Solutions Key Circles

CHAPTER 11 Solutions Key Circles ARE YOU READY? PAGE 74 1. C. E. B 4. A 5. total # of students = 19 + 08 + 16 + 184 = 800 ( 19 100% = 4% 800) 6. 16 100% = 7% 800 7. 08 _ + 16 100% = 5% 800 8. 11%(400,000) = 44,000 9. 7%(400,000) = 108,000 10. 19% + 1% = % 11. %(400,000) = 18,000 1. 11y - 8 = 8y + 1 y - 8 = 1 y = 9 y = 14. z + 0 = 10z - 15 0 = 9z - 15 45 = 9z z = 5 16. -x - 16 = x + 6-16 = x + 6 - = x x = - 18. 17 = x - 49 = x x = ±7 0. 4 x + 1 = 7 x 1 = x 4 = x x = ± 1. 1x + = 10 + x 11x + = 10 11x = - x = -1 15. 4y + 18 = 10y + 15 18 = 6y + 15 = 6y y = _ 17. -x - 11 = -x - 1 x - 11 = -1 x = 10 19. + y = 18 y = 16 y = ±4 1. 188-6 x = 8-6 x = -150 x = 5 x = ±5 11-1 LINES THAT INTERSECT CIRCLES, PAGES 746 754 CHECK IT OUT! PAGES 747 750 1. chords: QR, ST ; tangent: UV ; radii: PQ, PS, PT ; secant: ST ; diameter: ST. radius of circle C: - = 1 radius of circle D: 5 - = point of tangency: (, -1) equation of tangent line: y = -1. 1 Understand the Problem The answer will be the length of an imaginary segment from the summit of Mt. Kilimanjaro to the Earth s horizon. Make a Plan Let C be the center of the Earth, E be the summit of Mt. Kilimanjaro, and H be a point on the horizon. Find the length of EH, which is tangent to circle C at H. By Thm. 11-1-1, EH CH. So CHE is a right. Solve ED = 19,40 ft = 19,40.66 mi 580 EC = CD + ED 4000 +.66 = 400.66 mi E C E H + C H 400.66 E H + 4000 9,9.40 E H 171 mi EH 4 Look Back The problem asks for the distance to the nearest mile. Check that the answer is reasonable by using the Pythagorean Thm. Is 171 + 4000 4004? Yes, 16,09,41 16,0,016. 4a. By Thm. 11-1-, RS = RT x 4 = x - 6. x = 4x - 5. -x = -5. x = 8.4 RS = (8.4) =.1 THINK AND DISCUSS, PAGE 750 1. 4 lines b. By Thm.11-1-, RS = RT n + = n - 1 = n - 1 4 = n RS = (4) + = 7. No; if line is tangent to the circle with the larger radius, it will not intersect the circle with the smaller radius. If the line is tangent tothe circle with the smaller radius, it will intersect the circle with the larger radius at points.. No; a circle consists only of those points which are a given distance from the center. 4. By Thm. 11-1-1, m PQR = 90. So by Triangle Sum Theorem m PRQ = 180 - (90 + 59) = 1. Copyright by Holt, Rinehart and Winston. 7 Holt Geometry

5. 9. By Thm. 11-1-, JK = JL 4x - 1 = x + 9 x - 1 = 9 x = 10 x = 5 JK = 4(5) - 1 = 19 10. By Thm. 11-1-, ST = SU y - 4 = _ 4 y 4y - 16 = y y - 16 = 0 y = 16 ST = (16) - 4 = 1 EXERCISES, PAGES 751 754 GUIDED PRACTICE, PAGE 751 1. secant. concentric. congruent 4. chord: EF ; tangent: m; radii: DE, DF ; secant: l; diameter: EF 5. chord: QS ; tangent: ST ; radii: PQ, PR, PS ; secant: QS ; diameter: QS 6. radius of circle A: 4-1 = radius of circle B: 4 - = point of tangency: (-1, 4) equation of tangent line: y = 4 7. radius of circle R: 4 - = radius of circle S: 4 - = point of tangency: (1, ) equation of tangent line: x = 1 8. 1 Understand the Problem The answer will be the length of an imaginary segment from the ISS to the Earth s horizon. Make a Plan Let C be the center of the Earth, let E be ISS, and let H be the point on the horizon. Find the length of EH, which is tangent to circle C at H. By Thm. 11-1-1, EH CH. So CHE is a right. Solve EC = CD + ED 4000 + 40 = 440 mi E C E H + C H 440 E H + 4000 1,977,60 E H 1406 mi EH 4 Look Back The problem asks for the distance to the nearest mile. Check that answer is reasonable by using the Pythagorean Thm. Is 1406 + 4000 440? Yes, 17,976,86 17,977,600. PRACTICE AND PROBLEM SOLVING, PAGES 75 754 11. chords: RS, VW ; tangent: l; radii: PV, PW ; secant: VW ; diameter: VW 1. chords: AC, DE ; tangent: CF ; radii: BA, BC ; secant: DE ; diameter: AC 1. radius of circle C: - 0 = radius of circle D: 4-0 = 4 point of tangency: (-4, 0) equation of tangent line: x = -4 14. radius of circle M: - = 1 radius of circle N: 5 - = point of tangency: (, 1) equation of tangent line: y = 1 15. 1 Understand the Problem The answer will be the length of an imaginary segment from the summit of Olympus Mons to Mars horizon. Make a Plan Let C be the center of Mars, let E be summit of Olympus Mons, and let H be a point on the horizon. Find the length of EH, which is tangent to circle C at H. By Thm. 11-1-1, EH CH. So triangle CHE is a right triangle. Solve EC = CD + ED 97 + 5 = 4 km E C E H + C H 4 E H + 97 170,475 E H 41 km EH 4 Look Back The problem asks for the distance to the nearest km. Check that the answer is reasonable by using the Pythagorean Thm. Is 41 + 97 4? Yes, 11,710,178 11,710,084. 16. By Thm. 11-1-, AB = AC x = 8x Since x 0, x = 8 x = 4 AB = (4 ) = 18. S (true if circles are identical) 17. By Thm. 11-1-, RS = RT y = y 7 7y = y Since y 0, 7 = y RT = (7) 7 = 7 Copyright by Holt, Rinehart and Winston. 74 Holt Geometry

19. N 0. N 1. A. S (if chord passes through center). AC 5. AC 4. PA, PB, PC, PD 6. By Thm. 11-1-1, Thm 11-1-, the definition of a circle, and SAS, PQR PQS; so PQ bisects RPS. Therefore, m QPR = (4) = 1. By Thm 11-1-1, PQR is a right. By the Sum, m PQR + m PRQ + m QPR = 180 m PQR + 90 + 1 = 180 m PQR = 180 - (90 + 1) = 69 m PQS = m PQR = (69) = 18 7. By Thm 11-1-1, m R = m S = 90. By Quad. Sum Thm., m P + m Q + m R + m S = 60 x + x + 90 + 90 = 60 4x + 180 = 60 4x = 180 x = 45 m P = 45 8a. The perpendicular segment from a point to a line is the shortest segment from the point to the line. b. B c. radius d. line l AB 9. Let E be any point on the line m other than D. It is given that line m CD. So CDE is a right with hypotenuse CE. Therefore, CE > CD. Since CD is a radius, E must lie in exterior of circle C. Thus D is only a point on the line m that is also on circle C. So the line m is tangent to circle C. 0. Since points determine a line, draw auxiliary segments PA, PB, and PC. Since AB and AC are tangents to circle P, AB PB and AC PC. So ABP and ACP are right. PB PC since they are both radii of circle P, and PA PA by Reflex. Prop. of. Therefore, ABP ACP by HL and AB AC by CPCTC. 1. QR = QS = 5 QT = QR + RT (ST + 5 ) = 5 + 1 ST + 5 = 1 ST = 8. AB = AD = x AC = AE + x - 5 = x + DE + - 5 = + DE 41 = + DE DE = 18. JK = JL and JL = JM, so, JK = JM JK = JM 6y - = 0 - y 8y = y = 4 JL = JM = 0 - (4) = 4. Point of tangency must be (x, ), where x - = ± x = 5 or -1. Possible points of tangency are (5, ) and (-1, ). 5a. BCDE is a rectangle; by Thm. 11-1-1, BCD and EDC are right. It is given that DEB is a right. CBE must also be a right by Quad. Sum Thm. Thus, BCDE has 4 right and is a rectangle. b. BE = CE = 17 in. AE = AD - DE = AD - BC = 5 - = in. c. AB = AE + BE = + 17 AB = 9 AB = 9 17.1 in. 6. Not possible; if it were possible, XBC would contain right. which contradicts Sum Thm. 7. By Thm 11-1-1, R and S are right. By Quad. Sum Thm., P + Q + R + S = 60 P + Q + 90 + 90 = 60 P + Q = 180 By definition, P and Q are supplementary angles. TEST PREP, PAGE 754 8. C A D = A B + B D = 10 + = 109 AD = 109 10.4 cm 9. G - - (-4) =. So, (, -4) lies on circle P; y = -4 meets circle P only at (, -4). So it is tangent to circle P. 40. B π(5 ) _ = 5π π(6 ) 6π = 5 6 CHALLENGE AND EXTEND, PAGE 754 41. Since points determine a line, draw auxiliary segments GJ and GK. It is given that GH JK, so, GHJ and GHK are right. Therefore, GHJ and GHK are right. GH GH by Reflex. Prop. of, and GJ GK because they are radii of circle G. Thus GHJ GHK by HL, and JH KH by CPCTC. 4. By Thm. 11-1-1, C and D are right. So BCDE is a rectangle, CE = DB =, and BE = DC = 1. Therefore, ABE is a right with leg lengths 5 - = and 1. So AB = A E + B E = + 1 = 15 = 17 4. Draw a segment from X to the center C of the wheel. XYC is a right angle and m YXC = (70) = 5. So tan 5 = 1 XY XY = 1 18.6 in. tan 5 Copyright by Holt, Rinehart and Winston. 75 Holt Geometry

