Chapter 6 Geometry Homework Answers Sec 6 1

Chapter 6 Geometry Homework Answers

Sec 6. 1 1. x = 165°, definition of measure of an 11. The central angle of the smaller circle is arc. larger, because the chord is closer to the center. 2. z = 84°, Chord Arcs Conj. 12. M (-4, 3), N (-4, -3), O (4, -3) 3. w = 70° Chord Central Angles Conj. 15. ≈ 13. 8 cm 4. y = 96°, Chord Arcs Conj. 5. 8 cm, Chord Distance to Center Conj. 16. 1) Given; 3) All radii of a circle are congruent; 4) ΔBOA ê ΔCOD, SSS 6. 20 cm, Perpendicular to a Chord Conj. Congruence Conjecture; 5) ‹COD, 7. AC = 68°; ‹B = 34° (Since Δ OBC is CPCTC isosceles, ‹B = ‹C, ‹B + ‹C = 68°, and 17. They can draw two chords and locate therefore ‹B = 34°. ) the intersection of their perpendicular 8. The length of the chord is greater than bisectors. The radius is just over 5 km. the length of the diameter. 18. y = ⅟₇ x; (0, 0) is a point on this line. 9. The perpendicular bisector of the 19. a) rhombus; b) rectangle; c) kite; d) segment does not pass through the parallelogram center of the circle. 10. The longer chord is closer to the center; the longest chord, which is the diameter, passes through the center.

Sec 6. 2 1. 50° 2. 55° 3. 30° 4. 105° 5. 76 6. Answers will vary. 17. y = -⁸⁄₁₅ x + ²⁸⁹⁄₁₅ 18. 45° 19. Angles A and B must be right angles, but this would make the sum of the angle measures in the quadrilateral shown greater than 360°. 21. 1) Given; 2) Definition of perpendicular lines; 3) Definition of right angle; 4) Substitution property; 5) All radii or a circle are congruent; 6) Definition of isosceles triangle; 7) D, Isosceles Triangle Conjecture; 8) ΔODR, SAA; 9) DR, CPCTC; 10) Definition of bisect 22. a) False; b) False, isosceles trapezoid; c) False, rectangle

Sec 6. 3 1. 65° 2. 30° 3. 70° 4. 50° 5. 140°, 42° 6. 90°, 100° 7. 50° 8. 148° 9. 44° 10. 142° 11. 120°, 60° 12. 140°, 111° 13. 71°, 41° 14. 180° 15. 75° 16. The two inscribed angles intercept the same arc, so they should be congruent. 22. a = 108°; b = 72°; c = 36°; d = 108°; e = 108°; f = 72°; g = 108°; h = 90°; l = 36°; m = 18°; n = 54°; p = 36°

Sec 6. 4 6. True. Opposite angles of a parallelogram 1. See proof. are congruent. If it is inscribed in a circle, 2. Proof: m ‹ ACD = ½m. AD = m ‹ ABD by the opposite angles are also Inscribed Angle Conjecture. Therefore, ‹ supplementary. So they are right angles, ACD ê ‹ ABD. and the parallelogram is indeed 3. Proof: By the Inscribed Angle Conjecture, equiangular, or a rectangle. m ‹ ACB = ½m. ADB = ½ (180°) = 90°. 7. True. m. GA = m. ET by Parallel Lines Therefore, ‹ ACB is a right angle. Intercepted Arcs Conjecture. Therefore, 4. Proof: By the Inscribed Angle Conjecture, the chords that intercept these arcs are m ‹ C = ½m. YLI and m ‹ L = ½m. YCI = congruent (Converse of the Chord Arcs ½(360° - m. YLI) = 180° - m ‹ C. Therefore Conjecture), that is GA ê ET. Therefore ‹ L and ‹ C are supplementary. (A similar GATE is isosceles. proof can be used to show that ‹ I and ‹ Y 8. a. S b. A c. N d. A e. S are supplementary. ) 9. The base angles of the isosceles triangle have a measure of 39°. Because the 5. Proof: ‹ 1 ê ‹ 2 by AIA. m. BC = 2 m ‹ 2 = 2 m corresponding angles are congruent, m is ‹ 1 = m. AD by the Inscribed Angle parallel to n. Conjecture. Therefore BC ê AD.

