing Company. All rights reserved.
Practice Level A
answers
Substitution and Transitive Properties of Equality
can be used to show m∠ TJU 5 m∠ TKV.
Therefore, j i k by the Corresponding Angles
Converse.
2. Answers will vary but AM p MB should equal
ZM p MY.
3. Products are equal; If two chords intersect in
a circle, then the product of the lengths of the
segments of one chord is equal to the product of
the lengths of the other chord.
1. a. 40 ft b. 58 ft 2. a. Check sketches; 368,
368, 728, 728 b. 368, 368 3. Answers will vary.
4. a. 468 b. 468, 1348, 1348 c. 628 5. 5
6. a. 308 b. 1508 c. 4 in., 2 in.; Because the side
length of the square is 6 in. and 1 in. on each side
of the square is the cardboard, then the circle has a
diameter of 4 in., and the radius is 2 in.
Challenge Practice
1. 42.58
C   C 
1
C 1 m​AD ​
C) 
m∠ CPB 5 ​ 2 ​ (m​BC ​
Thus, m∠ APD 5 m∠ CPB. Therefore,
C 5 m​CB ​
C . 2​Ï5 ​ 128
m​AD ​
1
2. m∠ APD 5 }
​ 2 ​ (m​AD ​1 m​BC ​) and
} 
}
3.
 
4.
Lesson 10.6 Find Segment
Lengths in Circles
Teaching Guide
1 Sample answer:
Y
11. RT 5 35, TV 5 45 12. RT 5 36, TV 5 27
13. 15 14. 12 15. 4 16. 18 17. 30 18. 50
19. 4 20. 8 21. 10 22. 8 23. 4 24. 6
25. 15 26. 8 27. 4 28. 20.125 in. 29. 140 in.
30. 14.2 ft
Practice Level C
1. 3.7 2. 2.3 3. 7.4 4. 2.5 5. 1 6. 3.9
7. 14.3 8. 5 9. 10 10. 3 11. 6 12. 5
13. Sample answer: When you use the theorem
to solve for x and y you get x 5 26 and y 5 39.
These segments are not possible in the given
diagram, so Thm. 14 cannot be applied.
14. AP 5 3, PQ 5 5, QB 5 7, PD 5 18, EQ 5 4
OQ
OP
15. 908 16. }
​ OQ  ​ 5 }
​ OR  ​ 17. Sample answer: By
the Segments of Chords Thm., OP(OR) 5 OS(OQ).
But OS 5 OQ, so OP(OR) 5 (OQ)2, and by the
OP
OQ
Division Prop. of Equality, }
​ OQ  ​ 5 }
​ OR  ​.
18. Sample answer: By the Segments of Secants
and Tangents Thm., OP(OQ) 5 (OT )2 and
OR(OS) 5 (OT)2. Therefore, OP(OQ) 5 OR(OS)
by the Transitive Prop. of Equality.
A
M
Z
B
Geometry
Chapter Resource Book
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