SPIRAL REVIEW, PAGE 754 44. 14 + 6.5h > 1.5 + 6.5h 1.5 > 0.5h 6 > h Since h is positive, 0 < h < 6. LM + PR 45. P = = _ 10 + (16 + 4) LR 10 + 6 + 4 + 16 + 4 = 0 40 = 4 46. P = LP LR = 10 + 6 + 4 = 0 40 40 = _ MN + PR 6 + (16 + 4) 47. P = = = 6 LR 40 40 = 1 0 48. P = QR LR = 4 40 = 1 10 CONNECTING GEOMETRY TO DATA ANALYSIS: CIRCLE GRAPHS, PAGE 755 TRY THIS, PAGE 755 1. Step 1 Add all the amounts. 18 + 10 + 8 = 6 Step Write each part as a fraction of the whole. novels: 18 reference: 10 textbooks: 8 6 6 Step Multiply each fraction by 60 to calculate central measure. novels: 18 (60) = 180 reference: 10 (60) = 100 6 6 textbooks: 8 (60) = 80 6 Step 4 Match a circle graph to the data. The data match graph D.. Step 1 Add all the amounts. 450 + 10 + 900 + 0 = 1800 Step Write each part as a fraction of the whole. travel: 450 meals: 10 lodging: 900 1800 1800 1800 other: 0 1800 Step Multiply each fraction by 60 to calculate central measure. travel: 450 (60) = 90 meals: 10 (60) = 4 1800 1800 lodging: 900 (60) = 180 other: 0 (60) = 66 1800 1800 Step 4 Match a circle graph to the data. The data match graph C.. Step 1 Add all the amounts. 190 + 75 + 10 + 50 = 75 Step Write each part as a fraction of the whole. food: 190 health: 75 training: 10 other: 50 75 75 75 6 75 Step Multiply each fraction by 60 to calculate central measure. food: 190 (60) 9 health: 75 (60) 184 75 75 training: 10 (60) 59 other: 50 (60) 4 75 75 Step 4 Match a circle graph to the data. The data match graph B. 11- ARCS AND CHORDS, PAGES 756 76 CHECK IT OUT! PAGES 756 759 1a. m FMC = (0.0 + 0.09 + 0.10 + 0.11)60 = 108 b. m AHB = (1-0.5)60 = 70 c. m EMD = (0.10)60 = 6 a. m JPK = 5 (Vert. Thm.) m JK = 5 m KPL + m LPM + m MPN = 180 m KPL + 40 + 5 = 180 m KPL = 115 m KL = 115 m JKL = m JK + m KL = 5 + 115 = 140 b. m LK = m KL = 115 m KPN = 180 m KJN = 180 m LJN = m LK + m KJN = 180 + 115 a. m RPT = m SPT RT = ST 6x = 0-4x 10x = 0 x = RT = 6() = 1 b. m CAD = m EBF (11--()) mcd = mef 5y = 0y - 0 0 = 5y y = 4 mcd = m CAD = 5(4) = 100 4. Step 1 Draw radius PQ. PQ = 10 + 10 = 0 Step Use Pythagorean and 11--. P T + Q T = P Q 10 + Q T = 0 Q T = 00 QT = 00 = 10 Step Find QR. QR = (10 ) = 0 4.6 Copyright by Holt, Rinehart and Winston. 76 Holt Geometry

THINK AND DISCUSS, PAGE 759 1. The arc measures between 90 and 180.. if arcs are on different circles with different radii. EXERCISES, PAGES 760 76 GUIDED PRACTICE, PAGE 760 1. semicircle. Vertex is the center of the circle.. major arc 4. minor arc 5. m PAQ = 0.45(60) = 16 6. m VAU = 0.07(60) = 5. 7. m SAQ = (0.06 + 0.11)60 = 61. 8. m UT = m UAT = 0.1(60) = 6 9. m RQ = m RAQ = 0.11(60) = 9.6 10. m UPT = (1-0.1)60 = 4 11. m DE = m DAE = 90 m EF = m EAF = m BAC = 90-51 = 9 m DF = m DE + m EF = 90 + 9 = 19 1. m DEB = m DAE + m EAB = 90 + 180 = 70 1. m HGJ + m JGL = m HGL 7 + m JGL = 180 m JGL = 108 m JL = 108 14. m HLK = m HGL + m LGK = 180 + 0 = 10 15. QR = RS (Thm. 11--(1)) 8y - 8 = 6y y = 8 y = 4 QR = 8(4) - 8 = 4 16. m CAD = m EBF (Thm. 11--()) 45-6x = -9x x = -45 x = -15 m EBF = -9(-15) = 15 17. Step 1 Draw radius PR. PR = 5 + 8 = 1 Step Use the Pythagorean Thm. and Thm. 11--. Let the intersection of PQ and RS be T. P T + R T = P R 5 + R T = 1 RT = 1 Step Find RS. RS = (1) = 4 18. Step 1 Draw radius CE. CE = 50 + 0 = 70 Step Use the Pythagorean Thm. and Thm. 11--. Let the intersection of CD and EF be G. C G + E G = C E 50 + E G = 70 RG = 400 = 0 6 Step Find EF. EF = (0 6 ) = 40 6 98.0 PRACTICE AND PROBLEM SOLVING, PAGES 761 76 19. m ADB = 5 (60) = 5 (60) 1. 5 + 9 + 9 10 0. m ADC = 9 (60) = 101.4 10 1. m AB = m ADB 1.. m BC = m BDC = 9 (60) 16. 10. m ACB = 60 - m ADB 60-1. = 7.7 4. m CAB = 60 - m BDC 60-16. =.7 5. m MP = m MJP = m MJQ - m PJQ = 180-8 = 15 6. mqnl = mqnm + mml = m QJM + m MJL = 180 + 8 = 08 7. m WT = m WS + m ST = m WXS + m SXT = 55 + 100 = 155 8. m WTV = m WS + m STV = m WXS + m SXV = 55 + 180 = 5 9. m CAD = m EBF (Thm. 11--()) 10x - 6 = 7x x = 6 x = 1 m CAD = 10(1) - 6 = 147 0. m JK = m LM (Thm. 11--()) 4y + y = y + 68 y = 17 m JK = 4(17) + 17 = 85 Copyright by Holt, Rinehart and Winston. 77 Holt Geometry

