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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 (mAG 2 mCE ), you can use the given C C C 5 2mCE C to substitute: information mAG 1 C 2 mCE C) 5 12 mCE C . Because in m∠ D 5 2 (2mCE 1 C 2 mBF C) , you the small circle m∠ D 5 2 (mBHF 1 C 2 mBF C) 5 can write m∠ D 5 2 (3608 2 mBF C C . Therefore, 12 mCE C 1808 2 BF 5 1808 2 mBF which gives mC CE 1 2mC 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; 2Ï 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 mAD C) m∠ CPB 5 2 (mBC Thus, m∠ APD 5 m∠ CPB. Therefore, C 5 mCB C . 2Ï5 128 mAD 1 2. m∠ APD 5 } 2 (mAD 1 mBC ) 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 A53