Cpm Geometry Answers

Selected Answersfor Core Connections Geometry Lesson 1.1.1 1-3. Shapes (a), (c), (d), and (e) are rectangles. 1-4. a: 40 b: –6 c: 7 d: 59 1-5. a: 3 b: 5 c: 6 d: 2 1-6. a: 22a + 28 b: !23x ! 17 c: x 2 + 5x d: x 2 + 8x 1-7. Possibilities: Goes to bank, gets money from parent, gets paid; buys lunch, goes shopping, pays a bill, … Lesson 1.1.2 1-14. Answers vary. Possible responses include “How many sides does it have?”, “Does it have a right angle?”, “Are any sides parallel?” 1-15. Answers vary. Possible responses include “They have 3 sides of equal length” and “They have 3 angles of equal measure.” 1-16. a: 3 b: 2 c: 4 1-17. a: x = –7 b: c = 4.5 c: x = 16 d: k = –7 1-17. a: 12 b: 35 c: 24 d: 7 Lesson 1.1.3 1-25. c is correct; x = 7 1-26. No – if the points are collinear then they will not form a triangle. 1-27. See answers in bold below. y = x ! 3 x 3 –1 0 2 –5 –2 1 y 0 –4 –3 –1 –8 –5 –2 1-28. a: 55.5 sq. units b: 42 sq. units 1-29. a: 5 b: 12 c: –7 2 Core Connections Geometry Lesson 1.1.4 1-32. a: x = 9 24 = 3 8 = 0.375 b: No solution c: x ≈ 6.44 d: x = 0.5 1-33. Yes, his plants will be dead. If his plants are indoors, they will be dead because he will be gone for 2 weeks and so he did not water them at least once a week. If he left them outdoors, they will still be dead because it has not rained for 2 weeks, so he needed to water them once a week as well. 1-34. a: y = 2 3 x!4 b: y = ! 52 + 27 1-35. a: 6x + 6 b: 6x + 6 = 78 , so x = 12 and the rectangle is 15 cm by 24 cm. c: ( 2 !12 ) ( 2 + 3 ) = 360 1-36. a: ! 53 b: 6 3 = 2 1 =2 c: ! 63 = ! 12 d: 0 7 =0 Lesson 1.1.5 1-42. a: 100° b: 170° c: 50° 1-43. a: Yes, it is correct because the two angles make up a 90° angle. b: x = 33! so one angle is 33 – 10 = 23° while the other is 2(33) + 1 = 67°. 1-44. Perimeter: 74 centimeters, Area: 231 cm 2 1-45. a: y = 5 b: r = 12 c: a = 6 d: m = 5 1-46. While there are an infinite number of rectangles, possible dimensions with integral measurements are: 1 by 24 (perimeter = 50 units), 2 by 12 (perimeter = 28 units), 3 by 8 (perimeter = 22 units), and 4 by 6 (perimeter = 20 units). Selected Answers 3 Yes. a: Reflection b: Translation (or two reflections over parallel lines) c: Rotation or rotation and translation d: Rotation or rotation and translation depending on point of rotation e: Reflection f: Reflection and then translation or rotation or both 1-65. The flag would need to be a rectangle. no 1-64. 1-66.2. 19 + 7x + 10x + 3 = 52 .4 1-58. a: 27 48 ! 56. so 5(9) – 2 = 43 miles –4 y = !4x + 5 1-57.Lesson 1. and the cylinder’s radius would match the width of the rectangle. and 23. y y = 3x ! 3 4 1-55.25 f: x = –3. so x = 2. a: Area ≈ 16 square units b: Area ≈ 15 square units 1-67.2 1-63. See graph at right. a: x = 3. yes. The height of the cylinder would match the height of the rectangle along the pole. 5x ! 2 + 2x + 6 = 67 .1 1-54.2. Lesson 1. Side lengths are 19. 10.75 b: x = 3 c: x = 0 d: x = 3 e: x ≈ 372.3% b: 130 = 13 ! 8% 10 1 y = ! 23 x + 3 c: 0 d: 5 ! 56% 8 –4 4 x 1-56. x = 9. a: y = –4 b: y = 25 c: y = –2 4 Core Connections Geometry . 14 d: a = 22 1-87. 3). we can use the theoretical probability computation in the Math Notes box in Lesson 1.Lesson 1. a: There are 10 combinations: a & b.1.7 1-78. b & e. See answers in bold below.2. y = 3x + 2 x –3 –2 –1 0 1 2 3 4 y –7 –4 –1 2 5 8 11 14 Lesson 1.2. –14 Selected Answers 5 .2. a: y = 5 2 x!3 b: y = 3 2 x+7 1-78. b: A´ (4. C 1-79. c: 3 10 d: 9 10 e: The outcomes that satisfy part (d) include the outcomes that satisfy part (c).75 b: x = !94 c: x ! 1. so each pair of consecutive sides is perpendicular and forms 90° angles. –1). a & c. b & d. a: p = !5 b: w = 20 c: x = 14 d: y = 1. –1) 1-86. c & e. a: Yes b: 5 c: 5 d: 5 e: 5 1-77. but there are others on the part (d) list as well. a & e. d & e b: Yes. because they all have a slope of 4 . 3 e: y = ! 3 4 x + 16 1-89. a: Yes. It has four sides. B´ (6. –5). b & c. c & d. D´ (–4.3 2 3 1 4 1-76. C´ (–2. a: y = 4 3 x!2 b: The resulting line coincides with the original line. y = 4 3 x!2 c: The image is parallel. y = 4 3 x!7 d: They are parallel. a & d. If the outcomes are equally likely. a: x = !4. 1-88.4 1-85. mAB = mCD = 12 and mBC = mAD = !2 . 1-95. –7) d: (–52. a: 10 square units b: 20 square units c: 208. a: (9. 6 Core Connections Geometry . –3) c: (–2.Lesson 1. c: There are 6 lines of symmetry. 1 535 ! 0.2.0019 . No. a: b: A B P l c: d: C D Q m 1-98. a: The orientation of the hexagon does not change. 3) b: (3. through opposite vertices and through the midpoints of opposite sides. 1483) 1-96. b: The orientation of the hexagon does not change. this probability is very small.5 1-94.680 square units 1-97. a: One possibility: 4(5x + 2) = 48 b: x = 2 c: 12 !12 = 144 square units 1-107. b: Solution should be any value of 45k where k is an integer. a: b: A B P l c: d: D C Q m Selected Answers 7 .6 1-105. c: circle 1-109.Lesson 1. while (b) and (c) are parallel. 1-106.2. (a) and (b) are perpendicular. a: 4 52 = 13 1 b: 13 52 = 1 4 c: 1 52 d: 39 52 = 3 4 1-108. a: It looks the same as the original. Lesson 1. –3) b: The vertices are: (6. M. while Robert’s is 53 . (2. 3). E. K. G. U. T. Y Rotation only: N. 1-116. (5. W. R 1-127. Carol: only inside circle #2. C. 2). Q.3. Robert has a greater 4 5 chance. D. J.2 1-123. Scalene: (b) 1-124. B. Bob: outside both circles.1 1-112. a: heart b: square c: hexagon d: Answer vary. O. 1-113. a: isosceles triangle b: pentagon c: parallelogram d: obtuse scalene triangle e: isosceles right triangle f: trapezoid 1-126. L. –3) Lesson 1. Pedro: only inside circle #1. X Outside both regions: F. I.3. a: (–6. Therefore. V. Z Intersection only: H. 6) c: (8. Isosceles: (a) and (c). 1-114. a person would need to have long hair and study a lot for class. P. a: 7 ≈ 44% b: 9 ≈ 56% 16 16 c: 1 16 ≈ 6% d: 6 16 = 3 8 ≈ 38% 1-125. Reflection only: A. S. Sandy is more likely to select a shape with two sides that are parallel. a: x = ! 9 33 = ! 11 3 b: x = 5 and x = ! 23 c: x = 1 d: x = 12 13 1-115. a: x = !2 b: x = 3 2 = 1 12 c: x = 3 d: No solution 8 Core Connections Geometry . b: Sandy (Sandy’s probability = 1 while Robert’s is 0) c: Sandy’s probability = 43 . Therefore. In order to belong to the intersection of both circles. while Robert’s is 3 . a: Sandy’s probability = 2 . (– 4. a: m!B = m!C because the line of symmetry must pass through A (according to the marked sides of equal length) and these angles are on opposite sides of the line of symmetry. 5). a: Square b: (– 4. 3x + 5° = 5x ! 57° . Lesson 2. b: They form a “Z. units c: 33x 2 ! 50x + 8 sq.1 2-8. a: 33 sq. No. a: Vertical.1.1. c: 71° + x = 180° . a: Vertical angles. and the right could be “quadrilateral”. equal measure. 0). The left circle could be “equilateral”. 2x + 4x + 150° = 180° . cm b: 33x sq. Assuming this.2 2-19. (1. because 1 ! 3 " 1 . x = 31º b: Straight angle pair.Lesson 2. 0) 2-22. Answers vary. they have equal measure. parallelogram and square 2 6 2-10. a: Isosceles triangle b: Equilateral triangle c: Parallelogram 2-11. supplementary. units 2-9. and a rectangle to the right circle. m!B = 12 (124°) = 62° . a: 1 b: 2 . 2-23. x = 109° 2-21. b: Since they are equal. 5).” Selected Answers 9 . KITE ISOSCELES TRIANGLE REGULAR HEXAGON RHOMBUS 2-12. you could add an equilateral hexagon to the left. (1. a rhombus to the intersection. y = x ! 1 . x = 5º 2-20. The slopes are 12 and ! 23 . 2-34. because we do not know if BD $ EG .1. Since the slopes are not opposite reciprocals. the lines must not be perpendicular. a: x = 17. there is no solution. They used different units.600 square units. –1) 2-44. subtract the x.5 (corresponding angles) b: x = 5 (multiple relationships possible) 2-35. 2-43. 23 students. a: We do not know if the angle measures are equal. yes 2-32. (3. a: (–2. a: 105° 75° 105° 75° b: 30° 75° 105° 75° 105° 70° 80° 80° 70° 85° 30° 95° 95° 85° 85° 95° 95° 85° 100° 80° 70° 110° 80° 100° 110° 70° 2-42. so 7 . b: The diagram does not have parallel line marks. 10 Core Connections Geometry . a: an isosceles triangle b: a rectangle Lesson 2. 2-45. –1). (7. so they do not intersect.Lesson 2. 3) b: (–2.3 2-31.4 2-41. 3).1. !##" !##" 2-33. Therefore.and y-coordinates to find the length of the two sides. a: 12 boys b: 22 girls c: 2 d: 7 boys left. The lines are parallel. 3 23 2-36. a: 20 square units b: 2. x = 7º 2-56. a: 15° b: x = 12° .Lesson 2. c = 32°. a: x = 4 and y = 18 b: x = !13 and y = 6 2-58. 2-66. " 6) 2-69. "7) b: B!!(8. a: It should be a triangle with horizontal base of length 4 and vertical base of length 3. a: x = 10 units b: x = 6 c: x = 20! d: x = 10! 2-57. 2-68. m!D = 4(12°) + 2° = 50° c: It is equilateral. b: ! 4 3 c: Any equation of the form y = ! 3 4 x+b. d = 32° 2-67. a: Yes. Reasoning will vary. "1) . The acute and isosceles triangles. b: y = 1 2 x+5 c: Any equation of the form y = !2x + b for all real b-values. because the slopes are opposite reciprocals.1 2-65. "3) . a: A!