hapter10 10.1 Start Thinking 6. no; is not a right triangle because the side lengths do not satisf the Pthagorean Theorem (Thm. 9.1). 1. (3, ) 7. es; is a right triangle because the side lengths satisf the Pthagorean Theorem (Thm. 9.1).. r = 9. r = 0 (, 3) Sample answer: no; It does not pass through the center. 10. 5 11. 1. Sample answer: 3 and. (3, ) Sample answer: radius; It connects the center of the circle with a point on the circle. 3. Sample answer: es; no 10.1 arm Up 1. r = 5. r = 9 3. r = 6 10.1 umulative Review arm Up 1. es; obtuse. es; obtuse 3. es; right. es; acute 5. es; obtuse 6. no 10.1 Practice 1.., D 3. D, H. H (5, 0) 5. EG 13. a. 0 ft; the Eternal Tangent ongruence Theorem (Thm. 10.), the sidewalks are the same length. b. 60 ft 10.1 Practice 1. E, EF. F, D. D 5. G, H 3. F 6. es; is a right triangle because the side lengths satisf the Pthagorean Theorem (Thm. 9.1). 7. no; is not a right triangle because the side lengths do not satisf the Pthagorean Theorem (Thm. 9.1).. r = 1 9. r = 1. 10. 5 11. 3 and 7 1. when the two circles are concentric; There are no points of intersection and no segment joining the centers of the circles. 100 Geometr nswers opright ig Ideas Learning, LL
13. a. about 19. ft b. E = and DE = D, so D = D b the SSS Similarit Theorem (Thm..). 10.1 Enrichment and Etension 1. 9. 1 0.3 3. 19.6. It is given that IM and JL are tangent segments. The intersect at point K. ecause tangent segments from a common point to a circle are congruent, KI = KL and KM = KJ. the ddition Propert of Equalit, KI + KM = KL + KJ. The Segment ddition Postulate (Post. 1.) shows that IM = KI + KM and JL = KL + KJ. So, b the Transitive Propert of Equalit, IM = JL and so IM JL b the definition of congruent segments. 10.1 Puzzle Time HE S SERVING PI 10. Start Thinking 60 min or 1 h 1. 10. 70 3. 60. 5 5. 6. 31 10. arm Up 1. 9%. 119 3. 35 10. umulative Review arm Up 7. no; The are arcs of the same circle, but mstv = 10 and muvt = 150.. a. 5 b. 1. 10. Practice 1. semicircle; 10. minor arc; 7 3. major arc; 6. minor arc; 5. es; The are arcs of the same circle and m = md. 6. no; NM and OP have the same angle measure, but the are arcs of circles that are not congruent. 7. es; The are arcs of the same circle and m = md =...5 9. a. 135 b. 5 10. a. 170 b. 3 sec 10. Enrichment and Etension 1. 1.6 in.. about 19.1 cm 3. a. 6 times b. 60.% 1. 13 7. 1 3. 6. 3. 10. Practice 1. minor arc; 55. major arc; 5 3. semicircle; 10. minor arc; 10 5. a. 3 b. 0 c. 105 d. 60 6. es; The are arcs of congruent circles and mef = mgh. a. 7 b. about 5.9 in. c. about 9. in. d. about 59. in. e. 1 10 10 nd cos sin n n = 5. 73 6. 1 7. 16 10. Puzzle Time THE R PPLE opright ig Ideas Learning, LL Geometr nswers 101
10.3 Start Thinking 1. sometimes true; If a chord passes through the center of the circle, then it is a diameter.. alwas true; definition, a chord is a segment whose endpoints are on a circle and a diameter alwas satisfies this definition. 3. sometimes true; ecause a radius is half the measure of the diameter, it is possible to draw infinitel man chords within the circle that have a measure equal to the radius. However, there are also infinitel man chords that do not have the same measure as the radius. For eample, all the diameters do not have the same measure.. never true; It is possible for a chord to have the same measure as a diameter, but it will never be longer. diameter is the longest possible chord in a circle. 