1. AC = AB =.4 + 1.7 = 4.1 Let AB and CD meet at E. A E + CE = AC. 4 + C E = 4. 1 CE = 11.05 CD = 11.05 6.6. PR = PQ = () = 6 Let PQ and RS meet at T. P T + R T = P R + R T = 6 RT = 7 = RS = ( ) = 6 10.4. F; the measures between 0 and 180. So it could be right or obtuse. 4. F; Endpts. of a diameter determine arcs measuring exactly 180. 5. T (Thm. 11--4) 6. Check students graphs. 7. Let m AEB = x, m BEC = 4x, and m CED = 5x. m AEB + m BEC + m CED = 180 x + 4x + 5x = 180 1x = 18 x = 15 m AEB = (15) = 45 m BEC = 4(15) = 60 m CED = 5(15) = 75 8. m JL + m JK + m KL = 60 7x - 18 + 4x - + 6x + 6 = 60 17x = 74 x = m JL = 7() - 18 = 16 9. m QPR = 180 10x = 180 x = 18 m SPT = 6(18) = 108 40. Statements Reasons 1. BC DE 1. Given.. AB AD and AC AE. All radii of a circle are.. BAC DAE. SSS 4. BAC DAE 4. CPCTC 5. m BAC = m DAE 5. 6. m BC = m DE 7. BC DE 6. Definition of arc measures 7. Definition of arcs 41. Statements Reasons 1. BC DE 1. Given. m BC = m DE. Definition of arcs. m BAC = m DAE. Definition of arc measures 4. BAC DAE 4. 4. Statements Reasons 1. CD EF 1. Given. Draw radii CE and CF. points determine. a line.. CE CF. All radii of a circle are congruent. 4. CM CM 4. Reflex. Prop. of 5. CMF and CME 5. Def. of are rt.. 6. CMF and 6. Def. of a rt. CME are rt.. 7. CMF CME 7. HL Steps, 4 8. FM EM 8. CPCTC 9. CD bisects EF 9. Def. of a bisector 10. FCD ECD 10. CPCTC 11. m FCD = m ECD 11. Def. of 1. m FD = m ED 1. Def. of arc measures 1. FD ED 1. Def. of arcs 14. CD bisects EF. 14. Def. of a bisector 4. Statements Reasons 1. JK is the bis. of GH 1. Given. A is equidistant from G and H.. Def. of the center of circle. A lies on the bis. of GH. Perpendicular. Bisector Theorem 4. JK is a diameter of 4. Def. of diam. circle A. 44. The circle is divided into eight sectors, each with central measure 45. So possible measures of the central congruent are multiples of 45 between 0(45) = 0 and 8(45) = 60. So there are three different sizes of angles: 15, 90, and 45. 45. Solution A is incorrect because it assumes that BGC is a right. 46. To make a circle graph, draw a circle and then draw central that measure 0.4(60) = 144, 0.5(60) = 16, 0.15(60) = 54, and 0.1(60) = 6. 47a. AC = (7) = 1.5 in. AD = AB - DB = 1.5-7 = 6.5 in. b. C D + A D = A C C D + 6. 5 = 1. 5 CD = 140 = 5 11.8 in. c. By Theorem 11--, AB bisects CE. So CE = CD = 4 5.7 in. TEST PREP, PAGE 76 48. D m WT = 90 + 18 = 108 m UW = 90 49. F CE = (10) = 5 A E + 5 = 6 AE = 11. m VR = 180-41 = 19 m TV = 180-18 = 16 Copyright by Holt, Rinehart and Winston. 78 Holt Geometry

50. 90 AP is a horizontal radius, and BP is a vertical radius. So m AB = m APB = 90. CHALLENGE AND EXTEND, PAGE 76 51. AD = AB = 4 + = 6 cos BAD = 4_ = _ 6 m BD = m BAD = co s -1 ( _ ) 48. 5. points determine distinct arcs. points determine 6 arcs. 4 points determine 1 arcs. 5 points determine 0 arcs. n points determine n(n - 1) arcs. 5a. π 180. So π 90, π 60, and π 4 45 b. 15 15 180 (π) = π 4 70 70 180 (π) = π SPIRAL REVIEW, PAGE 76 54. (x) ( y )( - y ) ( 7 x ) ( y ) ( 6 x y 4 9 y ) 56. ( - r s ) ( t s ) - t s ( 9 t s 4) -18 t 5 s 6 55. a 4 b (-a) -4 a 4 b ( 1 a -4 16 ) 1 b 16 57. = 1 + 7 = + 4 1 = 7 + 6 1 = 1 + 8 1 + 10 = 1 58. C, E, G, I, K, M 59. 6 = 1 + 5 15 = 6 + 9 15 + 1 = 8 60. NPQ and NMQ are right (Thm. 11-1-). So NMQ = 90. 61. PQ = MQ (Thm. 11-1-) x = 4x - 9 9 = x x = 4.5 MQ = 4(4.5) - 9 = 9 CONSTRUCTION, PAGE 76 1. O is on the bisector of PQ. So by Conv. of Bisector, OP = OQ. Similarly, O is on bisector of QR. So OQ = OR. Thus, OP, OQ, and OR are radii of circle O, and circle O contains Q and R. 11- SECTOR AREA AND ARC LENGTH, PAGES 764 769 CHECK IT OUT! PAGES 764 766 1a. A = π r ( 60) m = π(1 ) ( 60) 90 = _ 4 π m 0.79 m b. A = π r ( 60) m = π(16 ) ( 60) = 5.6π in. 80.4 in.. A = π r ( 60) m = π(60 ) ( 60) 180 0,575 ft. Step 1 Find the area of sector RST. A = π r ( 60) m = π(4 ) ( 60) 90 = 4π m Step Find the area of RST. A = bh = (4)(4) = 8 m Step area of segment = area of sector RST - area of RST = 4π - 8 4.57 m 60) = π(6) ( 60) 40 = 4 π m 4.19 m 4a. L = πr ( m b. L = πr ( 60) m = π(4) ( 15 = π cm 9.4 cm 60) THINK AND DISCUSS, PAGE 766 1. An arc measure is measured in degrees. An arc length is measured in linear units.. the radius and central of the sector. EXERCISES, PAGES 767 769 GUIDED PRACTICE, PAGE 767 1. segment. A = π r ( 60) m = π(6 ) ( 60) 90 = 9π m 8.7 m. A = π r ( 60) m = π(8 ) ( 15 60) = 4π cm 75.40 cm 4. A = π r ( 60) m = π( ) ( 60) 0 = 9 π ft 0.70 ft 5. A = π r ( 60) m = π( ) ( 150 60) = 15 4 π mi 1 mi Copyright by Holt, Rinehart and Winston. 79 Holt Geometry

6. Step 1 Find area of sector ABC. A = π r ( 60) m = π( ) ( 60) 90 = 9 4 π in. Step Find area of ABC. A = bh = ()() = 9_ in. Step area of segment = area of sector ABC - area of ABC = 9_ π - 9_.57 in. 4 7. Step 1 Find the area of sector DEF. A = π r ( 60) m = π(0 ) ( 60) = 00 π m Step Find the area of DEF. bh = (0)(10 ) = 100 m A = Step area of segment = area of sector DEF - area of DEF = 00 π - 100 6. m 8. Step 1 Find the area of sector ABC. A = π r ( 60) m = π(6 ) ( 60) 45 = 9 π cm Step Find the area of ABC. bh = (6)( ) = 9 cm A = Step area of segment = area of sector ABC - area of ABC = 9 π - 9 1.41 cm 9. L = πr ( m 60) = π(16) ( 60) 45 = 4π ft 1.57 ft 10. L = πr ( 60) m = π(9) ( 10 = 6π m 18.85 m 60) 11. L = πr ( 60) m = π(6) ( 60) 0 = π in..09 in. PRACTICE AND PROBLEM SOLVING, PAGES 767 769 1. A = π(0 ) ( 150 60) = 500 π m 5.60 m 1. A = π(9 ) ( 100 60) = 45 π in 70.69 in. 14. A = π( ) ( 60) 47 = 47 90 π ft 1.64 ft 15. A = π(0 ) ( 180 = 00π 68 in. 60) 16. ABC is a 45-45 -90 triangle. So central angle measures 90. area of segment = area of sector ABC - area of ABC = π(10 ) ( 60) 90 - _ (10)(10) = 5π - 50 8.54 m 17. m KLM = m KM = 10 area of segment = area of sector KLM - area of KLM = π(5 ) ( 10 60) - ( _ ) 5 (5 ) = 5 π - 5 4 15.5 in. 18. area of segment = area of sector RST - area of RST = π(1 ) ( 60) - _ ( (1) ) = _ 6 π - _ 4 0.09 in. 19. L = π(5) ( 50 60) = 5 π mm 4.6 mm 18 0. L = π(1.5) ( 60) 160 = 4 π m 4.19 m 1. L = π() ( 60) 9 = 1 π ft 0.1 ft 10. P = π() ( 180 60) ( + π(1) ( 60) 180 ) = 6π 18.8 in.. never 4. sometimes (if radii of arcs are equal) 5. always 7. L = πr ( m 6. A = π r ( 60) m 60) 9π = π r ( 60) 90 8π = πr ( 60) 10 6 = r 4π = πr r = 6 r = 1 8a. L ( 7 ) (7) ( 60) 90 = 11 in. b. L = π(7) ( 60) 90 = 7 π 10.9955749 in. c. overestimate, since L < 10.996 < 11 9a. L = π(.5) ( 60) 90 = 5 π.9 ft 4 b. 4.5 = π(.5) ( 60) m 9 10π = m 60 m = 4 π 10 0. The area of sector BAC is 45 60 = _ area of circle A. 8 So if area of circle A is 4 in, the area of the sector will automatically be _ (4) = in. 8 So we need only to solve π r = 4 for r. π r = 4 r = 4 π r = 4 π.76 in. 1. If the length of the arc is L and its degree measure is m, then L = πr ( 60) m 60L = πrm r = _ 60L πm = _ 180L πm. Copyright by Holt, Rinehart and Winston. 80 Holt Geometry

