<strong>Chapter</strong> <strong>11</strong> <strong>Answers</strong> (continued) <strong>11</strong>. y x <strong>Chapter</strong> Test, Form B 1. C(0, 0); r = 9 2. C(-5, 2); r = 4 3. C(4, -8); r = "2 4. (x - 1) 2 + (y - 3) 2 = 4 5. (x - 1) 2 + y 2 = 1 6. (x + 3) 2 + (y - 1) 2 = 16 7. C = 25.1; A = 50.3 8. (x - 5) 2 + (y - 1) 2 = 89 9. y C(2, 1) r 4 12. y x x 10. 90 <strong>11</strong>. y 13. x 5 y x x 12. y 14. y y 4 y 6 15. 120 16. 105 17. <strong>11</strong>8.0 18. 40 19. 70 20. <strong>11</strong>5 21. 6 22. <strong>11</strong>.8 23. 5.5 24. 45 25. 35 26. 80 27. 5.8 28. 5.7 29. x = 63; y = 71 30. x = 87; y = 95; w = 85; z = 93 31. x = 120; y = 64; z = 176 32. 44 x 13. y y 2 x x © Pearson Education, Inc. All rights reserved. 36 <strong>Answers</strong> Geometry <strong>Chapter</strong> <strong>11</strong>
<strong>Chapter</strong> <strong>11</strong> <strong>Answers</strong> (continued) © Pearson Education, Inc. All rights reserved. 14. x 7 x 3 y 15. <strong>11</strong>0.3 16. 77.4 17. 120 18. 42 19. 80 20. x = 125; y = 90 21. 8.4 22. 20 23. 2 24. 40 25. 145 26. 74 27. 4.5 28. 3.5 29. x = 60; y = 80 30. x = 88; y = 93; z = 92; w = 87 31. x = <strong>11</strong>6; y = 80; z = 164 32. 45.5 Alternative Assessment, Form C TASK 1: Scoring Guide (a) center (2, 5), radius = 4 (b) The distance from point (x, y) on a circle to the center (2, 5) is given by the formula D = "(x 2 2) 2 1 (y 2 5) 2 .In any circle, the distance, or radius, is the same for all points on the circle. Therefore the distance formula, when applied to the circle in this problem, becomes 4 = "(x 2 2) 2 1 (y 2 5) 2 for all points (x, y) on the circle. Squaring both sides of this equation yields the equation of the circle, 16 = (x - 2) 2 + (y - 5) 2 . 3 Student gives accurate answers and a correct explanation. 2 Student gives answers or an explanation that may contain minor errors. 1 Student gives wrong answers or an incomplete or inaccurate explanation. 0 Student makes little or no effort. TASK 2: Scoring Guide (a) Construct the perpendicular bisector of two chords. The point at which they meet is the center of the circle. (b) Because the pentagon is regular, all chords are congruent. Therefore the corresponding arcs are congruent. Because there are 5 0 congruent arcs in the circle, each arc, and in particular AB , has measure 72. To find 0 AB, call the center O, and consider triangle AOB. Because AB has measure 72, so does angle AOB. Because OA = OB, AOB is isosceles. Therefore, the bisector of angle AOB is perpendicular to (and bisects) chord AB, forming a 36°-54°-90° triangle. Then 1 sin 36 2AB AB = = , yielding AB = 7.05. OA 12 3 Student devises a correct method and gives a correct answer and a valid explanation. 2 Student devises a method, gives an answer, and gives an explanation that may contain some errors. 1 Student gives a method, an answer, and an explanation that may contain major errors or omissions. 0 Student makes little or no effort. x TASK 3: Scoring Guide (a) Angles A and C are inscribed angles. Each is inscribed in a different arc, but together the arc in which they are inscribed comprise the entire circle. Therefore, 1 mA + mC = 2(360) = 180. And, because ABCD is a parallelogram, angles A and C are congruent. Finally, if angles are congruent and supplementary, then they are right. (b) The diagonals of ABCD bisect each other because ABCD is a parallelogram. Suppose that they intersect at point X. Then AX = CX and BX = DX. Also, because ABCD is inscribed in a circle, the diagonals are chords, and therefore AX ? CX = BX ? DX. Substituting yields AX 2 = BX 2 , and therefore AX = BX = CX = DX. Then AC = BD. (c) Either (a) or (b) allows you to conclude that ABCD must be a rectangle. 3 Student gives valid and accurate arguments and explanations. 2 Student gives arguments that, although basically valid, may contain minor flaws. 1 Student gives arguments containing major flaws. 0 Student makes little or no attempt. TASK 4: Scoring Guide (a) Because the circumscribing lines are tangent to the circle, AB = AX, CB = CY, and DX = DY. Then, because it is given that AB = BC, we have AX = CY. Adding segments together yields DA 0 = DC, showing that 1 ACD is isosceles. Secondly, because XY has measure 100, XBY has measure 1 260. Then mD = 2(260 - 100) = 80, and because ACD is isosceles, mA = mC = 50. Finally, AC = 20 AB because AB = BC = 10, and, by trigonometry, cos 50 = AD, AB 10 AD = cos 50 = 0.6428 = 15.56. And again, AD = CD. (b) Because angle A has measure 50 and line AB is tangent at B, triangle AOB is a 25°-65°-90° triangle. Therefore, by OB trigonometry, tan 25 = AB,OB= 10 ? tan 25 = 4.66. 3 Student gives accurate answers and explanations. 2 Student gives answers and explanations that may contain minor errors. 1 Student gives answers and explanations that contain significant errors. 0 Student makes little or no attempt. Cumulative Review 1. B 2. D 3. E 4. A 5. D 6. B 7. A 8. A 9. D 10. B <strong>11</strong>. C 12. C 13. D 14. C 15. B 16. "89 17. Check students’ work. 18. A tangent intersects a circle at exactly one point. A secant intersects a circle at two points. 19. Check students’ work. 20. This follows from the Pythagorean Theorem (c 2 = a 2 + b 2 ) or the fact that the longest side of a triangle is opposite the largest angle. 21. If a triangle is not a right triangle, then it is acute. 22. 135 Geometry <strong>Chapter</strong> <strong>11</strong> <strong>Answers</strong> 37
Change language