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Answers to Exercises CHAPTER 6 • CHAPTER 6 CHAPTER 6 • CHAPTER LESSON 6.1 1. 3. 5. 6. 50° 30° 76 in. possible answer: 2. 55° 4. 105° 11. Construct a line and label a point T on it. On the line, mark off two points, L and M, each on the same side of T at distances r and t from T, respectively. Construct circle L with radius r. Construct circle M with radius t. r L M T t Answers to Exercises Target 7. Possible answer: The perpendicular to the tangent passes through the center of the circle. Use the T-square to find two diameters of the Frisbee. The intersection of these two lines is the center. . Construct a line through point T 8. Construct OT . perpendicular to OT r T O , OY , and OZ . Construct tangents 9. Construct OX through points X, Y, and Z. Z O t Y X 10. Construct tangent circles M and N. Construct equilateral MNP. Construct circle P with radius s. M s s N P 74 ANSWERS TO EXERCISES 12. Start with the construction from Exercise 11. , mark off length TK so that TK s On line LM and K is on the opposite side of T from L and M. Construct circle K with radius s. r L M T t K s 13. Sample answer: If the three points do not lie on the same semicircle, the tangents form a circumscribed triangle. If two of the three points are on opposite sides of a diameter, the tangent lines to those two points are parallel. If the points lie on the same semicircle, they form a triangle outside the circle, with one side touching (called an exscribed triangle). 14. sample answer: internally tangent: wheels on a roller-coaster car in a loop, one bubble inside another; externally tangent: touching coins, a snowman, a computer mouse ball and its roller balls 15. Constructions will vary. 16. 360° 30° 90° 90° 150° and 150° % 41.6 360° 17. 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°. 18a. rhombus 18b. rectangle 18c. kite 18d. parallelogram 19. 78° 20. 9.7 km 21. x 55° 55° 180° and 40° y y 180°, so x y 70° 11 22. 2 1 LESSON 6.2 5 cm 5 cm 15. M(4, 3), N(4, 3), O(4, 3) 16. The center of the circle is the circumcenter of the triangle. Possible construction: 17. possible construction: O 18. 13.8 cm 19. They can draw two chords and locate the intersection of their perpendicular bisectors. The radius is just over 5 km. 20. See flowchart below. 20. (Lesson 6.2) 1 AB CD ? Given ᎏ 2 AO CO All radii of a circle are congruent 3 AOB COD 4 ? ᎏ ? ᎏ ? SSS Congruence ᎏ Conjecture 5 COD ? AOB ᎏ ? CPCTC ᎏ BO DO ? All radii of a circle are congruent ᎏ ANSWERS TO EXERCISES 75 Answers to Exercises 1. 165°, definition of measure of an arc 2. 84°, Chord Arcs Conj. 3. 70°, Chord Central Angles Conj. 4. 8 cm, Chord Distance to Center Conj. 68°; mB 34° (Because OBC is 5. mAC isosceles, mB mC, mB mC 68°, and therefore mB 34°.) 6. w 115°, x 115°, y 65°; Chord Arcs Conjecture 7. 20 cm, Perpendicular to a Chord Conj. 8. w 110°, x 48°, y 82°, z 120°; definition of arc measure 9. x 96°, Chord Arcs Conjecture; y 96°, Chord Central Angles Conjecture; z 42°, Isosceles Triangle Conjecture and Triangle Sum Conjecture. 10. x 66°, y 48°, z 66°; Corresponding Angles Conjecture, Isosceles Triangle Conjecture, Linear Pair Conjecture 11. The length of the chord is greater than the length of the diameter. 12. The perpendicular bisector of the segment does not pass through the center of the circle. 13. The longer chord is closer to the center; the longest chord, which is the diameter, passes through the center. 14. The central angle of the smaller circle is larger, because the chord is closer to the center. 1 21. y 7 x; (0, 0) is a point on this line. 22a. true C A X B Answers to Exercises D is the perpendicular Possible explanation: If AB , then every point on AB is bisector of CD equidistant from endpoints C and D. Therefore AD and BC BD . Because CD is not the AC , C is not equidistant perpendicular bisector of AB from A and B. Likewise, D is not equidistant from and BC are not congruent, and A and B. So, AC and BD are not congruent. Thus ACBD AD has exactly two pairs of consecutive congruent sides, so it is a kite. 22b. false, isosceles trapezoid 22c. false, rectangle 23. 