10.3 arm Up 1. 6. 3 3 3. 11 10.3 umulative Review arm Up 1. Given is the midpoint of E and D, ou can conclude that E and D. ecause E and D are vertical angles, ou can conclude that the are also congruent. Then b the SS ongruence Theorem (Thm. 5.5), ou can conclude that E D.. ou are given DE ED and. Then if ou conclude DE DE b the Refleive Propert of Segment ongruence (Thm..1), ou have ED DE b the S ongruence Theorem (Thm. 5.11). 10.3 Practice 1. 115. 160 3. 11. 65 5. 6. a. es; is a perpendicular bisector of MN. b. no; is not perpendicular to MN. 7. 1. 6 9. 6 10 19units 10. D 10.3 Practice 1. In a circle, if two chords are congruent, then their corresponding minor arcs are congruent.. 10 3. md = me = 110. 100 5. 7 6. 11 7. 3. es; is a perpendicular bisector of QR. 9. about 1. units 10. about 30. units 11. Sample answer: STTEMENTS 1. PQ is the diameter of U. PT QS 1. Sample answer: ou could also use the SS ongruence Theorem (Thm. 5.5). PT QS, so m PUT m QUS b the ongruent entral ngles Theorem (Thm. 10.). 10.3 Enrichment and Etension 1. 60. 19. 3. 53.1. 90 5. 103.5 6. 10 7. no; no; Sample answer:. 30 units RESONS 1. Given. PT QS. ongruent orresponding hords Theorem (Thm. 10.6) 3. UP UQ UT US 3. Definition of radius of circle. PUT QUS. SSS ongruence Theorem (Thm. 5.) P 1.5 0 O Q 10.3 Puzzle Time EUSE IT NTED THE SHOOL TO HVE LITTLE SPIRIT 10 Geometr nswers opright ig Ideas Learning, LL
10. Start Thinking m M = m ; m = m ; Sample answer: ecause M and M are radii of the same circle, we can conclude that the are congruent. ith this information, we can conclude that b the ase ngles Theorem (Thm. 5.6); m M = m + m b the Eterior ngle Theorem (5.). ecause, b substitution m M = m + m or m M = m. ecause m M = m,thenm = m or 1 m m. = 10. arm Up 1. m = 100, m D = 13. m = 97, m = 50, m = 33 3. m P = 115, m Q = 115, m R = 65, m S = 65 10. umulative Review arm Up 1. 1 square units. about 39.3 square units 3. about 15.9 square units 10. Practice 1. 0. 1 3. 5. ; Sample answer: RQS and RPS are inscribed angles that intercept the same arc, so the angles are congruent b the Inscribed ngles of a ircle Theorem (Thm. 10.11). 5. = 110, = 67 6. = 99, = 90 7. = 39, = 9. Opposite angles should be supplementar, not congruent; m = 95 9. a. 6.3 b. 3.1 c. acute, scalene; Sample answer: ecause m = 3.6, m = 6.3, and m = 3.1, has three acute angles and no congruent sides. 10. Practice 1. 90. 3. 5. 5. 5 6. 7. 96. 10 9. = 1, = 3 10. = 7, = 90 11. = 16, = 1 1. Sample answer: STTEMENTS 13. es; Sample answer: D and intercept the same arc, so the angles are congruent b the Inscribed ngles of a ircle Theorem (Thm. 10.11). 1. es; Sample answer: m = 60 b the Measure of an Inscribed ngle Theorem (Thm. 10.10) and m = 90 b the Triangle Sum Theorem (Thm. 5.1). is a right triangle with hpotenuse. So, is a diameter of the circle b the Inscribed Right Triangle Theorem (Thm. 10.1). 10. Enrichment and Etension 1. m 1 = m = 5, m = 0, m 3 = 70. 7.70 3. m 1= 60, m = 60, m 3= 10, m = 30. m 1= 0, m = 5, m 3= 0 5. 6. 7. 5, 135, 75, 105 10. Puzzle Time IT USED ITS HED RESONS 1. P 1. Given. ED E. Vertical ngles ongruence Theorem (Thm..6) 3. D D 3. Inscribed ngles of a ircle Theorem (Thm. 10.11).. ED ES. Similarit Theorem (Thm..3) opright ig Ideas Learning, LL Geometr nswers 103
