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Chords and Arcs GEOMETRY LESSON 11-2

Find AB.

QS = QR + RS QS = 7 + 7 QS = 14 AB = QS

AB = 14


Chords and Arcs GEOMETRY LESSON 11-2

P and Q are points on . O. The distance from O to PQ is 15 in., and PQ = 16 in. Find the radius of . O. Draw a diagram to represent the situation. The distance from the center of . O to PQ is measured along a perpendicular line. PM =

1 PQ 2

PM = 1 (16) = 8 2

OP 2 = PM 2 + OM 2 r 2 = 82 + 152 r 2 = 289 r = 17 The radius of . O is 17 in.


Chords and Arcs GEOMETRY LESSON 11-2

Pages 593-596 Exercises 1. BC

YZ; BC

YZ

2. ET GH JN ET GH JN TFE HFG JKN MKL 3. 14

ML ML;

8. 10

13. 8.9

9. Answers may vary. Samples are given. a. CE b. DE c. CEB d. DEA

14. 12.5

10. the center of the circle

18. 90

15. 9.9 16. 20.8 17. 108

4. 2 5. 7

19. about 123.9

6. 50

11. 6

7. 8

12. 5.4


Chords and Arcs GEOMETRY LESSON 11-2

20. She can draw 2 chords, and their bisectors, of the partial circle. The intersection pt. of the bisectors will be the center and she can then measure the radius. 21. 12 cm 22. 6 in.

23. a. PL b. PM c. All radii of a circle are . d. LPN e. SAS f. CPCTC 24. a. All radii of a circle are . b. AB CD c. Given d. SSS e. AEB CED f. central s have arcs.

25. a. All radii of a . are b. Given

.

c. Def. of d. Reflexive Prop. of e. HL Thm.

f. CPCTC g. CPCTC h. central arcs.

s

have

26. about 13.9 cm 27. He doesn’t know that the chords are equidistant from the center.


Chords and Arcs GEOMETRY LESSON 11-2

28. Check students’ work. 29. Circles can have chords or central s without having both. 30. 5 in. 31. 10 cm 32. 10 ft

33. 8 34. 3.5 cm, 15.5 cm

35. 1. . P with QS (Given)

RT

2. mQS = m QPS and mRT = m RPT (Arc measure = central measure.) 3. mQS = mRT (Def. of ) 4. QPS RPT (Subst.)

36. X is equidist. from W and Y, since XW and XY are radii. So, X is on the bis. of WY by the Conv. of the Bis. Thm. But is the bis. of WY, so contains X. 37. All radii of . O are , so AOB COD by SSS. A C by CPCTC. Also, OEA OFC since both are rt. s . Thus, OEA OFC by AAS, and OE OF by CPCTC.


Chords and Arcs GEOMETRY LESSON 11-2

38. 1. . A with CE BD (Given) 2. CF CF (Refl. Prop. of ) 3. BF FD (A diameter to a chord bisects the chord.) 4.

CFB and CFD are rt. s (Def. of ). 5. CFB CFD (SAS) 6. BC CD (CPCTC) 7. BC DC ( chords have arcs.)

39. 1661 gal

43. C

40. Let O be the center of the circles, and P be the pt. of tangency of the larger circle’s chord to the smaller circle. Then OP is to the chord, and therefore bisects it. So P is the mdpt. of the chord.

44. D

41. B 42. G

45. C 46. [2] PN is a radius of . N. Thus, PN = 13 cm. PX = 1 PQ 2 = 9, PNX is a rt. , so NX 9.38 cm. Thus, MN = 18.8 cm, to the nearest tenth. [1] incorrect length OR incorrect explanation


Chords and Arcs GEOMETRY LESSON 11-2

47. 40

48. 5.5 49. about 42 ft 50. 18.4 in. and 7.6 in.


Chords and Arcs GEOMETRY LESSON 11-2

For Exercises 1–5, use the diagram of . L.

1. If YM and ZN are congruent chords, what can you conclude? YM ZN; YLM ZLN 2. If YM and ZN are congruent chords, explain why you cannot conclude that LV = LC. You do not know whether LV and LC are perpendicular to the chords. 3. Suppose that YM has length 12 in., and its distance from point L is 5 in. Find the radius of . L to the nearest tenth. 7.8 in. For Exercises 4 and 5, suppose that LV YM, YV = 11 cm, and . L has a diameter of 26 cm. 4. Find YM. 22 cm

5. Find LV to the nearest tenth. 6.9 cm


geo book 11.2 key