140°, 160°, 60°; 180° minus the measure of the intercepted arc 76 ANSWERS TO EXERCISES 24. Using the Tangent Conjecture, 2 and 4 are right angles, so m2 m4 180°. According to the Quadrilateral Sum Conjecture, m1 m2 m3 m4 360°, so by the Subtraction Property of Equality, m1 m3 180°, or m1 180° m3. The measure of a central angle equals the measure of . its arc, so by substitution, m1 180° mAB 25. the circumcenter of the triangle formed by the three light switches 26. 7 27. Station Beta is closer. Downed aircraft N 4.6 mi 48 7.2 mi 72 38 Alpha 28. D Beta 312 8.2 mi LESSON 6.3 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° 23. (2, 1); Possible method: Plot the three points. Construct the midpoint and the perpendicular bisector of the segments connecting two different pairs of points. The center is the point of intersection of the two lines. To check, construct the circle through the three given points. 24. See flowchart below. 25. Start with an equilateral triangle whose vertices are the centers of the three congruent circles. Then locate the incenter/circumcenter/orthocenter/ centroid (all the same point because the triangle is equilateral) to find the center of the larger circle. To find the radius, construct a segment from the incenter of the triangle through the vertex of the triangle to a point on the circle. It works on acute and right triangles. 20. The camera can be placed anywhere on the major arc (measuring 268°) of a circle such that the row of students is a chord intersecting the circle to form a minor arc measuring 92°. This illustrates the conjecture that inscribed angles that intercept the same arc are congruent (Inscribed Angles Intercepting Arcs Conjecture). 21. two congruent externally tangent circles with half the diameter of the original circle 26. mA 60°, mB 36°, mC 90°; 60° 36° 90° 180° 24. (Lesson 6.3) 1 OR CD 2 ORC and ORD are right angles ? Definition of ᎏ perpendicular lines 3 6 OCD is isosceles 7 C ? D ᎏ ? Given ᎏ 5 OC OD ? ᎏ All radii of a circle are congruent ? Definition of ᎏ isosceles triangle mORC ⫽ 90 mORD ⫽ 90 ? Definition of ᎏ right angle ? ᎏ 7. Isosceles Triangle Conjecture 4 mORC ⫽ mORD (ORC ORD) Substitution ? ᎏ property 8 OCR ? ᎏ ? ᎏ ODR SAA 9 CR ? DR ᎏ ? ᎏ CPCTC 10 OR bisects CD ? Definition of bisect ᎏ ANSWERS TO EXERCISES 77 Answers to Exercises 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. 17. BFE DFA (Vertical Angles Conjecture). BGD FHD (all right angles congruent). Therefore, B D (Third Angle Conjecture). , mD 1 mEC , AC EC mB 21 mAC 2 18. Possible answer: Place the corner so that it is an inscribed angle. Trace the inscribed angle. Use the side of the paper to construct the hypotenuse of the right triangle (which is the diameter). Repeat the process. The place where the two diameters intersect is the center. 19. possible answer: Answers to Exercises LESSON 6.4 mABD by the 1. Proof: mACD 21mAD Inscribed Angle Conjecture. ACD ABD. 2. Proof: By the Inscribed Angle Conjecture, mACB 21mAD B 21(180°) 90°. ACB is a right angle. 3. Proof: By the Inscribed Angle Conjecture, and mL 1mYCI 1(360° mC 21mYLI 2 2 1 mYLI ) 180° 2mYLI 180° mC. L and C are supplementary. (A similar proof can be used to show that I and Y are supplementary.) 2m2 4. Proof: 1 2 by AIA. mBC 2m1 mAD by the Inscribed Angle Conjecture. AD . BC 5. True. Opposite angles of a parallelogram are congruent. If it is inscribed in a circle, the opposite angles are also supplementary. So they are right angles, and the parallelogram is indeed equiangular, or a rectangle. and DB . Diagonal RG 6. Construct diagonals RG is the perpendicular bisector of BD by the Kite Diagonal Bisector Conjecture. Because kite BRDG and RG are chords of is inscribed in the circle, BD the circle. By the Perpendicular Bisector of a Chord Conjecture, the perpendicular bisector of BD passes through the center of the circle. Because RG is a chord of the circle that passes through the is a center, by the definition of diameter, RG diameter. , , and AR . Because they 7. Draw in radii GR ER , TR TR . are radii of the same circle, GR ER and AR By the Parallel Lines Intercepted Arcs Conjecture, ET . Their central angles must also be GA congruent, so GRA ERT. Thus GRA ET by CPCTC. GATE is ERT by SAS, so GA an isosceles trapezoid. 