TEST PREP, PAGE 769. B A = π(8 ) ( 60) 90. G = 16π A = π(8) ( 60) 90 = 4π 4. 4.98 A = π(1 ) ( 60) 5 4.98 CHALLENGE AND EXTEND, PAGE 769 5. A = π(5 ) ( 60) 40 - π( ) ( 60) 40 = 5 9 π - 4 9 π = 7 π 6a. V = Bh = ( π(4 ) ( 60) 0 )() = 4π 1.6 in. b. B = ( π(4 ) ( 60) 0 ) = 8 π 60) )() = 4 + π π + 4 + π 8.7 in. L = ()(4) + ( π(4) ( 0 A = 8 7a. A( ) = π( ) = 4π A(red) = 4π(1 ) ( 60) 45 = _ π π P(red) = 4π = _ 8 60) - π(1 ) ( 45 _ π P(blue) = 4π = 8 b. A(blue) = 4 ( π( ) ( 45 c. P(red or blue) = SPIRAL REVIEW, PAGE 769 8. 8x - y = 6 y = 8x - 6 y = 4x - The line is. 40. y = mx + 1 0 = m(4) + 1 m = - _ 4 The line is. 4. S = 4π = 4π r r = 1 r = 1 cm C = π(1) = π cm π + _ π 4π = _ 60) ) = 1-9. slope = - 0 π = The line is neither nor. 41. V = 4_ π( ) = 6π cm 4. m KLJ = m KLH - (m GLJ - m GLH) 10x - 8 = 180 - (90 - (x + )) 10x - 8 = 90 + x + 8x = 10 x = 15 m KLJ = 10(15) - 8 = 1 44. m KJ = m KLJ = 1 45. m JFH = m JF + m FG + m GH = m JLF + m FLG + m GLH = 180 + 90 + (15) + = 0 11A MULTI-STEP TEST PREP, PAGE 770 1. Let l represent the length of the stand. Radius r = 14 in. Form a right with leg lengths 14-10 = 4 in. and l in., and hypotenuse length 14 in. 4 + ( l ) = 14 16 + 4 l = 196 4 l = 180 l = 70 l = 70 = 1 5 6.8 in. AD with BE. Let E be on ABE is a right. So A E + BE = AB (4 - ) + CD = 15. L = πr ( m 4. L = πr ( 60) m 4 π = π() ( 60) m _ = m 60 m = 60 = 10 CD and BE = CD. C D = 1 CD = 1 14.9 in. 60) = π(4) ( 60) = 4 π 4. in. 11A READY TO GO ON? PAGE 771 1. chord: PR ; tangent: m; radii: QP, QR, QS ; secant: PR ; diameter: PR. chords: BD, BE ; tangent: BC ; radii: AB, AE ; secant: BD ; diameter: BE. Let x be the distance to the horizon. The building forms a right with leg lengths x mi and 4000 mi, and hypotenuse length (4000 + 580) 7 mi. x + (4000) = ( 4000 + 7 x 1109.11 x mi 580) 4. m BC + m CD = m BD m BD + m CAD = m BAD m BD + 49 = 90 m BD = 41 5. m BED = m BFE + m ED = m BAE + m EAD = 180 + 90 = 70 6. m SR + m RQ = m SQ m SR + m RPQ = m SPQ m SR + 71 = 180 m SR = 109 Copyright by Holt, Rinehart and Winston. 81 Holt Geometry

7. m SQU + m UT + m TS = 60 m SQU + m UPT + m TPS = 60 m SQU + 40 + 71 = 60 m SQU = 49 8. Let JK = x; then x + 4 = (4 + ) x = x = JK = 11.5 9. Let XY = x; then x + 4 = 8 x = 48 x = 4 XY = 8 1.9 10. A = π( ) ( 60) 80 = 968 π 8 cm 9 11. AB = π(4) ( 150 60) = 10 π ft 10.47 ft 1. EF = π(.4) ( 60) 75 = π cm.14 cm 1. arc length = π(5) ( 60) 44 = 11 π in..84 in. 9 11. arc length = π(46) ( 180 = 46π m 144.51 m 60) 11-4 INSCRIBED ANGLES, PAGES 77 779 CHECK IT OUT! PAGES 77 775 1a. m ABC = m ADC 15 = m ADC m ADC = 70. m ABD = m m BAC = m BC 60 = m BC m BC = 10 a. ABC is a right angle m ABC = 90 8z - 6 = 90 8z = 96 z = 1 AD = (86) = 4 b. m DAE = mde = (7) = 6 b. m EDF = m EGF x + = 75 - x 4x = 7 x = 18 m EDF = (18) + = 9 4. Step 1 Find the value of x. m K + m M = 180 + 6x + 4x - 1 = 180 10x = 160 x = 16 Step Find the measure of each. m K = + 6(16) = 19 m L = 9(16) = 7 m M = 4(16) - 1 = 51 m J + m L = 180 m J + 7 = 180 m J = 108 THINK AND DISCUSS, PAGE 775 1. No; a quadrilateral can be inscribed in a circle if and only if its opposite are supplementary.. An arc that is of a circle measures 90. If the arc 4 measures 90, then the measure of the inscribed is (90) = 45.. EXERCISES, PAGES 776 779 GUIDED PRACTICE, PAGE 776 1. inscribed. m DEF = m DF = (78) = 9 4. m JNL = m JKL 10 = m JKL m JKL = 04 6. m QTR + m Q + m R = 180 m QTR + m RS + m S = 180 (90) + 5 = 180 m QTR + m QTR + 70 = 180 m QTR = 110. m EFG = m EG 9 = m EG m EG = 58 5. m LKM = m LM = (5) = 6 7. DEF is a right 4x 5 = 90 4x = 450 x = 11.5 8. FHG is a right. So FGH is a 45-45 -90 triangle. m GFH = 45 y + 6 = 45 y = 9 y = 1 9. m XYZ = m XWZ 7y - = 4 + 6y y = 7 m XYZ = 7(7) - = 46 Copyright by Holt, Rinehart and Winston. 8 Holt Geometry

10. Step 1 Find the value of x. m P + m R = 180 5x + 0 + 7x - 8 = 180 1x = 168 x = 14 Step Find the measures. m P = 5(14) + 0 = 90 m Q = 10(14) = 140 m R = 7(14) - 8 = 90 m S + m Q = 180 m S + 140 = 180 m S = 40 11. Step 1 Find the value of z. m A + m C = 180 4z - 10 + 10 + 5z = 180 9z = 180 z = 0 Step Find the measures. m A = 4(0) - 10 = 70 m B = 6(0) - 5 = 115 m C = 10 + 5(0) = 110 m B + m D = 180 115 + m D = 180 m D = 65 PRACTICE AND PROBLEM SOLVING, PAGES 776 778 1. m MNL = m ML 4 = m ML m ML = 86 14. m EJH = m EGH 19 = m EGH m EGH = 78 16. m ADC = m ADB + m BDC = mab + m BEC = (44) + 40 = 6 17. m SRT = 90 y - 18 = 90 y = 108 y = 6 y = ±6 19. m ADB = m ACB x = 10x x = 10 (x 0) x = 5 m ADB = m AB (5 ) = m AB 50 = m AB m AB = (50) = 100 1. m KMN = m KN = (95) = 47.5 15. m GFH = m GH = (95.) = 47.6 18. JKL is a 0-60 -90 6z - 4 = 60 6z = 64 z = 64 = 10 _ 6 0. MPN and MNP are complementary. m MPN + m MNP = 90 11x + x - 10 = 90 0x - 0 = 70 0x = 00 x = 15 m MPN = 11(15) = 55 1. Step 1 Find the value of x. m B + m D = 180 x + x + 0 = 180 4 x 4 = 150 x = 600 x = 00 Step Find the measures. m B = 00 = 100 m D = 00 + 0 = 80 4 m E = 00-59 = 141 m C + m E = 180 m C + 141 = 180 m C = 9. Step 1 Find the value of x. m U + m W = 180 14 + 4x + 6x - 14 = 180 10x = 180 x = 18 Step Find the value of y. m T + m V = 180 1y - 5 + 15y - 4 = 180 7y = 189 y = 7 Step Find the measures. m T = 1(7) - 5 = 79 m U = 14 + 4(18) = 86 m V = 15(7) - 4 = 101 m W = 6(18) - 14 = 94. always 4. never 5. sometimes (if opposite are supplementary) AC = m ADC = (11) = 56 6. m ABC = m 7. Let S be any point on the major arc from P to R. m PQR + m PSR = 180 m PQR + m PQR = 180 m PQR + (10) = 180 m PQR = 180-65 = 115 8. By the definition of an arc measure, m JK = m JHK. Also, the measure of an inscribed in a circle is half the measure of the intercepted arc. So m JLK = m JK. Multiplying both sides of equation by gives m JLK = m JK. By substitution, m JHK = m JLK. BC = ( (60) 6 ) = 0 mcae = ( 4_ (60) 6 ) = 10 c. FBC is inscribed in a semicircle. So it must be a right ; therefore, FBC is a right. (Also, m CFD = 0. So, FBC is a 0-60 -90.) 9a. m BAC = m b. m CDE = Copyright by Holt, Rinehart and Winston. 8 Holt Geometry