("6. and C !("5.1.5 2-55. Selected Answers 11 . a = 118° .2. 2-59. B!("2. 2 Lesson 2. 13) c: A!!!(3. b = 118°. 30 Lesson 2. c: No. straight angle is 180° c: x = 20° . a: 8x + 13 b: 2x + 3 c: 3x 2 ! 5x ! 12 c: 13x 2 + 30x 12 Core Connections Geometry . it would take 10 months for Sarita to catch up to Berti. she needs to attempt about 67 pancakes. a: x = 8° . sum of angles is 180° d: x = 60° . then Marla has a piano lesson.5(8)(17) = 68 sq. a: 8x 2 ! 26x ! 7 b: 10x 2 + 31x ! 14 c: 4x 2 ! 47x + 33 d: !6x + 17x ! 5 2 2-75. 2-86. it must be a rectangle. Since a square has four sides and four right angles. you have changed your height. then Mr. so the shaded area is the other half.3 2-85. a: 22 35 = 62. sum of angles is 180° 2-77.2.2 2-74. you measure from the very bottom to the very top. b: Diagram (1) is correct.Lesson 2. a: Because when you are not standing up straight. then the result is even. 2-89. No. Daniel is correct because the definition of a rectangle is a quadrilateral with four right angles.). right angle is 90° b: x = 20° . area = 28 square feet 2-76.9% b: 22 35 = 42 ? . b: If you add two even numbers together. 2-87. c: She should add three banana pancakes to make the probability of banana 3 . in. The unshaded triangle is half the area of the rectangle (0. a: If it rains. 2-88. and you will not get a true measure of your height. c: If it is Tuesday. 2-78.2. Spelling is unhappy. 1 2-104.5(10)4 = 20 sq.5(16 + 8)6 = 72 sq ft 2-95. in. a: 15x 2 + 21x b: x 2 + 5x + 6 c: 3x 2 ! x ! 10 c: 10x 2 ! 3x ! 4 2-96. c: 0. (–3. y y a: It is a trapezoid because it has exactly one pair of parallel sides. See graph at right.5 units Selected Answers 13 . C !("5.4 units 106° 74° 74° 2-107. "4) . d: The solution should be (–2. (2): (2.2. See diagram at right. b: A!(7. 106° 2-108. –3) 2-97. "2) c: 10 square units 2-98.!2). q: y = ! 12 x + 3 c: The slopes indicate that the lines are perpendicular. 4). 0) and (0.!"1). 2-106. b: A!("2.!6) x x c: A!!(1.!2) and C !!("2. 106° 74° b: 20 square units 74° 106° c: ≈ 23. C !(2. B!(8.!5) d: 1 2 (3)(2 + 7) = 13. 10 units 2-105. D!(3. –6) b: p: y = 2x + 8 . a: 12 52 = 13 3 b: 20 52 = 13 5 c: 2 52 = 1 26 d: 0 Lesson 2.3. a: 72 = 49 sq cm b: 0.! B!("5.!0). a: Right triangle. 3). a: Isosceles Trapezoid because two sides are parallel and the other two sides are the same length. slopes are opposite reciprocals.4 2-94.Lesson 2. See graph at right. "2) . "4) . D!("2. a: (1): (5. 2) 2-114. AB = 5. b: If a shape is a rectangle. Using the Pythagorean Theorem. ≈ 110° b: Obtuse scalene triangle 3-8. Triangle Angle Sum Theorem b: x = 23° . 14) d: (2. a: x = 28. c: A = 24 sq. relationships used varies c: x = 68° . BC = 4. a: x = 18 b: x = 3 c: x = 6 d: x = 2 3-7. a: D is not similar.40 units c: One is a ratio (slope) while the other is a length (distance). 3-9. y= 4 5 x + 95 b: MU = 41 ! 6. corresponding angles are congruent because the lines are parallel and base angles of an isosceles triangle are congruent. 5) c: (–12. P = 24 units 3-6. 14 Core Connections Geometry . AC = 3 b: A!B! = 100 = 10 units. a: If a shape is an equilateral triangle. a: 7 b: 3 c: 5 8 8 8 2-115. then the area of the shape is half the sum of its bases multiplied by its height. and A!C ! = 6 units. a: (–2.Lesson 2.5° . 5) b: (1.2 2-113.1 3-5. then the shape is a parallelogram. Height = 12 feet. then it has 120° rotation symmetry. area = 1 2 (12)(12 + 23) = 210 sq. feet 2-116.3. B!C ! = 8 units. ≈ 40°. a: triangle inequality b: Pythagorean Theorem c: base angles not equal 3-10. 2-117. a: ≈ 30°. 5" and 21" Lesson 3. a: 4 5 .1. c: If a shape is a trapezoid. units. yes 3-20. a: If the lines have the same slope. c: If lines have slopes 2 3 and ! 23 . a: The 15 corresponds to the 6. int.5 ratios are possible. Yes they are parallel because they have the same slope: ! 53 . a rhombus does not have to have four right angles.1. No. units and A(new) = 72 sq. b: It is 1:1 because it is congruent. so therefore there is not enough evidence that a rhombus is a rectangle. P(original) = 18 units and P(new) = 36 units.2 3-18.1. then the slope is undefined. 3-19. Lesson 3. x = 10° . and the side corresponding to the 11 must become 5. Result should be 12 units tall and 16 units wide. y = 61° 3-23.Lesson 3. a square is a rectangle because it has four right angles. a: 6x 2 ! 8x b: 2x 2 + x ! 15 c: 4x 2 ! 25 d: 2x 3 ! 5x 2 ! 3x 3-22.8 c: y = 3 + 53 x 3-33.5 3-32. 3-34. 25 10 = 2. One possibility: 15 20 b: 25 and 10.5 sq.5. so the side corresponding to the 16 must become 8. Notice that the enlarged area is 4 times greater. b: If a line is vertical. The enlarged perimeter is 2 times greater. the conclusion is not based on the facts. 3-21. then they are perpendicular. angles b: vertical angles c: corresponding angles d: supplementary and/or adjacent angles Selected Answers 15 . 3-31. The enlarged area is not 2 times greater. units. while the 20 corresponds to the 8.4 b: m = 22 c: t = 12.3 3-29. A(original) = 18 sq. As stated in Fact #2.5 . a: alt. a: Zoom factor: 0. this is not convincing. a: x = 42 5 = 8. However. While the facts are each correct. P = 8 + 34 ! 13. 3-30. then they are parallel. The sides are only half as long.5 d: x = 3 2 = 1.5. units. a: y = 3 ! 53 x b: A = 7. Multiple equivalent 6 = 8 = 2. 6). 6) b: Perimeters = 28 and 56 units.5 = y b: They will be 3 years old. 3-44. D(2. a: f = 9 b: g = 18 c: h = 70 3 3-42. – 4). a: Frank: 0.Lesson 3. Alice: 0. E(–6. inches. a: The coordinates of the image are A(–6.40x + 1. 3-43. C(10. 12). 3 < x < 13 but some of these lengths are not practical. a: If a rectangle has base x and height 2x. – 4).25x + 1. a: 180° ! 38° ! 63° = 79° and 180° ! 38° ! 79° = 63° . then the area is 864 sq. B(10. then the perimeter is 2x + 6y .95 = y . 3-46. then the area is 2x 2 .1. all unmarked angles are the same since the difference with 180° will be the same.4 3-41. units 16 Core Connections Geometry . b: Upon inspection. In theory. b: If a rectangle has base x and height 3y. 3-45. c: If a rectangle has base of 2 feet and a height of 3 feet. areas = 52 and 208 sq. corresponding angles are equal. 9 units. Note that you cannot apply zoom factor to angles. Dilate from right vertex. Squares or other regular polygons are also always similar. units 3-58.Lesson 3. 3-54. iii: 12 3-57.1 3-53. translate so that one pair of corresponding vertices coincide. a: One strategy: Translate one so that the centers coincide. since corresponding angles are not equal. ii: 12 . See graph at right. This can be repeated for each of the possible bus numbers. and dilate. Perimeter = 44. only trees must be green according to the statement. c: No. rotate so that rays coincide. 3-56. b: Equilateral triangles. since all trees are green and the oak is a tree. AA ~ since all angles are 60°. a: Yes. zoom factor of 2. a: There are 12 combinations. AA ~. d: No. which from part (b) of 3-54 were similar because they have equal angle measures.5. b: Yes. b: No. the second statement reverses the first. 9 8 1 y b: i: 12 . One way to systematically list them all is to list a bus number (such as 41) and then match it with each possible activity. c: Yes. Area = 94 sq. Then dilate so that the radius is the same as the other circle. 3-55. a: ABCD ~ EVOL x b: RIGHT ~ RONGW c: One possible answer: ΔTAC ~ ΔGDO Selected Answers 17 . a: Yes.2. a: Possible b: Not possible because the sum of the measures of an obtuse and right angle is more than 180°. d: Rotation. a: x = 20 mm b: w = 91 mm 3-66. dilated by factor of 2. and “Andrea’s flowers must be closed up” in the right oval. c: Not possible because a triangle with sides of equal length obviously cannot have sides of different lengths.2. a: Reflection. and translation b: Not enough information to determine similarity. This reasoning is incorrect. and reduced by zoom factor of 0. and translation. 8 4 3-67.Lesson 3. c: Rotation. d: Possible 18 Core Connections Geometry . a: 12 b: 8 3-68. 3-70. 3-69. reflection.5 (or translation and reflection instead of rotation and reflection). b: Possible c: Impossible: rejected because the sum of the angles is 179°.2 3-65. Rewrite “it is raining” in the lower left oval. a: Impossible: can be rejected using Triangle Inequality or Pythagorean Theorem. rotation. P = 6 + 20 ! 10.3 3-76. $400. Missing side length of second rectangle must be 9" because the area is 36 sq. 3-80. $1500. –2) b: (–4. It is the center of the figure. 3). and $3000. units.47 units c: y = !2x + 7 3-78.Lesson 3. 3-77. Selected Answers 19 . a: See tree diagram at right. or the midpoint of each diagonal.2. therefore. the rectangles are similar. a: (5. no—the spinners are independent $3000 2 2 $1500 keep e: 6 . 2) c: (3. a: n = 32 b: m ! 