8. x 65°, y 40°, z 148°; half the measure of the intercepted arc 9. Because the radii of a circle are congruent, OB , so OAB is isosceles. By the Isosceles OA Triangle Conjecture, 2 3, so m2 m3. By the Triangle Sum Conjecture, m1 m2 m3 180°, so by substitution, m1 2(m2) 180°, or m1 180° 2(m2). 78 ANSWERS TO EXERCISES , so BC By the Tangent Conjecture, OB mOBC 90°. By Angle Addition, m2 m4 mOBC, so m2 m4 90°, or m2 90° m4. Substitute 90° m4 for m2 in the earlier equation: m1 180° 2(90° m4). Simplifying gives m1 2(m4). m1, substitution gives mAB Because mAB . 2(m4), or m4 21mAB 10a. S; An equilateral triangle is equiangular, but a rhombus is not equiangular. 10b. A 10c. N; Only one diagonal is the perpendicular bisector of the other. 10d. A 10e. S; An equilateral triangle has rotational symmetry and three lines of symmetry; a parallelogram has rotational symmetry but no line of symmetry. 11. L 12. J 13. K 14. D 15. E 16. B 17. H 18. G 19. N 20. a b b a 180°, so a b 90° 2 21. 2 1 RS . OP OQ OR 22. It is given that PQ because all radii in a circle are congruent. OS Therefore, OPQ ORS by SSS and 2 1 by CPCTC. Now we move on to the smaller right triangles inside OPQ and ORS. It is given that PQ and OV RS . Therefore, OTQ and OT OVS are right angles by the definition of perpendicular lines and OTQ OVS because all right angles are congruent. Thus, OTQ OV by CPCTC. OVS by SAA and OT LESSON 6.5 1. d 5 cm 2. C 10 cm 12 3. r m 11 4. C 5.5 or 2 18. (Lesson 6.5) 1 2 ? MA MT ᎏ Given ? SA ST ᎏ Given 3 MAST is a kite ? Definition of kite ᎏ is the perpendicular bisector of AT MS 4 ? ᎏ ? Kite Diagonal Bisector Conjecture ᎏ ANSWERS TO EXERCISES 79 Answers to Exercises 5. C 12 cm 6. d 46 m 7. C 15.7 cm 8. C 25.1 cm 9. r 7.0 m 10. C 84.8 in. 11. 565 ft 12. C 6 cm 13. 16 in. 14. Trees grow more in years with more rain; 244 yr 15. 1399 tiles 16. g 40°, n 30°, x 70°; y 142°; z 110°; Conjecture: The measure of the angle formed by two intersecting chords is one-half the sum of the measures of the two intercepted arcs. In the mLG . diagrams, mAEN 21mAN and mLNG 1mLG 17. mAGN 21mAN 2 by the Inscribed Angle Conjecture. mAEN mAGN mLNG by the Triangle Exterior mLG Angle Conjecture. So, mAEN 21mAN by substitution and the distributive property. 18. See flowchart below. 19. b 90°, c 42°, d 70°, e 48°, f 132°, g 52° 5 150° 20. 360° or 12 21. The base angles of the isosceles triangle have a measure of 39°. Because the corresponding angles are congruent, m is parallel to n. 22. 10x 2y LESSON 6.6 Answers to Exercises 1. 4398 km/h 2. 11 m/s 3. 37,000,000 revolutions 4. 637 revolutions 5. Mama; C 50 in. 6. 168 cm 7. d 7.6 ft. The table will fit, but the chairs may be a little tight in a 12-by-14 ft room. 12 chairs 192 in., 12 spaces 96 in., C 288 in., d 91.7 in. 7.6 ft. 8. 0.35 ft/s 9. a 37.5°, s 17.5°, x 20°; y 80°; z 61°; Conjecture: The measure of an angle formed by two secants that intersect outside a circle is one-half the difference of the larger arc measure and the smaller arc measure. 80 ANSWERS TO EXERCISES and mSAN 1mSN by 10. mESA 21mEA 2 the Inscribed Angle Conjecture. mSAN mESA mECA by the Triangle Exterior Angle 1mEA mECA by Conjecture. So, 21mSN 2 substitution. With a little more algebra, mECA 1 mSN mEA . 2 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. 12. C 13. 38° 14. 48° 15. 30 cm 400 rev 2 . 2 6 ft 1 min 16. 1 min 1 rev 60 s 1089 ft/s 17. 12 cm third side 60 cm. This is based on the Triangle Inequality Conjecture. USING YOUR ALGEBRA SKILLS 6 50 40 Cost ($) 1 1. 2, 3 2. (3, 3) 2 3. 5, 3 4. (7, 2) 5. infinitely many solutions 6. no solution 7. 4. The lines intersect at the solution point, (7, 2). The circumcenter is 272 , 178 . Only two perpendicular bisectors are needed because the third bisector intersects at the same point. 