0. 1. Since any points determine a line, draw BX. Let D be a point where BX intercepts AC. By Case 1 of the Inscribed Thm., m ABD = m AD and m DBC = m DC. By Add., Distrib., and Trans. Props. of =, m ABD + m DBC = m AD + m DC = (m AD + m DC ). Thus, by Add. Post. and Arc Add. Post., m ABC = mac. Since any points determine a line, draw BX. Let D be pt. where BX intercepts ACB. By Case 1 of Inscribed Thm., m ABD = m AD and m CBD = m CD. By Subtr., Distrib., and Trans. Props. of =, m ABD - m CBD = m AD - m CD = (m AD - m CD ). Thus by Add. Post. and Arc Add. Post., m ABC = mac.. By the Inscribed Thm., m ACD = m AB and m ADB = m AB. By Substitution, m ACD = m ADB, and therefore, ACD ADB.. m J = m KLM = (16) = 108 m J + m L = 180 108 + m L = 180 m L = 7 m M = m JKL = (198) = 99 m K + m M = 180 m K + 99 = 180 m K = 81 4. PR is a diag. of PQRS. Q is an inscribed right angle. So its intercepted arc is a semicircle. Thus, PR is a diameter of the circle. 5a. AB + AC = BC, so by Conv. of Pythag. Thm., ABC is a right with right A. Since A is an inscribed right, it intercepts a semicircle. This means that BC is a diameter. b. m ABC = sin -1 - ( 14 Since m ABC = m 18) 51.1. AC, m AC = 10. 6. Draw a diagram through D and A. Label the intersection of BC and DE as F and the intersection of DA and BE as G. Since BC is a diameter of the circle, it is a bisector of chord DE. Thus, DF EF, and BFD and BFE are right. BF BF by Reflex. Prop. of. Thus, BFD BFE by SAS. BD BE by CPCTC. By Trans. Prop. of, BE ED. Thus, by definition, DBE is equilateral. 7. Agree; the opposite of a quadrilateral are congruent. So the opposite the 0 also measures 0. Since this pair of the opposite are not supplementary, the quadrilateral cannot be inscribed in a circle. 8. Check students constructions. TEST PREP, PAGE 778 9. D m BAC + m ABC + m ACB = 180 m BAC + mac + m ACB = 180 m BAC + (76) + 61 = 180 m BAC + 8 + 61 = 180 m BAC = 81 40. H m XY = m XCY = m XCZ m XY = m XCZ = 60 41. C Let m A = 4x and m C = 5x. m A + m C = 180 4x + 5x = 180 9x = 180 x = 0 m A = 4(0) = 80 4. F m STR = 180 - (m TRS + m TSR) = 180 - ( m PS + m QR ) = 180 - (4 + 56) = 8 QR = 56 m QPR = m m QPR = m m PQS = m QR = 56 PS = 4 CHALLENGE AND EXTEND, PAGE 779 4. If an is inscribed in a semicircle, the measure of the intercepted arc is 180. The measure of is (180) = 90. So the angle is a right. Conversely, if an inscribed angle is a right angle, then it measures 90 and its intercepted arc measures (90) = 180. An arc that measures 180 is a semicircle. Copyright by Holt, Rinehart and Winston. 84 Holt Geometry

44. Suppose the quadrilateral ABCD is inscribed in a circle. Then m A = m BCD and m C = m DAB. By Add., Distrib., and Trans. Prop. of =, m A + m C = m BCD + m DAB = (m BCD + m DAB ). m BCD + m DAB = 60. So by subst., m A + m C = (60) = 180. Thus, A and C are supplementary. A similar proof shows that B and D are supplementary. 45. RQ is a diameter. So P is a right. m PQ = m R = ta n -1 ( 7_ 14 ) 46. Draw AC and DE. m m CE = 19 m ACD = m AD = 6 m ABC + 19 + 6 = 180 m ABC = 15 m ABD + 15 = 180 m ABD = 55 47. Check students constructions. SPIRAL REVIEW, PAGE 779 48. a + b + c = 1 (1) 0a +.5b + 15c = 55 () 0a = 15c () Substitute () in (): 15c +.5b + 15c = 55 b + 4c = 4 (4) Substitute 1 () in (1): 15 c + b + c = 4 b + c = 4 (5) (5) - (4): 6b + 9c - 6b - 8c = 7-68 c = 4 Substitute in (): 0a = 15(4) = 60 a = Substitute in (5): b + (4) = 4 b = 1 b = 6 49. m = 50. m = - (-6) 8-4 = _ - - (-8) 0 - (-9) _ 6 = 1 7 = 6 9 = 8 = 5 51. m = 6 - (-14) = 0 11-5. By Vert. Thm., RWV SWT. So by Thm. 11-- (1), RV = ST z + 15 = 9z + 1 14 = 7z z = RV ST and RS VT. So by substitution, m ST + m VT = 60 m ST + m VT = 180 m ST + 1() + = 180 m ST + 64 = 180 m ST = 116 5. Let BD = x; then 1. 5 + x = (1 + 1.5 ).5 + x = 6.5 x = 4 x =. ABD has base () = 4 m and height 1.5 m. So A = bh = (4)(1.5) = m. CONSTRUCTION, PAGE 779 1. Yes; CR is a radius of circle C. If a line is tangent to a circle, then it is to the radius at the point of tangency. 11-5 TECHNOLOGY LAB: EXPLORE ANGLE RELATIONSHIPS IN CIRCLES, PAGES 780 781 TRY THIS, PAGE 781 1. The relationship is the same. Both types of have a measure equal to half the measure of the arc.. A tangent line must be to the radius at the point of tangency. If construction is done without this relationship, then tangent line created is not guaranteed to intersect circle at only one point. If a line is to a radius of a circle at a point on the circle, then line is tangent to circle.. The relationship remains the same. The measure of is half difference of intercepted arcs. 4. An whose vertex is on the circle (inscribed and created by a tangent and a secant intersecting at the point of tangency) will have a measure equal to half its intercepted arc. An whose vertex is inside the circle ( created by intersecting secants or chords) will have a measure equal to half sum of its intercepted arcs. An whose vertex is outside the circle ( created by secants and/or tangents that intersect outside circle) will have a measure equal to half its intercepted arc. 5. No; it is a means to discover relationships and make conjectures. Copyright by Holt, Rinehart and Winston. 85 Holt Geometry

11-5 ANGLE RELATIONSHIPS IN CIRCLES, PAGES 78 789 CHECK IT OUT! PAGES 78 785 1a. m STU = m ST = (166) = 8 a. m ABD = (m AD + m CE ) = (65 + 7) = (10) = 51 b. m RNP = (m MQ + m RP ) = (91 + 5) = (16) = 158 m RNM + m RNP = 180 m RNM + 158 = 180 m RNM =. m L = (m JN - m KN ) 5 = (8 - x) 50 = 8 - x x = 4. m ACB = = (m AEB - m AB ) (5-15) = 45 5. Step 1 Find m PR. m Q = (m MS - mpr) 6 = (80 - m PR ) 5 = 80 - m PR m PR = 8 Step Find m LP. m LP + m PR = mlr m LP + 8 = 100 m LP = 7 THINK AND DISCUSS, PAGE 786 b. m QSR = m SR 71 = m SR m SR = 14 1. For both chords and secants that intersect in the interior of a circle, the measure of formed is half the sum of the measures of their intercepted arcs.. EXERCISES, PAGES 786 789 GUIDED PRACTICE, PAGES 786 787 1. m DAB = m AB =. 7 = m AC (140) = 70 m AC = 54. 61 = m PN 4. m MNP = (8) m PN = 1 = 119 5. m STU = (m SU + m VW ) = = (104 + 0) (14) = 67 6. m HFG = (m EJ + m GH ) = (59 + ) = (8) = 41 7. m NPL = (m KM + m NL ) = (61 + 111) = (17) = 86 m NPK + m NPL = 180 m NPK + 86 = 180 m NPK = 94 (161-67) = 8. x = (94) = 47 9. x = (8-1) = (116) = 58 10. 7 = (x - 40) 54 = x - 40 x = 94 11. m S = (m ACB - m AB ) 8 = ((60 - x) - x) 76 = 60 - x x = 60-76 = 84 x = 14 1. m E = (m BF - m DF ) (150 - mdf) 100 = 150 - m DF m DF = 50 50 = 1. m BC + m CD + m DF + m BF = 60 64 + m CD + 50 + 150 = 60 m CD + 64 = 60 m CD = 96 14. m NPQ = (m JK + m PN ) 79 = (48 + m PN ) 158 = 48 + m PN m PN = 110 15. m KN + m PN + m JP + m JL = 60 m KN + 110 + 86 + 48 = 60 m KN + 44 = 60 m KN = 116 Copyright by Holt, Rinehart and Winston. 86 Holt Geometry

