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Name ——————————————————————— Date ———————————— Practice A LESSON 10.4 For use with pages 699–707 1. m∠A 2. m∠A 1588 LESSON 10.4 Find the indicated measure. 3. m∠B B A C B C C A C A 24� B C 4. mBC C 5. mBC 6. mBC A A B 768 C B A 7. m∠C C 8. m∠A C 9. mBC A C 328 B A 508 Copyright © Holt McDougal. All rights reserved. B C C 408 A 648 B B C Find the indicated measure in (M. 10. m∠PNO 11. m∠QNP 12. mPQ 13. mQO 14. m∠NMO 15. 16. m∠QMP 17. C N 1188 C C mNOP mC OQN M Q P O 688 Name two pairs of congruent angles. 18. X 19. E Y F W D C Z Geometry Chapter 10 Resource Book 263 Name ——————————————————————— Practice A LESSON LESSON 10.4 10.4 For use with pages 699–707 Date ———————————— continued Decide whether a circle can be circumscribed about the quadrilateral. 20. 21. 92� 22. 130� 120� 70� Find the values of the variables. 23. B A y8 x8 24. C 628 628 D O 25. N M x8 y8 E D y8 x8 1048 988 C 688 F P 27. 28. J E x8 R y8 U x8 y8 K M T F y8 S 408 758 1088 1148 528 x8 808 L H 1048 Find m∠ A and m∠ C. 29. B A 408 C 30. 31. B 318 458 B C (2x 1 8)8 A 328 D 738 D C A (3x 2 24)8 D 32. Construct the tangents to (Q that pass through a point C not on the circle. Label the points on the circle the tangents pass through R and S. Justify why aQRC and aQSC are right angles and } } why CR and CS are tangent to (Q. 264 Geometry Chapter 10 Resource Book 1 G # Copyright © Holt McDougal. All rights reserved. 1708 26. Lesson 10.3, continued 22. Sample answer: Because three points determine a plane, let P be the plane that contains X, Y, and Z. Let line j be the perpendicular } bisector of XY in P, and let line k be the } perpendicular bisector of YZ in P. Because X, Y, } } and Z are noncollinear points, XY and YZ are not collinear or parallel. Then j and k are not parallel either, which means that j and k intersect in some point C. By the Perpendicular Bisector Thm., CX 5 CY, and CY 5 CZ. Therefore, the circle } centered at C with radius CY contains X, Y, and Z. 23. Sample answer: If one chord is a perpendicular bisector of another chord, then the first chord is a diameter of the circle, so you can construct } } CK of DG and @##$ CL of EH perpendicular bisectors @##$ that contain diameters of the circle, where C is the } center of the (, K is the midpoint of DG, } and L is the midpoint of EH. Because > chords } } are equidistant from the center, CK > CL. } } CJ > CJ by Reflexive Prop. so nKJC > nLJC by HL. Therefore, ∠DJI > ∠HJI because corresponding parts of > ns are >. 24. no; The top board bisects the two boards it rests on whose inside edges form > 10-foot chords. However, the top board is not perpendicular to the other boards, so it does not lie on a diameter of the pool. Therefore, the center of the top board is not the center of the pool. Review for Mastery 1. 1008 2. 958 3. 708 } } 4. Step 1: Draw segments AB and AC as shown. B A C } Step 2: Draw perpendicular bisectors of AB and } AC. By Theorem 10.4, these are diameters of the circle containing A, B, and C. B } 1. 10 units 2. 3Ï 5 units 3. no; no; Sample answer: P 808 A 1.5 O 4. AC 5 AD 5 AE 5 EF. n ACD > n EAF } } (SAS congruence postulate).Therefore, CD > EF. 5. 184 6. 30 Lesson 10.4 Practice Level A 1. 798 2. 128 3. 908 4. 808 5. 1528 6. 1808 7. 408 8. 168 9. 528 10. 348 11. 318 12. 628 13. 1308 14. 1128 15. 1808 16. 628 17. 2488 18. ∠ C > ∠ D, ∠ E > ∠ F 19. ∠ W > ∠ Z, ∠ X > ∠Y 20. no; The 708 and 1308 angles are not supplementary. 21. yes 22. yes 23. x 5 118, y 5 118 24. x 5 112, y 5 76 25. x 5 90, y 5 82 26. x 5 105, y 5 75 27. x 5 66, y 5 66 28. x 5 91.5, y 5 106 29. m∠ A 5 408, m∠ C 5 328 30. m∠ A 5 738, m∠ C 5 318 31. m∠ A 5 458, m∠ C 5 728 ; ∠QRC and ∠QSC are right angles because they # are inscribed in 1 } semicircles. CR is perpendicular 3 } } to radius QR at its outer endpoint, and CS is } perpendicular to radius QS at its outer endpoint. } } So, by Theorem 10.1, CR and CS are tangent to (Q. 32. 2 1. B 2. 588 3. 1408 4. 468 5. 638 6. 288 C Step 3: Find the point where these bisectors intersect. This is the center of the circle containing points A, B, and C, so it is equidistant from each city. This is were the cellular tower should be located. B tower C A44 Challenge Practice Practice Level B A A 5. 808 6. 10 Geometry Chapter 10 Resource Book 7. 1238 8. 908 9. 428 10. 588 11. 488 12. 588 13. 428 14. 968 15. 1808 16. x 5 14, y 5 38 17. x 5 58, y 5 29 18. x 5 72, y 5 90 19. x 5 39, y 5 29 20. x 5 16, y 5 14 21. x 5 6, y 5 36.5 22. D 23. ∠ AED > ∠ BEC; Theorem 10.8; Angle-Angle Similarity Postulate Copyright © Holt McDougal. All rights reserved. ANSWERS 17. D 18. A 19. E 20. B 21. F