14. In fact. they are congruent because r = 1. a: y = 1 2 x+2 b: A = 4 sq.91 3-81. a: x = 51° alternate interior angles and Triangle Sum Theorem b: x = 43° circle has 360º c: x = 1 Pythagorean Theorem double $200 $100 3-79. in. Since angles are equal and ratios of corresponding side lengths are equal. 6 1 double d: . $200. $200. Missing side length of first rectangle must be 4 m because the perimeter is 26 m. because now the possible outcomes are $1500 $100. keep $100 double b: Yes $600 $300 1 3 keep $300 c: 6 . x = 18 b: 4. Another special case occurs when the resulting equality is always true. b: Must be similar by AA ~. the equations combine to create an impossible equality. 3-93. a: 3(4x ! 12) = 180° . creating infinite points of intersection. 3-92.Lesson 3. all side lengths have the same ratio. x ! 3. This reasoning is correct. such as 3 = 0. x = 72! d: 3x ! 2 = 2x + 9 . a: Not similar.4 3-88. This is the result when the two lines coincide. interior angles are different. such as 2 = 2.9 2 ! 3.79 c: x + (180° ! 51° ! 103°) + 82° = 180° . area = 60 square units. b: When solving a system of equations that has no solution. 3-91. height of triangle = 8 units.12 = x 2 .2. 3-90. c: Similar. x = 11 20 Core Connections Geometry . a: Scalene triangle b: Isosceles triangle c: Not possible d: Equilateral triangle 3-89. Perimeter = 10 + 10 + 4 + 3 + 4 + 3 + 4 = 38 units. a: The two equations should have the same slope but a different y-intercept. This forces the lines to be parallel and not intersect. b: 13+17 22+8+13+15+17 = 30 75 = 40% c: 3 ? = 5% . and dilate it by a factor of 0.2. Possible response: Rotate WXYZ clockwise. a: ! 56 b: LD = 61 ! 7. translate it to the left. a: 24 40 = 60% b: 18 x = 10 3 .2.6 3-108. a: SSS ~ and SAS ~ (if you show that the triangles are right triangles) b: AA ~ and SAS ~ c: None since there is not enough information. y = 7. !z = 48! because of reflection of !y or because of angle of incidence = angle of reflection with !x . a: y = ! 12 x + 4 b: y = 2x ! 1 c: y = 2 5 x + 57 d: C = 15 + 7(t ! 1) = 8 + 7t 3-112. h = 5 feet. a: 12x 2 ! 7x ! 10 b: 16x 2 ! 8x + 1 b: x = ! 59 c: x = 3 3-102.Lesson 3.2 feet 3-110.!y = 76° 3-109. 3-103. P = 48 units. units. perimeter ≈ 24. 3-111. a: 28 ? = 25 . Original: A = 135 sq.81 units c: Calculate ∆x and ∆y by determining the difference in the corresponding coordinates. units.!z = 9.5 3-99. !y = 48! because of vertical angles. x = 137°.2 miles 3-113. New: A = 15 sq. 3-104. There are 70 animals in the bin.6 Selected Answers 21 .5. You need a total of 60 animals in the bin.4. 3-100. ≈ 13. x = 60 3-101. P = 16 units Lesson 3. Since the slope ratio for 11° ≈ 0. AA ~. If h represents the number of hours and t represents the temperature. The slope ratio for 68° ≈ 2. side ratios not equal 12 64 ! 18 98 . x = 9. a: See flowchart at right.!y = 4. a: Yes. 4-20. So perimeter is 27 + 72 + 15 ! 28. 9 2 ! 32 = 72 .Lesson 4. Thus. x 95 ! 15 . AB ≈ 50 feet. then t = 77 + 3h and t = 92 ! 2h . b: Yes. 8 12 = 2 3 22 Core Connections Geometry . b: (x + 1)2 = x 2 + 2x + 1 c: x = 24 d: 56 units 4-9. a: side ratio = 4:1 b: perimeter ratio is 4:1 c: 28' 4-19. not enough angle values given.!z = 6 23 4-10. she used the Pythagorean Theorem. a: Yes. 5. 4-21.2 y 4-17.5. h = 3 hours and the temperature will be 86°F. b: No. c: Cannot tell. 6 2 ! 32 = 27 . 3. and 12. so BC ≈ 4 feet. y ! 175 4-18. P(15) = 0 d: The sum of 7. x ! 19 b: a = b = 45° c: 70 ! 52 . because the triangles are similar (AA ~) and the ratio of the corresponding side lengths is 1 (because AC = DF). 6. a: 12 b: Yes c: 6 12 = 12 .52 sq. P(10) = 3 36 .68 cm. a: ! = 11° . P(7) = 6 36 = 1 6 4-11. 4-22. The area is ( 27 + 72 )(3) ÷ 2 ! 20. c: P(even) = 18 36 . 7. Lesson 4. 8. 10. 11. 4.1.2. a: x = 11° b: x = 45° c: x = 30° d: x = 68° 4-7. cm.1 4-6. 9. a: 2. m∠A = m∠D m∠B = m∠E b: Yes. ΔABC ~ ΔDEF AA ~ 4-8.1. AB is actually longer. an = 5 + 7(n ! 1) = 7n ! 2 4-30. sum = 90°.3 4-27. because only 2 songs are left and only one is sung by Sapphire.1. equilateral triangle. c: 1 . since the slope ratio is greater than 1. 2 d: 1 .Lesson 4. the angle must be greater than 45°. 4-28. They both could be. a: Yes. then there is only 1 country song left to play out of 4 songs. c: Since the angle is greater than 45°. and equilateral hexagon. x must be less than 9. b: Isiah is correct. Therefore. an = 1+ 3(n ! 1) = 3n ! 2 b: neither c: G. the chance that the second song is a country song is 14 . an = 2 ! 2 n"1 = 2 n d: A. 4-31. possible solution: square. It depends on which angle is used as the slope angle. If the first song is a country song. m∠BAC = 68°. same 3 4-32. a: A. complementary Selected Answers 23 . a: 2 5 b: Yes. m∠ABC = 22°. Since the angle is less than 45°. 4-29. Answers vary. the slope ratio must be less than 1. then translate it. BADC. ADBC. x = 36 5 = 7. a: x2 + 182 = 302. b: ≈ 29. CDBA.4 4-39. a: G. ACDB.44 feet n"1 1n 4-51. CBDA. x = 24 b: 2x + 20° + 3x + 20° + x + 2x = 360° .524 b: p ! 3. a: 3x + 3° + x + 7° = 90° .1. x = 21 24 Core Connections Geometry . They are congruent. CABD. BCDA. x = 10 6 ! 1. a: Either 3 or 1 b: Either 9 or 1 3 9 4-48. P = 114. DABC.67 m c: A = 72 sq cm. an = 12 ! 12 = 2 b: A. an = !7. P = 58 feet Lesson 4. BCAD. DCAB. BACD. ACBD. Possible response: Reflect ΔADS across a vertical line.5 ! 2n 4-52. x = 20° b: 9x + 4° = 3x + 14° .Lesson 4.505 4-40.2 4-41. 24 possible ways: ABCD. 4-50. (a) and (d) are most likely independent.67 m2. x = 40° c: 5 12 = 3x . feet. CBAD. CDAB. DBAC. P = 52 cm b: A = 696.215 c: b ! 148. 4-44.5 4-47. DACB. ABDC. BDAC. DCBA 4-43. Her father’s eyes were ≈ 69. ADCB.5 ! 2(n ! 1) = !5. 10 2 + (x + 3)2 = 26 2 . a: t = 11. DBCA.67° 4-49. a: A = 144 cm2.126 inches high.1. CADB. BDCA. 4-42. P = 48 cm d: A = 130 sq. T e: The sample space remains the same. 4-61. a: 3. If the die is “fair.78 cm. y = 12 Selected Answers 25 . 4-60. a third dimension would be needed to represent H T the three coins with an area model. hypotenuse ≈ 30.22 cm Lesson 4.!0) and (3. b: The “Brian is always late on Tuesdays” and “Today is Tuesday” ovals should be next to each other.1 4-58. 4-71.!0) b: x = ! 12 or x = 3 .39 . Francis: y = x + 2. iii: 61 . See graph at right. so the perimeter ≈ 69. Yes. 375 b: 10. a: A tree diagram. they are similar due to AA ~ because m∠B = m∠E and m∠C = m∠C (triangles share an angle). –1250 4-74.2. H H T b: See tree diagram at right. 1 . 4-59. 75.44 cm.5 or –5 d: No real solution 4-63. 15. x ! 10. ii: 8 . ii: 4 + 4 + 4 = 12 . iii: 8 . a: (! 12 .667 c: x = 1. a: x ≈ 2.344 b: x ≈ 0. The probability the second contestant is a girl depends on whether the first contestant was a girl or not.” 4-73. they are not independent. both with arrows pointing to “Brian will be late today. 64 . 8 T H T H 1 3 7 3 T c: i: 8 .” each roll of the die is an independent event. iv: 8 T H d: They are both the same probability of 50%. 12 seconds 4-62. leg ≈ 29. 250. then today must be Tuesday. 6 4-72. iv: i: 125 12 125 125 125 125 125 125 4-70. Yes. a: Slope = 1 2 b: It must be parallel to or coincide with the line graphed at right. John: y = 3 4 x + 5 . –50.2. a: 10 20 = 1 2 b: 9 19 c: No.Lesson 4.2 H 4-69. a: It implies that because Brian is always late on Tuesday. so 170 x ! 15 . 4-86. the dimensions of the rectangle are 5 and 8 units. so x = !10 or 3. No.2. Triangle Inequality property prevents this because 7 + 10 < 20 and 20 ! 10 > 7 . Since x cannot be negative. (x + 2)(x + 5) = 40 . 170 11º ft b: Ratio for tan 11° ≈ 1 5 .5 4-100. $450 Lesson 4. The slope is ! 10 A few possible solutions: (5. x Alternatively. (25.2. a: See diagram at right. x = 3. (15.7x + 82. etc. y= 1 3 x+9 4-83. 8 4-82. Pythagorean Theorem b: x = 80° . and x ! 850 feet. a: x = 49 b: x = 2 c: x = 16 3 d: x = !5 or 1 4-85. 4-99. Therefore.4 4-95. a calculator could be used and x = 170 tan(11°) ! 875 feet. a: 5 ways b: 6 ways c: 11 d: 5 11 4-97. 79). 26 Core Connections Geometry . 4-84. Both equal 3 . a: Less than 45° b: Equal to 45° c: More than 45° 7 . Points will vary. Alternate interior angles and the Triangle Angle Sum 4-98. y = !0. 65).3 4-81. a: 3 b: 4 c: 24 36 36 36 4-96.Lesson 4. a: x = 13 . 72). x 2 + 7x ! 30 = 0 . Methods vary: θ = 68º (could be found using corresponding and supplementary angles). ΔABD ~ ΔEBC by AA ~. 4-113. a: Yes.5 4-110. b: Yes. 3) Selected Answers 27 . α = 85º (could be found using corresponding angles since lines are parallel. P(yellow) = 1 6 d: 2 3 e: You cannot move 1 6 + 16 + 13 = 2 3 or you can move 1 3 of the time and 1! 13 = 23 . C″(–6. Since DB = 9 units (by the Pythagorean Thm). –9). ≈ 1469. the common ratio is 1. 4-111. C′(–3.2. but you need to subtract the 52 number of cards that are both kings and clubs (1). a: P(K ) = 4 52 . B″(–3.27 feet 4-114. It assumes that everyone who likes bananas is a monkey. P(C) = 13 52 b: 16. a: y = 3 b: y = 9 4-112. a: A′(–3. mm 4-121. You can add the probabilities of king and club. 3) c: (9. 