Sec 6. 5 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. d = 5 cm C = 10π cm r = ¹²⁄π m C = 5. 5π C = 12π cm d = 46 m C ê 15. 7 cm C ê 25. 1 cm r ê 7. 0 m C ê 84. 8 in 565 ft C = 6π cm 16 in. 244 yr 15. 1399 tiles 16. Conjecture: The measure of the angle formed by two intersecting chords is equal to one-half the sum (average) of the measures of the two intercepted arcs. In diagrams, m ‹ NEA = ½ (m. AN + m. GL). 19. b = 90°, c = 42°, d = 70°, e = 48°, f = 132°, g = 52°

Sec 6. 6 12. C ê 4398 km/hr ê 11 m/sec 13. 38° 37, 000 revolutions 14. 48° ê 637 revolutions 15. 30 cm Mama; C ê 50 in. ê 168 cm m ‹ ECA = ½ (m. SN – m. EA) Conjecture: The measure of an angle formed by two intersecting secants through a circle is equal to one-half the difference of the larger arc measure and the smaller arc measure. 11. Both triangles are isosceles so the base angles in each triangle are congruent. But one of each base angle is part of a vertical pair. So, a = b by the Vertical Angles Conjecture and transitivity. 1. 2. 3. 4. 5. 6. 9.

Sec 6. 7 14. Outer horse ê 2. 5 m/sec, inner horse ê 1. ⁴π⁄3 in. 1. 9 m/sec. One horse has traveled 2. 8π m farther in the same amount of time 3. 14π cm (tangential velocity), but both horses 4. 9 m have rotated the same number of 5. 6π ft times (angular velocity). 6. 4π m 15. a = 70°, b = 110°, c = 110°, d 7. 27 in. = 70°, e = 20°, f = 20°, g = 90°, h = 70°, k = 20°, m = 20°, n = 20°, p = 140°, r= 8. 100 cm 80°, s = 100°, t = 80°, u = 120° 9. 217 m/min 10. ê 4200 miles 13. 18°/sec. No, the angular velocity is measured in degrees per second, so it is the same at every point on the carousel.

Ch 6 Review 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. The velocity vector is always perpendicular to the radius at the point of 17. tangency to the object’s circular path. Sample answer: An arc measure is between 0° and 360°. An arc length is 18. proportional to arc measure and depends on the radius of the circle. b = 55° 19. a = 65° c = 128° 20. e = 118° d = 91° f = 66° 125. 7 cm 42. 0 cm 15π cm 14π ft 2 · 57° +2 · 35° ≠ 180° 84° + 56° + 158° ≠ 360° m‹EKL = ½ m. EL= ½ (180°-108°) = 36° = m‹KLY. KE is parallel to YL by Converse of the Parallel Lines Conjecture. m. JI = 360°-56°-152° = m. ML. m‹JMI = ½ m. JI = ½ m. MI = m‹MJI. ΔJIM is isosceles. m. KIM = 2 m‹KEM = 140°. m. KI = 140°-70° = m. MI. m‹IKM = ½ m. MI = ½ m. KI = m‹IMK. ΔKIM is isosceles. Ertha can trace the incomplete circle on paper. She can lay the corner of the pad on the circle to trace an inscribed right angle. Then Ertha should mark the endpoints of the intercepted arc and use the pad to construct the hypotenuse of the right triangle, which is the diameter of the circle. Math is magical!

Ch 6 Review (cont) 34. False. 20° + 140° = 180°. An angle 44. with measure 140° is obtuse. 35. True 45. 36. False. See diagram. 46. 37. True 47. 38. True 39. True 48. 40. True 49. 41. False. (7 – 2) · 180° = 900°. It could have seven sides. 50. 42. False. The sum of the measures of any 51. triangle is 180°. 43. False. The sum of the measures of one set of 52. exterior angles for any polygon is 360°. The sum of the measures of the interior angles 53. of a triangle is 180° and of a quadrilateral is 360°. Neither is greater than 360°, so these are two counter examples. False. The consecutive angles between the bases are supplementary. False. 48° + 132° ≠ 180° False. Inscribed angles that intercept the same arc are congruent. False. The measure of an inscribed angle is half the measure of the arc. True False. AC and BD bisect each other but AC is not perpendicular to BD. (see diagram) False. It could be isosceles. False. 100°+100°+60° = 360° False. AB is not congruent to CD. (see diagram) False. The ratio of the circumference to the diameter is π.

Ch 6 Review (cont) 54. False. 24+24+48+48 ≠ 96. (see diagram) 55. True 56. This is a paradox. Isn’t Math fun? 57. a= 58°, b= 61°, c= 58°, d= 122°, e= 58°, f= 64°, g= 116°, h= 52°, i= 64°, k= 64°, l= 105°, m= 105°, n= 105°, p= 75°, q= 116°, r= 90°, s= 58°, t= 122°, u= 105°, v= 75°, w= 61°, x= 29°, y= 151° 62. m. AC = 84°, length of AC = 11. 2π ≈ 35. 2 in.