PRACTICE AND PROBLEM SOLVING, PAGES 787 788 16. m BCD = m BC = (11) = 56 17. m ABC = (60-11) = (48) = 14 18. m XZW = m XZ = (180) = 90 19. m XZV = m XZ + m ZV = 180 + m VXZ = 180 + (40) = 60 ST ) = 0. m QPR = (m QR + m 1. m ABC = 180 - m ABD = 180 - (m AD + m CE ) = 180 - (100 + 45) = 107.5. m MKJ = 180 - m MKL = 180 - (m JN + m ML ) = 180 - (8.5 + 51.5) = 15. x = (170 - (60 - (170 + 15))) (170-55) = 57.5 = 4. x = (0-140) = 40 5. x = ((180 - (0 + 104)) - 0) (56-0) = 18 = 6. m ABV = (180 - m 7. m DJH = (m DH + m EG ) 180-89 = (17 + m EG ) 91 = (17 + m EG ) 18 = 17 + m EG m EG = 45 8. m DJE = (m DE + m GH ) 89 = (m DE + 61) 178 = m DE + 61 m DE = 117 9. Step 1 Find m P. m P = m LR = (170) = 85 Step Find m QPR. m QPR + 50 + 85 = 180 m QPR = 45 Step Find m PR. m QPR = m PR 45 = m PR m PR = 90 0. 50 = (m PS - m ML ) = (m PM - m ML ) = LP m m LP = 100 (1 + 98) = 64.5 AB ) = (180-48) = 66 1. m ABC = m AB x = m AB m AB = x. m ABD + m ABC = 180 m ABD + x = 180 m ABD = (180 - x). m AEB + m AB = 60 m AEB + x = 60 m AEB = (60 - x) 4. 5. Since points determine a line, draw BD. By Ext. Thm., m ABD = m C + m BDC. So m C = m ABD - m BDC. m ABD = m AD by BD by Thm. Inscribed Thm., and m BDC = m 11-5-1. By substitution, m C = m AD - m BD. Thus, by Distrib. Prop. of =, m C = (m AD - m BD ). Since points determine a line, draw EG. By Ext. Thm., m DEG = m F + m EGF. So m F = m DEG - m EGF. m DEG = m EHG and m EGF = m EG by Theorem 11-5-1. By substitution, m F = m EHG - m EG. Thus, by EHF - m EG ). Distrib. Prop. of =, m F = (m 6. Since points determine a line, draw JM. By Ext. Thm., m JMN = m L + m KJM. So m L = m JMN - m KJM. m JMN = m JN and m KJM = m KM by Inscribed Thm. By substitution, m F = m JN - m KM. Thus, by Distrib. Prop. of =, m F = (m JN - m KM ). 7. m 1 > m because m 1 = (m AB + m CD ) and m = (m AB - m CD ). Since m CD > 0, the expression for m 1 is greater. Copyright by Holt, Rinehart and Winston. 87 Holt Geometry

8. When a tangent and secant intersect on the circle, the measure of formed is half the measure of the intercepted arc. When secants intersect inside the circle, the measure of each formed is half the sum of the measures of the intercepted arcs, or (90 + 90 ). When secants intersect inside the circle, the measure of each formed is half the difference of the measures of the intercepted arcs, or (70-90 ). 9. m AC + m AD + m CD = 60 x - 10 + x + 160 = 60 x = 10 x = 70 AD ) = m B = (m AC - m ((70) - 10-70) = 0 m C = m AD = (70) = 5 m A = 180 - (m B + m C) = 180 - (0 + 5) = 115 40. m B = (m AD - m DE ) 4x = (18x - 15 - (8x + 1)) 8x = 10x - 16 16 = x x = 8 m B = 4(8) = m AD = 18(8) - 15 = 19 m C = (m AED - m m A = 180 - (m B - m C) = 180 - ( + 51) = 97 41a. m BHC = m BC = AD ) = (1-19) = 51 6 (60) = 60 b. m EGD = (m DE - m DAE ) (60 + (60)) = 10 c. m CED = m EDF = (60) = 0 and m EGD > 90, so EGD is an obtuse isosceles. = TEST PREP, PAGE 789 4. C m DCE = (m AF + m DE ) = (58 + 100) = 79 4. J 44. 56 m JLK = (m MN - m JK ) 45 = (146 - m JK ) 90 = 146 - m JK m JK = 56 CHALLENGE AND EXTEND, PAGE 789 45. Case 1: Assume AB is a diameter of the circle. Then m AB = 180 and ABC is a right. Thus, m ABC = m AB. Case : Assume AB is not a diameter of the circle. Let X be the center of the circle and draw radii XA and XB. XA XB. So AXB is isosceles. Thus, XAB XBA, and m XBA + m AXB = 180. This means that m XBA = 90 - m AXB. By Thm. 11-1-1, XBC is a right. So m XBA + m ABC = 90 or m ABC = 90 - m XBA. By substituting, m ABC = 90 - (90 - m AXB) = m AXB. m AXB = m AB because AXB is a central. Thus, m ABC = m AB. 46. Since m WY = 90, m YXW = 90 because it is a central. By Thm. 11-1-1, XYZ and XWZ are right. The sum of measures of the of a quadrilateral is 60. So m WZY = 90. Thus, all 4 of WXYZ are right. So WXYZ is a rectangle. XY XW because they are radii. By Thm. 6-5-, WXYZ is a rhombus. Since WXYZ is a rectangle and a rhombus, it must also be a square by Theorem 6-5-6. (x - 1) = (14-50) 47. m V = x - 1 = 14-50 = 74 x = 74 + 1 = 95 48. Step 1 Find m CED. m DCE + m ECJ = 180 m DCE + 15 = 180 m DCE = 45 m CDE = m FDG = 8 m CED + m DCE + m CDE = 180 m CED + 45 + 8 = 180 m CED = 5 Step Find m GH. m CED = (m GH + mkl) 5 = (m GH + 7) 106 = m GH + 7 m GH = 79 SPIRAL REVIEW, PAGE 789 49. g(7) = (7 ) - 15(7) - 1 = 98-105 - 1 = -8; yes 50. f(7) = 9 - (7) = 9-1 = 8; no 51. - 7 8 (7) = - 49 ; no 5. V = 8 Bh = ap ) h ( = ( ) (4) ) (7) ( = 56 97.0 m Copyright by Holt, Rinehart and Winston. 88 Holt Geometry

5. L = πrl 60π = π(6)l l = 10 cm r + h = l 6 + h = 10 h = 8 cm V = π r h = π(6 ) (8) = 96π cm 10.6 cm 54. S = Pl + s 100 = (96)l + 576 64 = 48l l = 1 in. ( s ) + h = l 1 + h = 1 V = h = 5 in. Bh = (576)(5) = 960 in. 55. m BCA = m BA = (74) = 7 56. m DCB = m DB = m DAB = 67 m BDC = 90 ( BC is a diam.) m DBC = 180 - (90 + 67) = AC = (180-74) = 5 57. m ADC = m 11-6 TECHNOLOGY LAB: EXPLORE SEGMENT RELATIONSHIPS IN CIRCLES, PAGES 790 791 ACTIVITY 1, TRY THIS, PAGE 790 1.. CGF and EGD; FCG and DEG; CFG and EDG are by AA Post.. GC GF = GE ; GC GD = GE GF GD ACTIVITY, TRY THIS, PAGE 791 4. The products of the segment lengths for a tangent and a secant are similar to the products of the segment lengths for secants because for a tangent there is only 1 segment. Thus, the whole segment multiplied by the external segment becomes the square of the tangent seg. 5. tangent segments from same external point will have = lengths, so the circle segments are. 6. In diagram, the circle A is given with tangents DC and DE from D. Since points determine a line, draw radii AC, AE, and AD. AC AE because all radii are. AD AD by Reflex. Prop. of. ACD and AED are right because they are each formed by a radius and a tangent intersecting at point of tangency. Thus, ACD and AED are right. ACD AED by HL. Therefore, DC DE by CPCTC. ACTIVITY, TRY THIS, PAGE 791 7. DGE and FGC; GDE and GFC; GED and GCF; DGE and FGC are by AA. 8. When secants or chords of a circle intersect, 4 segments will be formed, each with the point of intersection as 1 endpoint. The product of segment lengths on 1 secant/chord will equal the product of segment lengths on the other secant/chord. If a secant and a tangent of a circle intersect on an exterior point, segments will be formed, each with exterior point as an endpoint. The product of the secant segment lengths will equal square of tangent segment length. 11-6 SEGMENT RELATIONSHIPS IN CIRCLES, PAGES 79 798 CHECK IT OUT! PAGES 79 794 1. AE EB = CE ED 6(5) = x(8) 0 = 8x x =.75 AB = 6 + 5 = 11 CD = (.75) + 8 = 11.75. original disk: PR = 11 in. New disk: AQ QB = PQ QR 6(6) = (QR) 6 = QR QR = 1 in. PR = 1 + = 15 in. change in PR = 15-11 = _ in.. GH GJ = GK GL 1(1 + z) = 9(9 + 0) 169 + 1z = 81 + 70 1z = 18 z = 14 GJ = 1 + (14) = 7 GL = 9 + 0 = 9 4. DE DF = DG 7(7 + y) = 10 49 + 7y = 100 7y = 51 y = 7 _ 7 Copyright by Holt, Rinehart and Winston. 89 Holt Geometry