6 " < x < 14 " 4-115.47 sq. a: 1 b: Intersection 12 c: No. 4-116. A ! 97. ≈ 26 years 4-118. P(K or C) = 52 4 + 13 ! 1 = 16 52 52 52 c: P(K or Q) = 52 8 = 2 . There is no overlap in the events so you can just add the 13 probabilities. AB ! 11. P(Q) = 4 52 . B′(9. –3).47 mm. 4-119. 52 = d: P(not a face card) = 1! 12 40 52 4-117.Lesson 4. LE = MS and LI = ES = MI 4-120. –6) b: A″(–3. 3). –3). If Aces are included. a: G. 52 = b: 1! 16 36 40 52 . then any one of the trig ratios or the Pythagorean Theorem to get the other. B (4 cards less than 5)! (4 suits) 5-10. Use either sine or cosine to get the first leg.60°.66 ft 5-12. an = 0 ! 50(n ! 1) = 50 ! 50n 5-20. a: sin 22° = 17 x b: x ! 6. a: 52 = 16 52 . Since the spins are independent.90° triangle is a counterexample) b: False (this is only true for rectangles and parallelograms) c: True 5-22.50 and y ≈ 8. x = 3 5-18. x ≈ 7. 12 52 . ΔABC ~ ΔEFD by SAS ~ 28 Core Connections Geometry .1. the probability of A and A is 14 ! 14 = 16 1 1 (80) = 5 .04 units.1 5-7. Region A is 14 of the circle. If Aces are not included. a: False (a rhombus and square are counterexamples) b: True c: False (it does not mention that the lines must be parallel) 5-9.92 feet 5-19.09 c: cos 60° = 6x . a: False (a 30°. 6. tan 49° = 7x . 52 .37 . an = 100( 10 1 )n!1 = 10 3!n b: A. a: 6x 2 ! x ! 2 b: 6x 3 ! x 2 ! 12x ! 5 c: !3xy + 3y 2 + 8x ! 8y d: x 2 ! 9y 2 5-23.Lesson 5.2 5-17. perimeter = 47. In 80 games. we expect A and A to occur 16 times. ≈ 26. c: P(red) + P(face) – P(red and face) = 26 52 + 12 52 ! 52 = 6 32 52 5-11. a: x = –3 b: m = 10 c: p = !4 or 2 3 d: x = 23 Lesson 5.1. 5-8. 5-21. area = 74 sq ft. 4 5-41. Since tan(33. a: x = ±5 b: All numbers c: x = 2 d: No solution 5-30. 5-43. y ! 23 x + 7 . remove the arrow connecting “Marcy likes chocolate” and “Whizzbangs are 100% chocolate. union b: 3 .3 5-29.21 m 5-34. a: 22 .Lesson 5.7°) ! 23 . 9. a: an = 108 + 12(n ! 1) = 96 + 12n b: an = 25 (2)n!1 = 15 (2)n c: an = 3741! 39(n ! 1) = 3780 ! 39n d: an = 117(0.55 sq m. Using cos A = 13 5 . a: cos!23° = 18 x or 0. perimeter ≈ 78.921 .1. ! 11. Area ≈ 294. a: 1 b: 20 36 36 5-33. sin A = 12 . 5-42.” 5-35. then sin 67° = cos 23° . a: It uses circular logic. They are all equilateral triangles. a: sin ! = b a b: tan ! = a b c: cos ! = a b 52 = 5-45. All of the triangles are similar. A ! 67.2)n!1 = 585(0.” Also.921 = x 18 b: Since 67° is complementary to 23°. or tan A = 12 .1.2)n 5-44.38 minutes Lesson 5. intersection c: 1! 22 30 52 52 52 5-46.4º 13 5 5-31. b: Reverse the arrow between“Marcy likes chocolate” and “Marcy likes Whizzbangs.5 seconds 5-32. Selected Answers 29 . So sin 67° ! 0. a: Answers vary.2. then dilate. a: A = 1 sq.Lesson 5.!x = 53º d: y = 3. P = 15 + 5 3 ! 23. 5-56.1 5-52. then dilate. m. x is even. a: an = 500 + 1500(n ! 1) = 1500n ! 1000 b: an = 30(5)n!1 = 6(5)n 5-58.45°. c: Similar: Reflect ΔMNP. 10. 24 square units 5-55.66 ft 5-53. a: Not similar b: Similar: Rotate ΔGHI. d: Similar: Rotate ΔTUV. sample responses: x < 3.90° triangle.!x = 47º c: y = 83º . etc. a: 4 2 units. b: The length of each leg is 6 units.!x = 53º b: y = 79º .65 square ft. P = 2 + 2 2 m b: A = 25 3 2 ! 21. translate. 5-57. a: y = 111º . translate. b: It is a trapezoid. Use the Pythagorean Theorem or that it is a 45°. translate. then dilate. 30 Core Connections Geometry .! x = 3 2 units 5-54.1% by using the Addition Rule. Since both triangles are isosceles. 5-70. m∠B = 35º. 6) 5-81. B! (–5. 14 27 = x 40 . ≈ 61° 5-78. x ! 20.3. m∠ACB = 110º.50 . we cannot tell if one is the reflection or the rotation of the other (after dilation). tan!52° = ba . so DC = EC and m∠ACB = m∠DCE therefore ΔABC ~ ΔDEC SAS ~. B!! (1. state that two pairs of corresponding angles have equal measure. 5-80.2 5-64. b: Impossible because the sum of the angles is more than 180°. It is not fair because the expected value is not 0. a: 1 b: 0 c: 3 d: 1 2 4 Lesson 5. m∠DCE = 110º b: Answers vary. m∠D = 35º. cos!38° = sin!52° 5-67. 3). 1). such as m∠A = m∠D and m∠B = m∠E to reach the conclusion that ΔABC ~ ΔDEC by AA ~ or AC = BC and DC = EC.74 inches 5-68. cos!52° = bc . Selected Answers 31 . William is correct. C ! (0. Once all the angles have been found. – 4) b: A!! (3. m∠E = 35º. cos!38° = ac . a: x = 16 5 b: No solution c: x = –11 or 3 d: x = 288 5-82. a: A! (–3. – 4).1 5-77. b is correct. 5-79. a: explicit b: an = !3 + 4(n ! 1) = 4n ! 7 c: a50 = 193 d: an = 3 ! 13 (n ! 1) = 3 13 ! 13 n 360 ) + $5( 360 ) + (!$6)( 360 ) = $1. a: 16 inches b: 4 yards and 4 2 yards c: 24 feet d: 10 meters and 10 3 meters 5-65. –6). a: Impossible because a leg is longer than the hypotenuse. AC BC c: They are both correct. a: m∠A = 35º. if two sides of a triangle are congruent.2. the angles opposite them must be equal. 5-66. C !! (1.Lesson 5. $3( 135 135 90 5-69. They must have equal length. sides opposite equal angles must be the same length. (!2. ≈ 10.6 mm 5-93.3.2 5-89. Since a side opposite a larger angle must be longer than a side opposite a smaller angle. red apple red apple yogurt 5-94. See tree diagram at right (an area model is not practical).5% yogurt green apple green apple chance of not getting three yogurts.25 green apple b: x = !10 or x = 10 red apple c: x = 1. a: x = 45 4 = 11. yogurt 100% ! 12.3 yogurt yogurt green apple d: No solution red apple 5-95. red apple P(three yogurts) = 12.5% = 87.!4) yogurt green apple green apple green apple red apple red apple yogurt green apple red apple yogurt yogurt green apple red apple red yogurt apple green apple green apple red apple red apple yogurt green apple red apple 32 Core Connections Geometry .6 yogurt yogurt green apple 5-93. 5-90.Lesson 5.5%. 72 square units 5-92. 12. 4 5-111. – 4) c: (3. tan !1 ( 43 ) " 36. b: ≈ 190 feet. a: 2x 2 + 6x b: 3x 2 ! 7x ! 6 c: x = 1 or 7 d: y = –3 or 5 1 1 5-105. Law of Cosines 5-112. a: (–1. –2) b: (4. 7 years Selected Answers 33 . 31 terms 5-116. 5-117. 5-113. It must be longer than 5 and shorter than 23 units. y ! 36. sin !1 7 8 " 61. a: Corresponding angles have equal measure.0º 5-104.73 5-101. a: 29° b: cos!29° = 42 .3. y = (tan 25°)x + 4 or y ! 0. a: 12 b: 3 Lesson 5.83 6 The game is not fair because the expected value is not zero. a: The diagram should be a triangle with sides marked 116 ft.87° 5-103.Lesson 5.3 y 5-100. 4) 5-102. and 224 ft. 3 12 (3) + 127 (!1) + 12 2 (10) = 11 " $1. so corresponding sides are proportional.3. b: The ratio of corresponding sides is constant.466x + 4 5-114. and the angle between them marked 58°. 5-115. Counterexample: A parallelogram without 4 right angles.75 .77 Bush Shed 5-138. Recursive t(0) = !2.3.Lesson 5. 0) 34 Core Connections Geometry . 5) 5-131. x ! 11. m∠DEC = 40º. a: P ≈ 40. Recursive t(0) = 6.27 cm. 5-128. A = 72 sq mm b: P = 30 feet. perimeter ≈ 24. or to use sine or cosine to solve using a trigonometric ratio. C(4. while the angle opposite the side of length 17 is approximately 99. so the perimeter is ≈ 25. so her expected winnings over 3 games are 3(3. yes. 5-132.2)n e: t(4) = 1620 Chain x x 5-137. using the 45°. The third side is 12. Methods include using the Pythagorean Theorem to set up the equation x 2 + x 2 = 16 2 . units.5 cm 5-135.3 units. 4). the parallelogram needs to have 4 right angles. Area ≈ 21. a: 5 + 20 + 37 ! 15.45°. A(2. a: ≈ 8. m∠ECD = 80º.11 c: (–2. a: Explicit t(n) = !2 + 3n . a: See diagram at right.55 units b: ! 31.45°. she should win enough tickets to get the panda bear. The expected value per throw is 14 (2) + 14 (3) + 12 (5) = 15 4 = 3. 5-129. m∠EBC = 60º. because to be a rectangle.25.64 cm b: PS = SR = 5.5 5-126.90° triangle shortcut to divide 16 by 2 . No. 2). a: m∠ABE = 80º.75) = 11. The angle opposite the side of length 10 is approximately 35. b: x = 103 3 ! 5.86 sq. A = 36 square feet 5-130. 5-127.32 mm.!t(n + 1) = 12 t(n) c: t(n) = 24 ! 7n d: t(n) = 5(1.59 units 5-136. m∠CEB = 80º. m∠BCE = 40º.55°. B(6. m∠BEA = 60º b: 360° 5-134.!t(n + 1) = t(n) + 3 b: Explicit t(n) = 6( 12 )n . y = 3 4 x+4 5-133.2 units long. ∠v & ∠w.Lesson 6.97 ft c: ≈ 15. 1) c: Reflection across the y-axis. 6-8. a: Alternate interior angles. 6-10. 6-9. B′(–5. and ∠t & ∠y 6-5. 8). –8).1.90° ratios c: ≈ 9. b = 15º. 3x + 1° + 52° = 180° . c: ∠u & ∠z. a: 25 units b: 56 sq. c = 68º. A Selected Answers 35 .9 . –1) b: A″(2. C′(–3.45° .2 in. Law of Cosines 6-16. 3: The fans will not buy the cupcakes because they are burned. a: 3 or 75% b: 3 or 15% 4 20 c: 1 or 100% c: (b) is an intersection. a: 8 cm b: ≈ 14. 6-6.2 6-14. C″(3. Pythagorean Theorem or 45°. a: A′(–2. d = 68º 6-15. a: ≈ 3. units and 350 sq. ∠s & ∠x.64 = 64% b: (0. B″(5. a: They are similar by SAS ~.1. 7).75. x = 127 3 ! 42.2) = 0.1 6-4. a = 97º. –7). because they are similar and the corresponding sides have a ratio of 1. 