10.5 Start Thinking Sample answer: Two chords intersect at the center of the circle. The circle is divided into four arcs, and opposite arcs are congruent. Two chords intersect within the circle, but not at the center. The circle is divided into four arcs. In the diagram of, none of the arcs have the same measure. In the diagram of M, md = me and the chords are congruent. Two chords intersect at a point on the circle. The circle is divided into three arcs. Of the three arcs, none ma be congruent, two ma be congruent, or all three ma be congruent. Two chords do not intersect. D F E D E D M M Q F D E 10.5 arm Up 1. 10. 7 3. 10.5 umulative Review arm Up 1. 1. 3. 10.5 Practice 1. 0. 10 3. 56. 133 5. 6. 35 7. 6. Sample answer: This finds the supplement of the angle labeled. The measure of the angle should be one-half the sum of the measures of the arcs intercepted b the angle and its vertical angle; m = 1 ( 66 + 66 ), so m = 6 6. 9. 1 10.5 Practice 1. 60. 30 3. 60. 60 5. 30 6. 60 7. D; The measure of is one-half the sum of the measures of the arcs intercepted b the angle and its vertical angle. So, 1 m = ( 75 + 15 ) = 100 90.. 50 9. 7 10. 70 11. a. 10 b. 100 c. 10 1. about 6. 10.5 Enrichment and Etension 1. a. 16 b. 196 c. d. 3 e. 6 f. 0 The circle is divided into four arcs. Of the four arcs, ou ma have none that are congruent, or two, three, or all four congruent. 10 Geometr nswers opright ig Ideas Learning, LL
. a. 60 b. 60 c..5 d. 1.15 e. ( 1.15 3) 1.95 f. (.5 +.5 3) 6.1 3. Sample answer: Draw chords RU and ST. It is given that RU ST. ecause congruent arcs have congruent chords, RU ST. It is given that RS TU. RUS, URT, TSU, and STR are all inscribed angles that intercept either RS or TU. So, all four angles have the same measure and are congruent. the SS ongruence Theorem (Thm. 5.5), QRU and QST are congruent triangles. lso, the base angles are all the same, so the are isosceles triangles. So RQ, UQ, SQ, and TQ are congruent because corresponding parts of congruent triangles are congruent. ongruent segments have equal lengths, so Q is equidistant from points R, U, S and T that lie on the circle. So, Q is the center of the circle. 10.5 Puzzle Time PRLLEL 10.6 Start Thinking 1. Sample answer: P 1 = 7.7, P1 = 11.6. Sample answer: P = 6., P = 13. 3. Sample answer: P 3 = 6.7, P3 = 13.3. Sample answer: P =.3, P = 10. Each pair of segments has approimatel the same period. 10.6 arm Up 1. = 16. = 1 3.. =, 5 5. = 6 ± 11 6. = 7,1 = 3,1 10.6 umulative Review arm Up 1. 6. 10.6 Practice 13 + 5 53 3. 3 10 1. 15. 3. 1. 5 5. 6 6. 7 7. 15. 1 9. 10. 11. 1. 7 13. about 1. ft 10.6 Practice 1. 10. 3.. 5. 6. 15 7. 9. 5 9. 30 10. about 0.1 in. 11. about 139. in. 10.6 Enrichment and Etension 1. = 16.5, D = 16.. 0.5 3. a. 60 b. Sample answer: FE b the Vertical ngles ongruence Theorem (Thm..6). ecause m = 60 and m EFD = 60, then EFD. Using the Similarit Theorem (Thm..3), FE. c. Sample answer: + 10 10 ; + = = 3 6 d. Sample answer: = ( + 16) e. =, = 6 f. 30;Sample answer: ecause F 1 FE and, = 6 = then E. = 1 Let E = and =. So, = 60 b the Segments of hords Theorem (Thm. 10.1), which implies = 30 and E = 30.. OT = OP OQ and OT = OR OS b the Segments of Secants and Tangents Theorem (Thm. 10.0). So, OP OQ = OR OS. opright ig Ideas Learning, LL Geometr nswers 105