9a. Plan A: y 4x 20; Plan B: y 7x; x 6 h 40 min, y $46.67 y 9b. y y 2x 16 5 30 Plan B 20 (0, 20) 10 2 5 ( 7, 2) x 10 4 6 Hours x The point of intersection shows when both plans cost the same, the answer for 9a. 9c. Plan B; 721 hours, with Plan A 10. (3, 1), (0, 3), (3, 1) y 12 x 32 –5 –10 y y 13 x 2 x 6. The lines are parallel (the slopes are the same, but the y-intercepts are different). y There is no solution. y = –2x + 9 5 x –5 –5 y = –2x – 1 –10 8. Possible solution using the equations y 21x 1 and y 32x 134 : 1 y 2x 1 2 14 1 3x 3 2x 1 4x 28 3x 6 22 7x 22 x 7 1 22 y 2 7 1 11a. y 43x 23 y x 12 11b. (6, 6) 11c. (6, 6); (6, 6) 11d. The diagonals intersect at their midpoints, which supports the conjecture that the diagonals of a parallelogram bisect each other. 2 13 12. y 3x 3 13. (4, 7) 3 15. 3, 2 69 6 14. 1, 4 7 16. The circumcenter is the midpoint of the hypotenuse. For a right triangle, the perpendicular bisectors of the legs are two of the midsegments of the triangle. As with any pair of triangle midsegments, they intersect at the midpoint of the third side. 18 y 7 ANSWERS TO EXERCISES 81 Answers to Exercises 5. The lines are the same. There are infinitely many solutions. 4 6 _23 , 46 _23 Plan A Answers to Exercises LESSON 6.7 4 1. 3 in. 2. 8 m 3. 14 cm 4. 9 m 5. 6 ft 6. 4 m 7. 27 in. 8. 100 cm 9. 217 m/min 10. 4200 mi 11. Desks are about 17 meters from the center. About four desks will fit because an arc with one-half the radius and the same central angle will be one-half as long as the outer arc. 12. The measure of the central angle is 7.2° because of the Corresponding Angles Conjecture. Therefore, 500 376.20 C, so C 25,000 mi. 13. 18°/s. No, the angular velocity is measured in degrees per second not in distance per second, so it is the same at every point on the carousel. 14. Outer horse 2.5 m/s, inner horse 1.9 m/s. One horse has traveled farther in the same amount of time (tangential velocity), but both horses have rotated the same number of times (angular velocity). 15. a 50°, b 75°, x 25°; y 45°; z 35°; Conjecture: The measure of the angle formed by an intersecting tangent and secant to a circle is one-half the difference of the larger intercepted arc measure and the smaller intercepted arc measure. 16. Let z represent the measure of the exterior . and tangent PB angle of PBA formed by AB 1 By the Tangent Chord Conjecture, z 2mAB . By . the Inscribed Angle Conjecture, mBAP 21mBC By the Triangle Exterior Angle Conjecture, z mBPA mBAP, or mBPA z mBAP. 82 ANSWERS TO EXERCISES 1mBC . By substitution, mBPA 21mAB 2 Using the distributive property, mBPA 1 (mAB mBC ). 2 17. a 70°, b 110°, c 110°, d 70°, e 20°, f 20°, g 90°, h 70°, k 20°, m 20°, n 20°, p 140°, r 80°, s 100°, t 80°, u 120° 18. possible answer: 19. 170° 8 289 20. y 1x 5 15 21. 45° 22. The sum of the lengths is 8 cm for Case 1, Case 2, Case 3, and Case 10. 23. 6 24. Yes, as long as the three points are noncollinear; possible answer: connect the points with segments, then find the point of concurrency of the perpendicular bisectors (same as circumcenter construction). CHAPTER 6 REVIEW 23. Sample answer: Construct a right angle and and RT with any label the vertex R. Mark off RE lengths. From point E, swing an arc with radius RT. From point T, swing an arc with radius RE. Label the intersection of the arcs as C. Construct the and RC . Their intersection is the diagonals ET center of the circumscribed circle. The circle’s radius is the distance from the center to a vertex. It is not possible to construct an inscribed circle in a rectangle unless it is a square. E C R T 24. Sample answer: Construct acute angle R. Mark off equal lengths RM and RH. From points M and H, swing arcs of lengths equal to RM. Label the intersection of the arcs as O. Construct RHOM. The intersection of the diagonals is the center of the inscribed circle. Construct a perpendicular to a side to find the radius. It is not possible to construct a circumscribed circle unless the rhombus is a square. H O R M 4 32 25. 