4: The team will not have enough money for uniforms. b: Vertical angles. b: Yes.33° 6-7. and (c) is a union.8)(0. tangent b: 7 2 ! 9.16 = 16% 6-19. A Lesson 6. 2: The cupcakes are burned.8) = 0.8)(0. a: (0. 6-18. 1a and 1b: statements ii and iv.54. units 6-17. ΔDEF ≅ ΔLJK. b: One possible answer.1% 56 36 Core Connections Geometry . the probability 176 of A and B (the overlap) was 0. b: Similar (AA ~) 6-24. c: KL ≈ 4. b: P(A or B) = P(A) + P(B) – P(A and B) ! 75% = 114 212 + 212 ! x " x # 5. followed by a rotation. b: 14 22 = 10 DE . c = 10 by substitution. a: Yes because of AAS ≅ or ASA ≅. a: P(A or B) = P(A) + P(B) – P(A and B) = 64 212 + 112 212 ! 0 = 212 " 83.3 units 6-27. a reflection across line segment JK and then a translation of ΔDEF to line up point J and point E. DE ≈ 15. 6-28.Lesson 6.3 6-23.1.0% .71 6-26. a: Not similar because there are not three pairs of corresponding angles that are congruent. a: y = 5 2 x!8 b: y = 3 2 x +1 6-25. ∠s = 6-35. and m∠C = 19º. x = 7. alt.1. then when lines are // . a: a = 123°. 6-37. ∠s = and corres. 4 10 = 5 x+5 .4 6-34. when lines are // . ∠s b: all = 98°. alt. so m∠A = 122º. int. corr. int. when lines are // . See sample responses below. Let B represent the measure of angle B. ∠s are =. ∠s =. alt. 3 6 2 parents b: 1 4 c: 1 9 + 16 + 16 + 14 = 25 36 ! 69% 1 3 1 9 1 18 1 6 niece 1 1 1 1 6-40. C 6 18 36 12 boyfriend 1 1 1 1 2 6 12 4 Selected Answers 37 .5 6-38. c = 57°. a: Similar (SSS ~) b: Similar (AA ~) 6-36. parents niece boyfriend 1 1 1 6-39. b = 123°. int. the lines are parallel. then vert. a: x = !4 and y = 0 b: No solution. Then (3B + 5º) + B + (B – 20º) = 180º. or vert. m∠B = 39º.Lesson 6. a: See possible area model at right. Reasoning can vary. suppl. ∠s. ∠s = c: g = h = 75°. suppl. ∠s are =. or corres. when lines are // . f = 127° . a: For left-hand triangle: c 2 = 9 + 36 ! 2 · 3 ·6 cos 60° . y-intercept: (0. given. then it is a square. 6-49. 6-48. 0). For right-hand triangle: c 2 = 36 + 27 ! 2 ·6 · 3 3 cos 30° . then it has three angles. Not always true. Always true. straight angle (with ∠a). b: Yes. e = 55° . when lines are parallel. b: Converse: If a polygon is a rectangle. 6-47. e: Converse: If vertical angles are congruent. They are congruent. c = 3 units. d = 53° . d: Converse: If a polygon is a triangle. sum must be equal to one. triangle angle sum. then two lines intersect. straight angle (with ∠g). h ≈ 385. a: y = 13 4 b: y = –2 c: 4 23 " d: x = 4 6-51. a: 3 b: 1 8 8 c: 3 d: 1 . b = 55° . Not always true.5 6-46.67 feet 38 Core Connections Geometry . c: Converse: If a polygon has four 90° angles. sin 40° = h 600 .Lesson 6. Not always true. then it is raining. Always true.196 units.1. then it is a rectangle. alternate interior angles are equal. c = 72° . a: Converse: If the ground is wet. alternate interior angles are equal. c = 3 3 ! 5. 6) 6-50. x-intercept: (4. so they are supplementary. 8 8 6-52. by SSS ≅ or SAS ≅. when lines are parallel. Justifications and order may vary: a = 53° . Lesson 6.2.1 6-55. a: x ! 45.56 b: x ! 10.63 c: x ! 265.48 d: x = 5 6-56. 9 square units; First find AC = 5 and then calculate 1 2 (5)(3.6) , or use BC as the base and calculate 12 (2 + 4)(3) . 6-57. a: m = 33 m, n = 36 m b: Area (small) = 378 cm2, perimeter (small) = 80 cm, area (big) = 850.5 m2, and perimeter (big) = 120 m 6-58. a: Similar because of AA ~. b: Neither because angles are not equal. c: Congruent because of ASA ≅ or AAS ≅. 6-59. a: ≈ 71.56° b: y = x + 3 c: (1, 4) 6-60. D Selected Answers 39 Lesson 6.2.2 6-62. a: Lines l and m are parallel because alternate interior angles are equal. b: Line n is perpendicular to line m because w + k = 180° and if w = k, then each is 90°. c: No special statements can be made because vertical angles are always equal. d: Lines l and m cannot be parallel because otherwise z + k = 180° . 6-63. a: ΔABC ~ ΔDEF (AA ~) b: ΔMON ≅ ΔPQR, (AAS ≅ or ASA ≅) c: Neither congruent nor similar because m!J " 62! . If m!J = 62! , then m!L = 180! " 2 # 62! = 56! . Since sin556° ! sin872° , this triangle cannot exist. 6-64. a: Converse: If the cat runs away frightened, then it knocked over the lamp. Not always true. b: Converse: If the chances of getting a 3 are 1 , then a 6-sided dice was rolled. Not 6 always true. c: Converse: If a triangle is a right triangle, then it has a 90° angle. Always true. 6-65. 19 4 6-66. D !##" !##" 6-67. a: It is a trapezoid. The slope of WZ equals the slope of XY . b: ≈ 18.3 units c: (–9, 1) d: 2 40 Core Connections Geometry Lesson 6.2.3 6-73. a: Yes, because parallel lines assure us that the alternate interior angles are congruent. Since corresponding angles in the triangles have equal measure, the triangles are similar by AA ~. b: x 20 = x+2 24 , x = 10 6-74. a: x = 4 b: x = 55° c: x = 23° and y = 43° d: x = 5.5 and y = 45.2 6-75. P(A or B) = P(A) + P(B) – P(A and B) = 4% + 12 % ! 12 % = 4% . If a refrigerator has a dent it also always has a paint blemish. 6-76. area ≈ 100.55 sq. yards; perimeter ≈ 43.36 yards 6-77. a: 288 feet by 256 feet b: area of shape = 59.5 square units; area of island = 60,928 square feet 6-78. C Lesson 6.2.4 6-83. a: Congruent (HL ≅ or SAS ≅) b: Congruent (AAS ≅) c: Not necessarily congruent. d: Congruent (SAS ≅) 6-84. a: x + 4x ! 2° = 90° , x = 18.4 , complementary angles b: 2m + 3° + m ! 1° + m + 9° = 180° , m = 42.25 , Triangle Angle Sum Theorem c: 7k ! 6° = 3k + 18° , k = 6 , vertical angles are equal d: x 8 , x ! 9.8 , corresponding parts of similar figures have equivalent ratios = 13 16 6-85. x = 11; m∠ABC = 114º 6-86. a: Converse: If a triangle is isosceles, then its base angles are congruent. Always true. b: Converse: If the sum of the angles in a figure is 180° , then the figure is a triangle. Always true. c: Converse: If my mom is happy, then I cleaned my room. Not always true. 6-87. 36% 6-88. D Selected Answers 41 c + t = 180º because if lines are parallel. these curves do not intersect. A quadrilateral. Triangle Angle Sum 6-98. Triangle Angle Sum Theorem. same-side interior angles are supplementary. c: x ! 10.5 c: y = x 1 3 d: y = 2 6-96.4 units and DF = 20 units 7-7.2. Isosceles triangle c: t = 9° and w = 131° . then corresponding angles are equal. Relationships used will vary. 42 Core Connections Geometry . a: (6. but may include alternate interior angles. 66° c: 180° – 88° = 92°. area = 660 mm2 7-9.1 b: Not enough information. a: x ! 8. then alternate interior angles are equal.94 . Triangle Angle Sum b: k = 5 . g + q = 180° because when lines are parallel. a: 5x + 3 = 4x + 9 because if lines are parallel. y = 24 c: x = 8 = 8 3 3 .5 6-94. parallel lines d: x ! 49. c = 26°. then same side interior angles are supplementary. 7-10. a = 26°. y = 9 3 b: x = 24 2 . y= 16 = 163 3 3 3 7-11. B Lesson 7. d = 117° 7-8. b: q = t because if lines are parallel.Lesson 6.1..1 7-6. etc. a: They are congruent by ASA ≅ or AAS ≅. 6-95. width = 60 mm. a: y = 6 5 x!3 b: y = ! 1 4 x + 4.67 6-97.50 b: Lose $12 6-99. x = 6º. a: x = 18. a: Lose $1. –13) b: Not possible. a: x = 15° . b = 65°. b: AC ≈ 9. m = 14° b: 3x + 38° + 7x ! 8° = 180° . M(0. line M: y = 2 3 x + 1 .1. 7-20. c: XY ! XY " . No d: (3. a: The triangles are similar by SSS ~. d: The triangles are congruent by AAS ≅ or ASA ≅. ΔABC ~ ΔHGJ because of SSS ~. point of intersection: (6.4 °. Line L: y = ! 16 x + 6 . 7) 7-30. a: No solution. They are equal. a: It is a parallelogram. 2) 7-33. lines are parallel. a: The 90° angle is reflected. b: They must be congruent because rigid transformations (such as reflection) do not alter shape or size of an object. 4) c: 5 units. and ∠YZX ≅ ∠Y′ZX 7-29. 7-16. Then. Then m∠YZY′ = 180º. BD :!y = !x + 5 . n = 9 units d: 2(3x + 12) = 11x ! 1 . Using the Pythagorean Theorem. 11) Selected Answers 43 . x = 15° c: 2(n + 4) = 3n ! 1 . a: 3m = 5m ! 28 . 7-32. Rotating about the midpoint of a base forms a hexagon (one convex and one non- convex). AB = 8 and JH = 5. it must be half of AB because C is the midpoint of AB . opposite sides are parallel.1. b: (0. ∠Y ≅ ∠Y′. YZ ! Y "Z . 5) 7-18. 7 cm 7-17. 3) and (4. c 2 and a 2 + b 2 7-31. since 3 6 = 4 8 = 10 5 . c: AC :!y = 1 2 x + 12 . c: Not enough information is provided. XZ ! XZ . b: 63. a: 10 units b: (–1.Lesson 7.2 7-15. ∠YXZ ≅ ∠Y′XZ. b: The triangles are similar by AA ~. Lesson 7. x = 5 units 7-19.3 (Day 1) 7-28. so m∠XYZ = 90º. Rotating the trapezoid about the midpoint of either of the non-base sides forms a parallelogram. 0).5° c: 5w + 36° + 3w = 180° .1. so y = 22. It has four sides of length 5 units. The graph is a parabola with roots (–3. diagonal is 50 !·! 2 = 100 = 10 units. n = 4° b: 7x ! 19° + 3x + 14° = 180° . a: It is a rhombus. k = 35 7-38. Side length = 50 units.24 m 44 Core Connections Geometry . d: (6. 7-35.3 (Day 2) 7-34. so x = 18. and y-intercept at (0.5) ! 19° . –3). a: P(scalene) = 1 4 b: P(isosceles) = 2 4 = 1 2 c: P(side of the triangle is 6 cm) = 24 = 1 2 7-37. w = 18° d: k 2 = 15 2 + 25 2 ! 