10.6 Puzzle Time EUSE THE ELLIPSES RE TOO EENTRI FOR THE IRLES 10.7 Start Thinking + = ; 10. center: ( 0, 3 ), radius: 6 + ( 3) = 10.7 arm Up 1. PQ =, midpoint = ( 0, ). 7 PQ = 53, midpoint =, 6 3. PQ = 10, midpoint = ( 5, ). 117 3 PQ =, midpoint = 1, 10.7 umulative Review arm Up 1. 63. 3. 13. 117 5. 10 10.7 Practice 1. + = 9 5 + 1 = 5. ( ) ( ) 3. + = 6 + + 5 =. ( ) 5. + = 5 6. ( ) ( ) 3 + + = 1 7.. 9. 11. Sample answer: The distance from point ( 3, 3) to the origin is 3, but the radius of the circle is, so the point does not lie on the circle. 1. a. from left to right, top row: + = 169, ( ) ( ) ( ) ( ) ( ) ( ) 57 + = 169, 6 + = 169; from left to right, bottom row:.5 + 31 = 169, ( ) ( ) ( ) ( ) 71.5 + 31 = 169 b. Sample answer: Subtract 3 from the radius to obtain 100 on the right side of each equation. 10.7 Practice 1. + = 9. ( ) ( ) 3 + = 3. ( ) ( ) + + 7 = 16 + 3 + = 5. ( ) 5. + = 1 6. ( ) ( ) + + 1 = 5 7. ( ) ( ) + = 169. center: ( 0, 0 ), radius: 10 + = 100 106 Geometr nswers opright ig Ideas Learning, LL
9. center: (, 9 ), radius: 10. center: ( 0, ), radius: 6 11. center: ( 1, 0 ), radius: 1. Sample answer: The statement is true. The distance from point ( 3, ) to the origin is 5, and the radius of the circle is 5, so the point lies on the circle. 13. Sample answer: The statement is false. The distance 1. a. ( ) + ( 9) = 1 from point (, 3 ) to the origin is 7, but the radius of the circle is 3, so the point does not lie on the circle. b. (, ) + ( + ) = 36 c. no; The point (, 5) is about 10. miles awa from the epicenter. opright ig Ideas Learning, LL ( 1) + = ( + 6) + ( ) = 16 ( ) + ( 1) = 5 ( + ) + ( + ) = 36 10.7 Enrichment and Etension 1. Sample answer: ( ) ( ). ( ) ( ) ( ) + 3 + + 3 = 9 3,, 3 + + = 6 + + 9.5 = 56.5 3. ( ). ( ) ( ) 1 + 19 = 56.5 5. a. h = 1 and h = 10 b. c. ( ) ( ) + + + = 16 and ( ) ( ) + + + = 6. a. ( ) ( z ) + 5 + + = 11 b. ( ) ( ) ( z ) 10 + + 6 + = 169 c. ( ) ( ) ( z ) + 1 + + + = 59 10.7 Puzzle Time OINIDE umulative Review 1. 10 + 1. j + j + 3 3. c + c 96. 5. + 1 + 110 6. s 7. 9. 11. q 5 17 1 m 1m + 0 + 11s + 30 q. 1 p 0 p 7 f 17f + 10. 15g + 6g 1. 5b 36b + 6 1k + 35k 56 13. = and = 9 1. = 10 and = 1 15. = 6 and = 1 16. = 7and = 6 17. = 1and = 7 1. = and = 19. = 10 and = 11 0. = and = 9 1. = 9and = 1. = 1and = 3. = and =. = and = 5 5. = 6and = 1 6. = and = Geometr nswers 107
7. = and = 7. = and = 7 9. = 3 30. = 10 31. = 11 3. = 1 33. = 6 3. = 35. = 133 36. = 6 37. = 19 3. = 175 39. 3 0. 5 1.. 3 3.. 1 5. 9 6. 6 7. 1. a. + 11 b. 6 c. 11 d. 17 9. ( 3, 3) 50. 5. 55. 15 7, 5 9, 53. 56. 11, 13 7, 5 1, 51. 5. 1 1, 11, 6 57. 75 5. 9 59. 33 60. 15 61. 365 6. 109 63. 569 6. 65. Sample answer:, D,and D 66. Sample answer: GFJ, GFH,and JFH 67. 107 6. 113 69. 130 70. 1 71. 3 65 7. 66 73. 7. 10 75. a. 1. in. b.. in. 76. a. 7.6 in. b. 17.6 in. hapter 11 11.1 Start Thinking 10π 31. cm 1. 5π 15.7 cm. 5 π 7.9 cm 3. 5 π 13.1 cm 6 11.1 arm Up 1. 10, π. 90,15π 3. 135,16π 11.1 umulative Review arm Up 1. 130. 5 3. 16 = 1 11.1 Practice 1. 1 m. about 169.6 ft 3. about 7.1 in.. 1. cm 5. Divide the circumference of the tree b π to find the diameter of the tree. ecause the diameter is 50 π 15.9 inches, which is less than 1 inches, the tree is not suitable for tapping. 6. about 6. cm 7. about 7.1 in.. about 7.33 ft 9. about 36.57 mm 10. about 6.5 in. 11. 3 π 1. 5 13. 157 ft 11.1 Practice 1. 36 m π. 10.π ft 3. about.0 cm. 160 5. 00 6. about 19.5 m 7. about.3 m. 0 9. about 3.1 m 10. about 11 11. about 5.03 ft 1. about 0.53 cm 13. 7 π 1 1. 150 15. a. 35.6 in. b. about 71 teeth 10 Geometr nswers opright ig Ideas Learning, LL