4x 3y 32, or y 3x 3 26. (3, 2) 27. d 0.318 m 28. Melanie: 151 m/min or 9 km/h; Melody: 94 m/min or 6 km/h. 2(6357) 2(6378) 29. 1.849 1.852 1.855 360 60 360 60 30. Possible location 7700 ft 5280 ft 5500 ft D T Possible location ANSWERS TO EXERCISES 83 Answers to Exercises 1. Answers will vary. 2. Draw two nonparallel chords. The intersection of their perpendicular bisectors is the center of the circle. Fold the paper so that two semicircles coincide. Repeat with two different semicircles. The center is the intersection of the two folds. Place the outside or inside corner of the L in the circle so that it is an inscribed right angle. Trace the sides of the corner. Draw the hypotenuse of the right triangle (which is the diameter of the circle). Repeat. The center is the intersection of the two diameters. 3. The velocity vector is always perpendicular to the radius at the point of tangency to the object’s circular path. 4. Sample answer: An arc measure is between 0° and 360°.An arc length is proportional to arc measure and depends on the radius of the circle. 5. 55° 6. 65° 7. 128° 8. 118° 9. 91° 10. 66° 11. 125.7 cm 12. 42.0 cm 13. 15 cm 14. 14 ft 15. 2 57° 2 35° 180° 16. 84° 56° 56° 158° 360° 1 (180° 108°) 36° 17. mEKL 21 mEL 2 YL by Converse of the Parallel mKLY. KE Lines Conjecture. 360° 56° 152° 152° mMI . 18. mJI 1 mMI mMJI. By the mJMI 21 mJI 2 Converse of the Isosceles Triangle Conjecture, JIM is isosceles. 2mKEM 140°. mKI 19. mKIM 140° 70° 70° mMI . mIKM 1 1 mMI mKI mIMK. By the Converse of the 2 2 Isosceles Triangle Conjecture, KIM is isosceles. 20. 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. 21. Sample answer: Construct perpendicular bisectors of two sides of the triangle. The point at which they intersect (the circumcenter) is the center of the circle. The distance from the circumcenter to each vertex is the radius. 22. Sample answer: Construct the incenter (from the angle bisectors) of the triangle. From the incenter, which is the center of the circle, construct a perpendicular to a side. The distance from the incenter to the foot of the perpendicular is the radius. 200 31. ft 63.7 ft 32. 8 m 25.1 m 8 33. The circumference is 346 0 2(45) 12; the diameter is 12 cm. 34. False. 20° 20° 140° 180°. An angle with measure 140° is obtuse. 35. true D 36. false C Answers to Exercises A 53. False. The ratio of the circumference to the diameter is . 24 cm 24 cm 54. false; 24 24 48 48 96 48 cm 55. true 56. This is a paradox. 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° 58. TAR YRA by SAS, TAE YR E by SAA 59. FTO YTO by SAA, SAS, or SSS; FLO YLO by SAA, SAS, or SSS; FTL YTL by SSS, SAS, or ASA 60. PTR ART by SSS or SAS; TPA RAP by SSS, SAS, SAA, or ASA; TLP RLA by SAA or ASA 61. ASA 84°, length of AC 11.2 35.2 in. 62. mAC 63. x 63°, y 27°, w 126° 64. sample answer: B 37. true 38. true 39. true 40. true 41. False. (7 2) 180° 900°. It could have seven sides. 42. False. The sum of the measures of any triangle is 180°. 43. False. The sum of the measures of one set of exterior angles for any polygon is 360°. The sum of the measures of the interior angles of a triangle is 180° and of a quadrilateral is 360°. Neither is greater than 360°, so these are two counterexamples. 44. False. The consecutive angles between the bases are supplementary. 45. False. 48° 48° 132° 180° 46. False. Inscribed angles that intercept the same arc are congruent. 47. False. The measure of an inscribed angle is half the measure of the arc. 48. true and BD bisect each other, but AC is 49. False. AC . not perpendicular to BD D C A B 48 cm 30 150 65. See table below. 66a. The circle with its contents has 3-fold rotational symmetry, the entire tile does not. 66b. No, it does not have reflectional symmetry. 67. 68. 9.375 cm 50. False. It could be isosceles. 51. False. 100° 100° 100° 60° 360° CD 52. false; AB C A 69. 90 90 70. D B 65. (Chapter 6 Review) 84 n 1 2 3 4 5 6 f (n) 5 1 3 7 11 15 ANSWERS TO EXERCISES ... n ... 20 . . . 9 4n . . . 71