2(15)(25) cos120° . Then 5y ! 2° = 7(18. 0) and (1. b: HJ : y = !2x + 8 and GI : y = 1 2 x+3 c: They are perpendicular. 7-39. a: 6n ! 3° = n + 17° . ≈ 35. –1) e: 20 square units 7-36.Lesson 7.5°. (a) and (c) are correct because if the triangles are congruent. Then ΔABE is a right triangle and AE = 15 2 ! 12 2 = 9 mm. Show that the lengths on both sides of the midpoint are equal and that (2. since the diagonals are perpendicular bisectors. AA ~ ΔFED ~ ΔBUG SSS ~ 7-47. Yes. 5) and (7. a: 360° ÷ 36° = 10 sides b: Regular decagon 7-44. Since alternate interior angles are congruent.83 . Selected Answers 45 .Lesson 7. 7-48. AC = 18 mm. therefore C is closer to B. then corresponding parts are congruent.32 . 7-46. AB = 40 ! 6.1. c: See below. she is correct.4 7-43. 3) . 7-45. then BE = 12 mm. 4) lies on the line that connects (–3. a: See flowchart at right. then AB // DE . If the diagonals intersect at E. Thus. BC = 34 ! 5. b: Not similar because corresponding sides do not have the same ratio. 7) = 0. 7-57.2.4 feet from the point on the street closest to the art museum.Lesson 7. ≈ 36.3)(0. a: x = 8. a: 360° ÷ 72° = 5 sides b: 360° ÷ 9 = 40° 7-56. 7-60. 46 Core Connections Geometry .7) = 0.5° b: x = 11 c: x = 14° 7-55.21 = 21% 7-59. a: Similar (SSS ~) b: Congruent (ASA ≅ or AAS ≅) c: Similar.1 7-54. d: Similar (AA ~) but not congruent since the two sides of length 12 are not corresponding. the triangles are similar (SSS ~). a: (0. then 3 pairs of corresponding sides have a common ratio. thus. Possible response: Rotate the second triangle 180° and then translate it to match the sides with the first triangle.7)(0. a: an = 20 + 20n = 40 + 20(n ! 1) b: an = 6( 12 )n = 3( 12 )n!1 7-58.49 = 49% b: (0. because if the Pythagorean Theorem is used to solve for each unknown side. 4(360°) = 144°.5(360°) = 180° c: Yes. !m = 54°. b: (1. Selected Answers 47 . The sum must be 100%. !r = 54°. there could be more than three sections to the spinner.8)3 = $614. white = 0. !z = 108° b: Possible response: !y and !z are supplementary interior same side angles. Given ∠AEB ≅ ∠CED ≅ ≅ Vertical angles Definition of Definition of are congruent. a: 50%. midpoint. midpoint. y = 4 7-66.40 c: 1200(0. b: central angle for red = 0. !y = 72°. 7-68. a: !a = 36°.2. a: 0. –5) E is a midpoint 7-69. but the ratio of the areas for each color must match the ratios for the spinner in part (b).2 7-65.8)!2 = $1875 7-67. blue = 0. x = 3 . 7-71. 5y + 2 = 22 . a: It is a parallelogram. ΔAEB ≅ ΔCED SAS ≅ 7-70.8 b: 1200(0.1(360°) = 36°.Lesson 7. 4x ! 1 = x + 8 . because MN // PQ and NP // MQ . b: (1) 2. a: Yes.2. 7-85. x = 5° Lesson 7.2 d: 5x + 6° = 2x + 21° . See answers in problem 7-75. Law of Cosines 7-86.4 7-83. c: No. a: Congruent (SSS ≅) b: Not enough information c: Congruent (ASA ≅) d: Congruent (HL ≅) 7-76.2. his probability is still 4 20 = 1 5 because the ratio of the shaded region to the whole sandbox is unchanged. 64° b: 4x ! 3° + 3x + 1° = 180° . a50 = !130 48 Core Connections Geometry . BD : y = ! 23 x + 9 7-80. !##" !##" b: AC : y = 43 x . a: 2x + 52° = 180° .73 cm. a: 4 b: 5 . HL ≅ b: 18°.3 feet. tangent b: x ≈ 7. 26° c: sin 77° x = sin 72° . (3) 7. 8 x ≈ 8. A = 24 square units 20 = 5 4 1 7-88.4 square units 7-84. you can subtract 1! 15 = 45 . 7-89.3 7-75.7 7-81. a: Congruent (SAS ≅) and x = 2 b: Congruent (HL ≅) and x = 32 7-87.3'. No.86 mi. 7-77. Law of Sines c: x ≈ 15. a: x ≈ 10.Lesson 7.2.2. a: Parallelogram because the opposite sides are parallel. 4 c: tan 18° = 4 AD . then 68 must be a little greater.3 units d: ≈ 49.2. a: 83° b: 92° 7-78. 36 3 ! 62. the perimeter of the triangle is ≈ 26.2 square units 7-79. AD ≈ 12. (4) 4. Since the sum of the probabilities of finding the ring and not finding the ring is 1. Using the Pythagorean Theorem and the Law of Cosines. a: 68 ≈ 8. (2) 9.1. since 64 = 8. T P 7-97. so ∠BAC ≅ ∠DCA (alt. the angle of depression equals the angle formed by the line of sight and the ground. 0) 21 7-98. alternate interior angles equal and ASA ≅ f: ΔDEF. SSS ≅ Selected Answers 49 . m∠a = 132º. ∠BCA ≅ ∠DAC ( !! s!"!!!parts ). x-intercept: (4. int. and m∠c = 120º. Thus. A b: Since corresponding parts of congruent triangles are 14 18 congruent. AC ! CA ! (Reflexive Property) so ΔABC ≅ ΔCDA (SAS ≅). m∠b = 108º. then the lines cut by the transversal are congruent). angles). Because alternate interior angles are congruent.85° . d: ΔTZY. Then tan!! = 5238 . angles are congruent. BC // AD (if alt. 2y + 7 = 21 and y = 7. m∠a + m∠b + m∠c = 360º 7-99. a: See diagram at right. ! " 53. 7-102. only angles are congruent. 15% 7-100.2. y-intercept: (0.5 7-96. int. SAS ≅ and vertical angles e: ΔGFE. 6). AB ! CD and AB // CD (given). HL ≅ c: No solution. AAS ≅ or ASA ≅ b: ΔSQR. 7-101.Lesson 7. a: ΔADC. 5 7-113. ∠ZYW ≅ ∠YWX (alt. etc. ! " # ! parts . The triangles described in (a). all sides congruent. d: Triangle is isosceles but the base angles are not equal. congruent adjacent sides. ! ’s). See reasons in bold below. 7-112. congruent angles. corresponding angle measures are equal.6 feet. 7-110. a: 12 b: 15 c: 15. 7-109. ΔXWM ≅ ΔZYM. 7-111. (b). alternate interior angle measures are equal. 4. a: The triangles should be ≅ by SSS ≅ but 80º ≠ 50º. ASA ! . YM ! WM and XM ! ZM . c: The triangles should be ≅ by SAS ≅ but 10 ≠ 12. c = d Vertical angle measures are equal. which will meet industry standards. Her sink should be located 3 23 feet from the right front edge of the counter. diagonals that bisect each other. b = c If lines are parallel. 3. 2 pairs of opposite sides that are congruent. 50 Core Connections Geometry . This problem is similar to the Interior Design problem (7-21). a = c Substitution 5. ∠YZX ≅ ∠WXZ. 6. ZY ! WX . f: The triangles should be ≅ by SAS ≅ but sides 13 ≠ 14. Typical responses: right angles. This will make the perimeter ≈ 25.2. a = d Substitution 7-114. b: The triangles should be ≅ by SAS ≅ but 80º ≠ 90º and 40º ≠ 50º. e: The large triangle is isosceles but base angles are not equal.Lesson 7. int. and (d) are isosceles. AD // EH and BF // CG Given 2.5 7-108. Statements Reasons 1. a = b If lines are parallel. congruent diagonals. b: 2HR = AK. Her conclusion in Statement #3 depends on Statement #4. and thus must follow it. –2) 7-120.3. ∠J ≅ ∠J. SAS ≅. a: (4. 7-123. Selected Answers 51 .5.5. HI + IJ = LK + KJ. all corresponding pairs of angles equal is not sufficient.1 7-119. a: ΔSHR ~ ΔSAK because ΔSHR can be dilated by a factor of 2.5) c: (1. SAS ≅ c: ΔHJK.Lesson 7. See answers in problem 7-122. No. d: Not ≅. 2SH = SA. ASA ≅ b: ΔEFG. 7-125. 7-124. a: ΔCED. SH = HA c: 6 units 7-121. b: Must be a quadrilateral with one pair of opposite sides that are parallel. vertical angles are equal . 1. 3) b: (–3. a: Must be a quadrilateral with all four sides of equal length. a: 4 b: 1 c: –8 7-122. Could be: isosceles trapezoid. and square. by SAS ~. a: It is a right triangle because the slopes of AB and AC ( 43 and ! 4 3 . and square.83 feet b: x ≈ 7 yds c: x ≈ 66. 7-132. See graph at right. Multiple answers are possible.57 feet 7-137. so x = 4 and GH = 4(4) ! 3 = 13 units. a: ≈ 23. Any order is valid as long as Statement #1 is first. rectangle.3. and Statement #4 follows both Statements #2 and #3. (Triangle Midsegment Theorem) d: 2(4x ! 3) = 3x + 14 . rectangle. 7-133. each side has the same length ( 29 units). b: B! is at (–2. a: Must be: trapezoid. parallelogram. a: 6 b: 3 c: – 6. 7-136. rhombus. b: ∠FGH ≅ ∠FIJ. Statements #2.42 d: ≈ 334.Lesson 7.5 7-134. 0). Therefore the diagonals are perpendicular. ∠FHG ≅ ∠FJI c: Yes. !##" !##" b: BD is y = x . b: Must be: parallelogram. a: Yes. respectively) are opposite reciprocals. It is an isosceles triangle because ∠B′AB must be a straight angle (because it is composed of two adjacent right angles) and because BC ! B"C (because reflections preserve length). and #5 are independent of each other and can be in any order as long as #2 and #3 follow Statement #1. 52 Core Connections Geometry . AC is y = 5 ! x c: The slopes are 1 and –1. because corresponding angles are congruent. Could be: rhombus.2 7-131. 7-135. Statement #6 is last. a: Yes. #3. The expected value is 14 ($3) + 43 ($1) = $1. 7-143. a: X and Y b: Y and Z 7-142. then its area equals its base times its height.5.3. 6) 7-141. 7-144. then the polygon is a parallelogram. Selected Answers 53 . square. then the polygon is a triangle.3 7-140. It cannot be 180° (because this polygon would only have 2 sides) or 13° (because 13 does not divide evenly into 360°). Could be: right trapezoid. a: Must be: none. rectangle. a: 360° ÷ 18° = 20 sides b: It can measure 90° (which forms a square). b: It could be a kite. This is always true. square. or square because the diagonals are perpendicular and at least one diagonal is bisected. Could be: kite. rhombus. rhombus. 