THINK AND DISCUSS, PAGE 795 1. Yes; in this case, chords intersect at center of the circle. So segments of the chords are all radii, and theorem simplifies to r = r... EXERCISES, PAGES 795 798 GUIDED PRACTICE, PAGES 795 796 1. tangent segment. HK KJ = LK KM 8() = 4(y) 4 = 4y y = 6 LM = 4 + (6) = 10 HJ = 8 + = 11 4. PS SQ = RS ST z(10) = 6(8) 10z = 48 z = 4.8 PQ = 10 + (4.8) = 14.8 RT = 6 + 8 = 14 6. AC BC = EC DC (x + 7.)7. = (9 + 7.)7. x + 7. = 9 + 7. x = 9 AC = (9) + 7. = 16. EC = 9 + 7. = 16. 7. PQ PR = PS PT 5(5 + y) = 6(6 + 7) 5 + 5y = 78 5y = 5 y = 10.6 PR = 5 + (10.6) = 15.6 PT = 6 + 7 = 1. AE EB = CE ED x(4) = 6(6) 4x = 6 x = 9 AB = (9) + 4 = 1 CD = 6 + 6 = 1 5. Let the diameter be d ft. GF FH = EF (d - EF) 5(5) = 0(d - 0) 65 = 0d - 400 5 = 0d d = 51 4 ft 8. GH GJ = GK GL 10(10 + 10.7) = 11.5(11.5 + x) 07 = 1.5 + 11.5x 74.75 = 11.5x x = 6.5 GJ = 10 + 10.7 = 0.7 GL = 11.5 + (6.5) = 18 9. AB AC = AD ( + 6) = x 16 = x x = 4 (since x > 0) 10. MP MQ = MN ( + y) = 4 9 + y = 16 y = 7 y = 11. RT RU = RS ( + 8) = x = x x = (since x > 0) PRACTICE AND PROBLEM SOLVING, PAGES 796 797 1. DH HE = FH HG (4) = (y) 1 = y y = 6 DE = + 4 = 7 FG = + (6) = 8 14. UY YV = WY YZ 8(5) = x(11) 40 = 11x x = 7 11 UV = 8 +5 = 1 WZ = 7 7 + 11 = 14 11 15. 590(590) = 5.4(d - 5.4) 48,100 = 5.4d - 50,805.16 98,905.16 = 5.4d d 1770 ft 11 16. AB AC = AD AE 5(5 + x) = 6(6 + 6) 5 + 5x = 7 5x = 47 x = 9.4 AC = 5 + (9.4) = 14.4 AE = 6 + 6 = 1 18. PQ PR = PS PT ( + 6) = ( + x) 7 = 4 + x = x x = 11.5 PR = + 6 = 9 PT = + (11.5) = 1.5 1. JK KL = MN KN 10(x) = 6(7) 10x = 4 x = 4. JL = 10 + (4.) = 14. MN = 6 + 7 = 1 17. HL JL = NL ML (y + 10)10 = (18 + 9)9 10y + 100 = 4 10y = 14 y = 14. HL = (14.) + 10 = 4. NL = 18 + 9 = 7 19. UW UX = UV 6(6 + 8) = z 84 = z z = 84 = 1 (since z > 0) Copyright by Holt, Rinehart and Winston. 90 Holt Geometry

0. BC BD = BA ( + x) = 5 4 + x = 5 x = 1 x = 10.5 1. GE GF = GH 8(8 + 1) = y 160 = y y = 4 10 (since y > 0) a. RM MS = PM MQ 10(MS) = 1(1) = 144 MS = 14.4 cm b. RS = RM + MS = 10 + 14.4 = 4.4 cm a. PM MQ = RM MS PM = 4(1-4) = 6 PM = 6 in. b. PQ = PM = (6) = 1 in. 4. Step 1 Find x. AB AC = AD AE 5(5 + 5.4) = 4(4 + x) 5 = 16 + 4x 6 = 4x x = 9 Step Find y. AB AC = AF 5 = y y = 5 = 1 5. Step 1 Find x. NQ QM = PQ QR 4(x) =.(10) = x = 8 Step Find y. KM KN = KL 6(6 + (8) + 4) = y 108 = y y = 108 = 6 6. SP = SE (SE + d) = 6000(14,000) = 84,000,000 SP = 84,000,000 = 000 1 9165 mi 7. Solution B is incorrect. The first step should be AC BC = DC, not AB BC = DC. 8. Since points determine a line, draw AC and BD. ACD DBA because they both intercept AD. CEA BED by Vert. Thm. Therefore, ECA EBD by AA. Corresponding sides are proportional. So AE = CE. By Cross Products ED EB Property, AE EB = CE ED. 9. Since points determine a line, draw AD and BD. m CAD = m BD by Inscribed Thm. m BDC = m BD by Thm. 11-5-1. Thus, CAD BDC. Also, C C by Reflex. Prop. of. Therefore, CAD CDB by AA. Corresponding sides are proportional. So AC = DC. By Cross Products DC BC Property, AC BC = DC. 0. Yes; PR PQ = PT PS, and it is given that PQ = PS. So PR = PT. Subtracting the segments from each of these shows that QR ST. 1. Method 1: By Secant-Tangent Product Theorem, BC = 1(4) = 48, and so BC = 48 = 4. Method : By Thm. 11-1-1, ABC is a right. By Pythagorean Thm., BC + 4 = 8. Thus BC = 64-16 = 48, and BC = 4. a. AE EC = BE DE 5.(4) = (DE) 0.8 = DE DE 6.9 cm b. diameter = BD = BE + DE + 6.9 9.9 cm c. OE = OB - BE (9.9) - 1.97 cm TEST PREP, PAGE 798. B PQ = PR PS = 6(6 + 8) = 84 PQ = 84 9. 4. F PT PU = PQ 7(7 + UT) = 84 7 + UT = 1 UT = 5 5. CE = ED = 6, and by Chord-Chord Product Thm., (EF) = 6(6) = 6. So EF = 1, FB = 1 + = 15, (15) = 7.5. and radius AB = CHALLENGE AND EXTEND, PAGE 798 6a. Step 1 Find the length y of the chord formed by KM. KL = 144 = 8(8 + y) 18 = 8 + y y = 10 Step Find the value of x. 6(6 + x) = 8(8 + (10)) 6 + 6x = 144 6x = 108 x = 18 b. KL + LM KM 1 + (6 + 18 ) (8 + 10 + 8 ) 70 > 676 KLM is acute. Copyright by Holt, Rinehart and Winston. 91 Holt Geometry

7. Let R be center of the circle; then PQR is a right. So PR = PQ + QR = 6 + 4 = 5 PR = 5 = 1 in. Let S be the intersection of PR with circle R, so that SR = 4 in.; then the distance from P to the circle = PS = 1-4. in. 8. By Chord-Chord Product Thm., (c + a)(c - a) = b b c - a = b c = a + b 9. GJ HJ = FJ (y + 6)y = 10 = 100 y + 6y - 100 = 0-6 ± 6-4(1)(-100) y = -6 + = 46 (since y > 0) = - + 109 7.44 SPIRAL REVIEW, PAGE 798 40. P = # favorable outcomes # trials 0.05 = 14 n n = _ 14 0.05 = 400 41. P = 6 = 0.7 = 7% 50 4. BA and CD 4. CD 45. A = π(1 ) ( 60) 55 = π ft 69.1 ft 46. L = π(1) ( 60) 55 = π ft 11.5 ft 44. BC 47. The area of sector YZX is 40π - π = 18π ft A = π r ( 60) m 18π = π(1 ) ( 60) m m 60 = 18 144 = _ 8 m = _ (60) = 45 8 11-7 CIRCLES IN THE COORDINATE PLANE, PAGES 799 805 CHECK IT OUT! PAGES 799 801 1a. (x - h ) + (y - k ) = r (x - 0 ) + (y - (-) ) = 8 x + (y + ) = 64 b. r = ( - ) + ( - (-1)) = 16 = 4 (x - ) + (y - (-1)) = 4 (x - ) + (y + 1 ) = 16 a. Step 1 Make a table of values. Since the radius is 9 =, use ± and values in between for x-values. x - - -1 0-1 - - y 0 ±. ±.8 ± ±.8 ±. 0 Step Plot points and connect them to form a circle. b. The equation of given circle can be written as (x - ) + (y - (-) ) = So h =, k = -, and r =. The center is (, -) and the radius is. Plot point (, -). Then graph a circle having this center and radius.. Step 1 Plot given points. Step Connect D, E, and F to form a. Step Find a point that is equidistant from points by constructing bisectors of sides of DEF. The bisectors of sides of DEF intersect at a point that is equidistant from D, E, and F. The intersection of the bisectors is P(, -1). P is the center of circle passing through D, E, and F. Copyright by Holt, Rinehart and Winston. 9 Holt Geometry