7-146. then its area equals one half its base times its height. c: “If a polygon is a triangle. This is always true. b: If the area of a polygon is the base times its height. so each player should pay $1. 7-145.50 so that there is no net gain or loss over many games. a: If a polygon is a parallelogram.50 per spin. a: They are perpendicular because their slopes are opposite reciprocals. 8) b: (6.” Arrow diagram: Polygon is a triangle  area of the polygon equals one-half base times height. d: If a polygon’s area is one-half its base times its height. a: (8. b: Must be: none.Lesson 7. 8-9. a: no b: yes c: no d: yes e: no f: yes g: yes h: no 8-8.9 units d: congruent (SAS ≅). b: The measure of an exterior angle of a triangle equals the sum of the measures of its remote interior angles.1 8-6.) c: True d: True e: False (counterexample is a parallelogram that is not a rhombus) 54 Core Connections Geometry . c: congruent (AAS ≅). x = 72° and y = 54° 8-10. a: congruent (SAS ≅). x + c = 180º (straight angle). c: a + b + c = 180º (the sum of the interior angles of a triangle is 180°). 360° ÷ 15° = 24 8-11. a + b + c = x + c (substitution) and a + b = x (subtracting c from both sides).1. x = 79° b: Cannot be determined.9° 8-12. x ≈ 60.Lesson 8. a: True b: False (counterexample is a quadrilateral without parallel sides. a: 110° b: 70° c: 48° d: 108° 8-7. therefore. x ≈ 5. m∠BCF = m∠EFC and 2. FC = FC 3. !! s!"!!!parts 8-23. (–3. a: Isosceles right triangle. ΔABC ≅ ΔDEF 6. because AC = BC and AC ! BC . If two lines cut by a transversal m∠EDF = m∠CAB are parallel. 1). AF + FC + CD = FC 4. Segment addition 6. (–5. Additive Property of Equality (adding the same amount to both sides of an equation keeps the equation true) 5. 7). and AF = DC 1.1. P ≈ 27. methods vary 8-18. 2) b: an = 27( 3 ) = 9( 3 ) 1 n 1 n!1 8-21. c = 96°. AC = DF 5. BC ! EF ! 7.5 square miles. d = 94°. A = 40. AB // DE . and (–6. b: 45°. then alternate interior angles are equal. b = 83°. 360° 8-19. a: an = !45 + 15n = !30 + 15(n ! 1) 8-22. a = 87°. BC // EF .7 miles 8-20. Reflexive Property 4. ASA ≅ 7. 3. B Selected Answers 55 . See answers in bold below. Given 2. Statements Reasons 1.Lesson 8.2 8-17. The other two angles must be equal since the triangle is isosceles.5 and the y-coordinate is 2 + 83 (8) = 5 . a: (6. 5) b: 3 8 c: Using the strategy developed in Lesson 7. (180° – 36°) ÷ 2 = 72°. Therefore. !x = 14 " 2 = 12 and !y = 10 " 2 = 8 . 8-33. P ≈ 108. a: The region can be rearranged into a rectangle with dimensions 14 and 7 units.3 8-29. so ΔPQR ≅ ΔSRQ (SSS ≅) and ∠P ≅ ∠S (≅ Δs → ≅ parts).5 = 145 square inches 8-32.3. a: A = 36 sq. c: 10!·!14.2. B 56 Core Connections Geometry .1. ft.5. a: Isosceles triangles b: The central vertex must be 360° ÷ 10 = 36°. x = 49° b: 2(71°) + x = 180° . QR = QR (Reflexive Property). cm. a: x + x + 82° = 180° . QP = RS and PR = SQ (given). b: (14)(7) = 98 square units 8-35. P = 28 ft b: A = 600 sq. Then the x-coordinate is 2 + 83 (12) = 6.3 cm 8-30. 8-31. x = 38° 8-34.Lesson 8. a: 1. Therefore. 8-41. All circles are similar. c: The ratio is 5 = 5.Lesson 8.4 8-40.1. a: Non-convex b: Convex c: Convex d: Non-convex 8-43. 1 d: The ratio of the areas is 25 1 = 25 . The reflections are all congruent triangles with equal area. a: 3 b: 15 c: 4 d: 9 8-45. A = 4.52 square inches.04)t c: ≈ $199. or use similarity transformations to justify the similarity.42) = 68. the ratio of the perimeters equals the zoom factor. 8-46.04 b: f (t) = 135000(1.833 8-42. the total area is (6)(11.9 units2 8-44. P = 70 cm b: The length of each side is 5 times the corresponding side in the floor plan. a: A = 192 cm2. The ratio of the areas equals the square of the zoom factor (52). Selected Answers 57 .800 cm2 and P = 350 cm. a: 64 units2 b: ≈ 27.0 units2 c: 8 3 ! 13. a: Equilateral triangle b: Rectangle c: Nonagon d: Rhombus or kite 8-62.25 units. Thus. So the y-coordinate of point C could be 6 3 or – 6 3 . 8-63. See the diagram for problem 8-30 for a similar diagram.5 8-53. the relationships in the figure are true. b: No. b: ≈ 32. 0). 8-64. a: Yes. 180º(n!2) b: Use 360° ÷ 20° = 18 sides or solve the equation n = 160° to find n = 18. as long as the two angles remain congruent.17 and –1. a: (–2.1° 8-55. ≈ 103. Therefore. 8-54. since the midpoint M of segment AB is (6. a: The interior and exterior angles must be supplementary. 180° – 20° = 160°. AB ! DC (given). they must intersect at the midpoint of BD . a: The perimeter of both triangles is ≈ 48. D 58 Core Connections Geometry .9° and ≈ 57.8 meters 8-58. AC = DB (≅ Δs → ≅ parts).1. The x-coordinate must lie on the perpendicular bisector of segment AB.84 b: w ≈ 2. MC = 12 2 ! 6 2 = 6 3 because of the Pythagorean Theorem. and ∠ABC ≅ ∠DCB (given). B 8-66. 0) and (3. 8-56. Therefore.57 c: No real solution 8-59.Lesson 8. a: 60° b: 82° c: 14° d: 117° 8-61. since BC = BC (Reflexive Property). Since the diagonals of a parallelogram bisect each other. 8-65.5. a: w = ± 17 5 ! ±!1. Thus. 21). the x-coordinate of point C must be 6. 0) b: The graph of y =!!(2x 2 ! x ! 15) would be the reflection of y = 2x 2 ! x ! 15 across the x-axis because each y-value would have its sign changed.2 mm2 8-57. E 8-60. A = 100 3 ! 173. then ΔABC ≅ ΔDCB (SAS ≅). they intersect at (6. ΔAMC is a right triangle and the hypotenuse must have a length of 12 units for ΔABC to be equilateral. 6. 8-75.2. 8-76. 8-77.7 units b: A = 306 units2. 8-74. a: A = 34 units2. Adding the rectangles makes the total area ≈ 41. 12 are factorable. ratio of the areas = 9 8-72.Lesson 8.4 ft2.4 ft2. x = 2. b: Reasoning will vary. yes. but now the first spinner is definitely more attractive. P ≈ 77 units c: ratio of the perimeters = 3. it is most likely you will earn more extra credit if the class spins the spinner with the options of 5 and 10 points.05% . 80 inches or ≈ 6. For example. k = 0.1 8-71. The area of the hexagon ≈ 23.5 or 6. there are two possible answers. a: Reasoning will vary. 10. 4 21 ! 19. 4x 2 = 2x 2 + 17x ! 30 .67 feet 8-73. P ≈ 25. B Selected Answers 59 . (or Reflexive angles are ≅. b: Answers vary. and 6. a: 5 + 1 = 6. // Given in diagram ∠A ≅ ∠C ∠ADB ≅ ∠CBD ≅ Given in If lines are //. a150 = !74 1 2 8-86. a: ≈ 403. ΔABC ≅ ΔDCB b: ASA ≅.2 8-83. Shared side diagram then alt. 8-87. One solution is 2. D 60 Core Connections Geometry .Lesson 8. so two sides will collapse on the third side.1 cm2 b: ≈ 100. a: 3 4 b: rp c: ar 2 8-84. Property) ΔADB ≅ ΔCBD AAS ≅ ≅ ≅ Δs → ≅ parts 8-88. 5.2.8 cm2 8-85. ΔABC ≅ ΔEDC 8-89. int. a: AAS ≅. 3.78. triangle area formula.25 = 2. ! So ΔABC ≅ ΔEDC (AAS ≅) and AB ! ED ( ! s " ! parts ).5 units 8-99. 5) b: (3. The area of the hexagon is 24 3 square units.2 8-105. or (33.3.23 square units. n = 59°. Area of the entire pentagon ≈ 172.Lesson 8. If lines are parallel and cut by a transversal. D Selected Answers 61 . b: x ≈ 5.2 units 3 8-107.3 units2 8-108.657 8-111. P ≈ 30.5° = x 7 b: x ≈ 3. 168° 8-96. (3. then same-side exterior angles are supplementary. a: 8 b: ≈ 18 c: (1. 30°-60°-90° pattern. angles are equal. 7) c: y = 4 3 x+3 d: 2 2 + 1. b: x = 33. so the side length of the square is 24 3 ! 6. a: (1.05) ! 103.5 2 = 6. a: x + x + 125º + 125º + 90º = 540º.1 8-93. a: cos!58. 8-95. d: 24 units. 11). 8-94. int. B Lesson 8. then alt. Law of Sines. a: x = 26. 33). c: No solution. x = 3° 8-98. 1) 8-110. (11. a: 20 b: ≈ 126. 8-106. 8-97.5 units b: A = 4 2 square units.5. BC ! DC and ∠A ≅ ∠E (given) and ∠BCA ≅ ∠DCE (vertical angles are ≅).05 square units. P ≈ 10. 8-109. so the shaded area ≈ 53 (172. x = 100° b: 6x + 18º = 2x + 30º.45 units. a: A = 42 square units. if lines are parallel and cut by a transversal. hypotenuse must be longest side. 3). a: x = 14 3 . a: C = 28 π units. Definition of a Circle (radii must be equal) 5. Reflexive Property 4.8 d: 2 and –2 8-121. c = 68° . a = 97° .Lesson 8.3. r = 50 units. d = 68° 8-122. a: (3) b: (1) c: (4) d: (2) 8-120. area = 2500 π units2 d: A = C2/4π units2 8-117. FC = FC 3. ΔAFC ≅ ΔBFC 5. AF ! FB ! 6. AB ! DE and DE is a diameter of !C . !! s!"!!!parts 8-119. Statements Reasons 1. AC = BC 4. ∠AFC and ∠BFC are right angles. A = 196 π units2 b: C = 10 π units.8 and –2. A = 25 π units2 c: d = 100 units. D 62 Core Connections Geometry .3 Day 1 8-116. b = 15 . Given 2. See bold answers in table below. 2. 1. Definition of Perpendicular 3. a: –3 b: –4 c: 2. HL ≅ 6. 57 sq.96 or approximately $2.3.3 Day 2 8-123. BC = 7 . A = ! " 45.6 square feet.84 square cm 144 b: 2πr = 18π. x = 35º 8-125.71 square units b: 36 square units. d: A circle 1 5 8-127.509. a: supplementary angles sum to 180º. more c: 24 3 ! 41. a: 8 b: 6 8-128.5 feet.4 square feet b: 110 + 60 π ≈ 298.388 c: Area is four times as large ≈ 24.Lesson 8. its area is greater than both the square and the equilateral Δ. perimeter is twice as long ≈ 597 units. x = 26º b: alternate exterior angles are congruent. x = 5º c: Triangle Angle Sum Theorem. A = 81π ≈ 254. r = 9. r = 12 ! . and ED = 6 .5 · 8 = $2387. a: 2πr = 24. a: CD = 22 .47 square cm 8-129. 