THINK AND DISCUSS, PAGE 801 1. x + y = r. First find the center by finding the midpoint of the diameter. By the Midpoint Formula, the center of the circle is (-1, 4). The radius is half the length of the diameter. So r =. The equation is (x + 1 ) + (y - 4 ) = 4.. No; a radius represents length, and length cannot be negative. 6. The equation of given circle can be written as (x - 1 ) + (y - (-) ) = So h = 1, k = -, and r =. The center is (1, -) and radius is. Plot point (1, -). Then graph a circle having this center and radius. 4. 7. The equation of given circle can be written as (x - (-) ) + (y - (-4) ) = 1 So h = -, k = -4, and r = 1. The center is (-, -4) and radius is 1. Plot point (-, -4). Then graph a circle having this center and radius 1. EXERCISES, PAGES 80 805 GUIDED PRACTICE, PAGE 80 1. (x - h ) + (y - k ) = r (x - ) + (y - (-5)) = 1 (x - ) + (y + 5 ) = 144. (x - h ) + (y - k ) = r (x - (-4)) + (y - 0 ) = 7 (x + 4 ) + y = 49. r = ( - 4 ) + (0-0 ) = 4 = (x - ) + (y - 0 ) = (x - ) + y = 4 4. r = ( - (-1) ) + (- - ) = 5 = 5 (x - (-1) ) + (y - ) = 5 (x + 1 ) + (y - ) = 5 5. The equation of given circle can be written as (x - ) + (y - ) = So h =, k =, and r =. The center is (, ) and radius is. Plot point (, ). Then graph a circle having this center and radius. 8. The equation of given circle can be written as (x - ) + (y - (-4) ) = So h =, k = -4, and r =. The center is (, -4) and radius is 4. Plot point (, -4). Then graph a circle having this center and radius 4. 9a. Step 1 Plot the given points. Step Connect A, B, and C to form a. Step Find a point that is equidistant from points by constructing bisectors of sides of ABC. The bisectors of sides of ABC intersect at a point that is equidistant from A, B, and C. The intersection of bisectors is P(-, ). P is center of circle passing through A, B, and C. b. There are approximately 10 units across the circle. So the diameter is approximately 10 ft. Copyright by Holt, Rinehart and Winston. 9 Holt Geometry

PRACTICE AND PROBLEM SOLVING, PAGES 80 804 10. (x - h ) + (y - k ) = r (x - (-1)) + (y - (-10)) = 8 (x + 1 ) + (y + 10 ) = 64 11. (x - h ) + (y - k ) = r (x - 1.5 ) + (y - (-.5)) = ( ) (x - 1.5 ) + (y +.5 ) = 1. r = ( - 1 ) + ( - 1 ) = (x - 1 ) + (y - 1 ) = ( ) (x - 1 ) + (y - 1 ) = 1. r = (-5-1 ) + (1 - (-)) = 45 = 5 (x - 1 ) + (y - (-)) = ( 5 ) (x - 1 ) + (y + ) = 45 14. The equation of given circle can be written as (x - 0 ) + (y - ) =. So h = 0, k =, and r =. The center is (0, ) and radius is. Plot point (0, ). Then graph a circle having this center and radius. 15. The equation of given circle can be written as (x - (-1) ) + (y - 0 ) = 4. So h = -1, k = 0, and r = 4. The center is (-1, 0) and radius is 4. Plot point (-1, 0). Then graph a circle having this center and radius 4. 16. The equation of given circle can be written as (x - 0 ) + (y - 0 ) = 10. So h = 0, k = 0, and r = 10. The center is (0, 0) and radius is 10. Then graph a circle having origin as center and radius 10. 17. The equation of given circle can be written as (x - 0 ) + (y - (-) ) =. So h = 0, k = -, and r =. The center is (0, -) and radius is. Plot point (0, -). Then graph a circle having this center and radius. 18a. Step 1 Plot the given points. Step Connect A, B, and C to form a. Step Find a point that is equidistant from the points by constructing bisectors of the sides of ABC. The bisectors of sides of ABC intersect at a point that is equidistant from A, B, and C. The intersection of bisectors is P(-1, -). P is center of the circle passing through A, B, and C. b. There are approximately 10 units across the circle. So the diameter is approximately 10 ft. 19. The circle has the center (1, -) and radius. (x - 1 ) + (y - (-) ) = (x - 1 ) + (y + ) = 4 0. The circle has the center (-1, 1) and radius 4. (x - (-1) ) + (y - 1 ) = 4 (x + 1 ) + (y - 1 ) = 16 1a. r = 4 + = 40 d = (40) = 80 units or 80 ft b. x + y = 40 x + y = 1600. F; r = 7. T; (-1 - ) + (- + ) = 9 4. F; center is (6, -4), in fourth quadrant. 5. T; (0, 4 ± ) lie on y-axis and. 6. F; the equation is x + y = = 9. Copyright by Holt, Rinehart and Winston. 94 Holt Geometry

7a. Possible answer: 8 units b. r =, so A = π( ) = 9π 8. units c. Check students work. 8. slope of radius from center (4, -6) to pt. (1, -10) is -10 - (-6) m = = -4 1-4 - = 4 tangent has slope -, and eqn. 4 y - (-10) = - 9a. r = E(-, 5 - ()) = E(-, -1) G(0 - (), ) = G(-6, ) b. d = () = 6 c. center is (0 -, ) = (-, ) (x - (-) ) + (y - ) = (x + ) + (y - ) = 9 0. (x - ) + (x + ) = 81 (x - ) + (x - (-) ) = 9 center (, -), radius 9 1. x + (y + 15 ) = 5 (x - 0 ) + (y - (-15) ) = 5 center (0, -15), radius 5 4 (x - 1) or y + 10 = - (x - 1) 4. (x + 1 ) + y = 7 (x - (-1)) + (y - 0 ) = ( 7 ) center (-1, 0), radius 7. r = ; A = π( ) = 9π; C = π() = 6π 4. r = 7 ; A = π ( 7 ) = 9π; C = π ( 7 ) = 7 π 5. r = ( - (-1) ) + (-1 - ) = 5 A = π(5 ) = 5π; C = π(5) = 10π 6. Graph is a single point, (0, 0). 7. The epicenter (x, y) is a solution of the equations of circles. Let 1 unit represent 100 mi. seismograph A: (x + ) + (y - ) = x + 4x + 4 + y - 4y + 4 = 9 x + 4x + y - 4y = 1 (1) seismograph B: (x - 4 ) + (y + 1 ) = 6 x - 8x + 16 + y + y + 1 = 6 x - 8x + y + y = 19 () seismograph C: (x - 1 ) + (y + 5 ) = 5 x - x + 1 + y + 10y + 5 = 5 x - x + y + 10y = -1 () (1) - (): 1x - 6y = -18 x + = y (4) (1) - (): 6x - 14y = x = 7y + 1 (5) (4) in (5): x = 7(x + ) + 1 x = 14x + -11x = x = - y = (-) + = -1 The location of epicenter is (-00, -100). 8. The circle has a radius of 5. So 5 is tangent to x-axis if the center has y-coordinate k = ±5. 9. d = (5 - (-)) + (- - (-)) = 8; r = 4 center = ( - + 5, _ - + (-) ) = (1, -) equation is (x - 1 ) + (y + ) = 16 40. The locus is a circle with a radius centered at (, ). 41. The point does not lie on circle P because it is not a solution to the equation (x - ) + (y - 1 ) = 9. Since ( - ) + ((-1) - 1 ) < 9, the point lies in the interior of circle P. TEST PREP, PAGE 804 4. C (-, 0) lies on the circle. 4. H (x - (-)) + (y - 5 ) = (1 - (-)) + (5-5 ) (x + ) + (y - 5 ) = 16 44. A distances from statues to fountain: (4 - (-1) ) + (- - (-) ) = 5 (-1 - (-1) ) + ( - (-) ) = 5 (-5 - (-1) ) + (-5 - (-) ) = 5 Copyright by Holt, Rinehart and Winston. 95 Holt Geometry