8-124. a: (55)(60) + 900! " 6127. units. a: 16 3 ! 27. x = 15º d: exterior angle equals sum of remote interior angles. 298. the perimeter is 22 + 14 + 12 = 48 cm b: 19(4) = 76 cm2 8-126. E Selected Answers 63 . Possible solution shown at right. shoes clothing tennis gender other long pants tennis shorts other male 0 dress or skirt tennis long pants other tennis female shorts other tennis dress or skirt other c: Using M or F for gender. The central angle is 360 ÷ 10 = 36°. MST. Assuming there are no hidden cubes. 2 0 1 b: Answers vary. a: 4 b: 196:1 c: 9:1 25 9-9.4 units2. and T or O for shoes. 9-10. or D for clothing.1 3 2 0 9-7. or. (36 ! 9" ) ÷ 2 # 3. V = 11 units3 1 1 1 9-8. FLO. and FDS}. –2) b: (–6. a: Solutions vary. – 4) c: ( 12 . The intersection is {MLT. S.1. 6 long pants P(long pants) = 12 12 12 12 2 # successes total # possible = 12 students = 12 . 2 + 2 + 1 + 1 = 1 . FLT. FLT} 9-13. A tree diagram follows part (b).Lesson 9. Since the perimeter is 100. Area = 769.!12) 9-12.86 units2 9-11. The right triangle has acute angles 18° and 72°. each side is 10. and L. a: (4. b: From the following tree diagram. A 64 Core Connections Geometry . FST. MLO. the union is {MLT. a: An area model is not possible because there are more than two events. a: ≈ 986. 9-21.271 or 72. a: 2 2 units b: (–1. a: 6 or –6 b: No solution because absolute value cannot be negative. c: 27 · 180° = 4860° 9-23.1. 986. ! (6)2 (14.9% decrease 9-25. By the Addition Rule. y) ! (x.25.44 cm2 9-35.5) = 522! in. As a midsegment. a regular 36-gon has interior angles of 170°. c: x = 3 or –17 9-38. 9-34. a: 151° F R T b: Yes. 5) d: (2.3 Selected Answers 65 . 522! !in. Yes.3. 240 cm3 9-27.1. 2 9-37.07 = 200 11 + 4 ! P(long and lost) . If the ratio of lengths is 0. SA = 1438. DE must be half the length of BC. ≈ 0.1!gal 1 231!in.16 square mm b: 400% is 4 times as large. by AAS ≅. (x. Base Area = 509. 0. Height = 5 cm. D Lesson 9. "y) c: (8. its area increases by a factor of 42. x = 31° b: 4x + 20° + 5x ! 2° = 180° . 9-36.Lesson 9.3 !·! 1!gallon " 7.16 · 16 ≈ 15.2 9-20. b: 42 units2 9-22.52 = 0.23 cm2.3 9-33. x = 18° 9-24. resulting in a probability of 200 1 % that the food took too long and the rider got lost.61 square mm.778.5. 24 square units. 6). –3) 9-26. a: See solutions at right. Therefore. a: 2x + 4x ! 3 + 7x ! 6 + 3x + 12 + x + 10 = 540 . then the ratio of areas is 0. 45 + 45 = 45 ! 15. If she needs the balloon to double in width.3 b: 324π · 27 = 27. 9-51. total distance = 20π ≈ 62.65 in. 9-46.5 in. and the shading should be to the right of the line. a: See diagram at right. c: The length of the base of the composite triangle must be 6 3 . then the volume will increase by a factor of 8. The smaller triangle has a base length 6 3 ! 5 " 5.07 9-49.3 9-47. b: V = 16 cubic units.Lesson 9. D 66 Core Connections Geometry .482.4 9-45.8 feet 9-48.6% 3 4 7 3 4 2 red oak 9 white oak 9 maple 9 granite 1 3 4 5 45 45 4 tile 5 9-52.2. x ≈ 8. A tree diagram would have worked as well. The line should be solid. See the area model below. a: SA = 180π ≈ 565. she needs 21 more to fill the balloon.9 in. a: x ≈ 10. SA = 52 square units F R T 9-50. V = 324π ≈ 1017.39 .3 b: No solution because the hypotenuse must be the longest side of a right triangle. Since she has already used 3 breaths.1. Circumference of each circle = 10π. That means the balloon requires 24 breaths to blow it up. 3 9-61. 9-63.026 fish per cubic inch.85 b: f (t) = 27000(0. a: 0. but the other is not for the same pair of lines cut by a transversal. It is the cube of the linear scale factor..4 in. so V = 7!·!13!·!6 3 = 546 3 ! 945. ratio = 8 = 23. One pair of alternate interior angles are equal.7 in 3 1 in 3 25 fish 45.5 9-56. ratio = 4.980 9-62.68 fish c: ! 12 1 foot ! 1 foot ! 1 foot " inch 12 inch 12 inch 945. a: It is possible.7 inches 3 25 fish 0. D Selected Answers 67 . 9-57.1.026 fish b: ! or about 0. or. a: False (isosceles trapezoid) b: True c: True d: False (parallelogram) 9-59.!! 12 c: x = 2 d: x = –2. 2(12x + 7) = 30x ! 4 . b: 2 c: 24 and 96. b: Not possible. 945. d: 6 and 48. c: Not possible. a: x = –2 b: x = 5.85)t c: $11. It is the square of the linear scale factor. the vertical angles are not equal. so x = 3 9-60.Lesson 9. a: Height of the tank = 6 3 ! 10.7 in 3 1 ft 3 9-58. Same-side interior angles should add up to 180°. 9-74. b: 3000 ft3 20' 3 c: 4. the diagonals bisect the angles of the quadrilateral. c: 32 3 ≈ 55.4 units2 9-69. See tree diagram below. a: Not congruent since side ratio ≠ 1 b: ΔABC ≅ ΔDBC (HL≅) c: ΔABC ≅ ΔMLK (SSS≅) 9-72.1 9-68. a: See diagram at right.2.2 butterflies per cubic foot because 625butterflies ! 0.Lesson 9.21 butterflies .4 units2 d: 32 3 ≈ 55. y = 6x ! 2 9-73. 4x ! 5° = 2x + 9° . x = 7° 9-71. Or.8 ft 10' d: 0. 15' 3000ft 3 3 ft 9-70. a: A rhombus b: They are perpendicular and intersect each other at their midpoints. B Bread Protein Condiment mayo salami plain mayo white turkey plain mayo ham plain mayo salami plain mayo grain turkey plain mayo ham plain 68 Core Connections Geometry . Connect the original point to each of the other two points. Possible response: Construct a circle of any radius. 9-80. 6) 9-86. 36! " 113. although not using Karen’s incremental strategy. the line segment got a bit longer. Area of whole circle = 16! m2.Lesson 9. regular heptagon: 7 " 128. C x Selected Answers 69 . 9-81. so A = 2 = 153 3 ! 265 m2 9-83. so the measure of the interior angle of a regular heptagon has greater measure.10 ft 3 (7!2)!·!180° 9-84. 0). Height = 9 3 m. Area for dog to roam = 240 (16! ) = 32! " 33. b: It got longer.51 m2 360 3 (20+14)(9 3) 9-82. Then choose a point on the circle and mark off two radius lengths in each direction along the circle circumference. One way: Construct a hypotenuse for a right isosceles triangle with leg lengths 1 cm.2. c: Although the change is minute.2 9-79.6° . x-intercept: (4. y 9-85. Equilateral triangle: 360° ÷ 3 sides = 120°. y-intercept: (0. d: It is possible. See graph at right. then the two sides opposite those angles are also congruent) and ΔDBG is isosceles (definition of an isosceles triangle). B 70 Core Connections Geometry . ΔABC ≅ ΔEDF (given).Lesson 9.25q = 2. a: See diagram below right.67 ft 3 9-96. AB ! BA (reflexive property). f (x) = 7. mat plan ! 9-92. b: ! 30177. DA ! CB and m∠DAB ≅ m∠CBA (given). Then ∠C ≅ ∠D ( !! s!"!!!parts ). Using equations: n + q = 14 and 0. 6 0 0 b: 24 cubic units 3 0 1 2 6 6 c: The volume of the new solid must be 18 the original.5) x 9-98. or approximately 1 goat per square meter. ! 9-97.3 9-91.2 m2 b: ≈ 36. 9-95.2 m 2 m2 10 9-94.1 . a: ≈ 74.05n + 0.2. so n = 3 and q = 11.2 meters 6 goats goats c: ! 0. so !DAB !!!CBA (SAS≅). so ∠B ≅ ∠D ( !! s!"!!!parts ). 9-93. so DG ! BG (if base angles of a triangle are congruent. so the reduced volume must be 3 cubic units. a: It has 8 sides.68(2. 74.90 . 5 2 )(6) # 55.1.5 square miles 9-107.3 in.!b = 1. 000(0.17 . 5). 10-10.2.98 cm 3 b: BA ≈ 19. She should then construct two more circles with radius k. 81. They intersect only once at (3. 9967 d: 60 = 25(b)10 .3 ft2.3) x x c: y = 42. 9% increase 9-108. C Lesson 10. a: 70° b: 50° c: 2x 10-8. a: b: 9-109.1 10-6. centered at R and S. Height = 6 3 . 81. !4 43 . ≈ 5626. The fourth vertex lies where these two circles intersect. arithmetic and geometric 10-11.2 ft3 9-104.000° y 9-106.3)(7) ≈ 135.3 cm 3 b: Answers vary.2 . so V ≈ (19. 81.4 9-103. She should construct an arc centered at P with radius k so that it intersects n and m each once (call these intersections R and S). 12 . a: 14! " 43.17. One possibility: It could represent a pencil sharpener. a: See graph at right. !7 83 . She is constructing an angle bisector. 9-105. 81 . !17.Lesson 9. perimeter = 15 + 12 + 12 + 12 + 15 + 16 (24!) = 54 + 4! " 66 in.11. neither c: 81. a: 17. D Selected Answers 71 .09. 10-7. 10-12. area = 15(6 3) + 16 (144!) = 90 3 + 24! " 231. b: f (x) = 10(2. a: V = (2)(5)(6) ! "(0.75)5 . !17. geometric b: 32. 1000 · 180° = 180. 10-9. p = 10 d: 18t = 360! . 10-24. so !POY " !EKY (SAS ! ). a: They are similar. a: 9 cubic units b: 10 cubic units c: Possible response: There are no “floating” blocks and that there are no hidden blocks behind the ones visible.2 10-18. c: 60 360 (28! ) " 14. a: 3 b: 6 c: 2 d: 1 e: 4 f: 5 ! is a dilation of AB 10-32. a: 5m + 1 = 3m + 9.6°. Central angle = 3. b: 108°. central angle d: 216°. D Lesson 10. PO ! EK because ! "s #! parts. yes 10-23. d: Either a left or back view would reveal any hidden blocks. ! b: They have the same measure because they have the same central angle. x = 17° c: ( p ! 2)2 + 6 2 = p 2 . However. !PYO " !EKY (arc measures are equal). OY ! KY ! EY ! PY (all radii are congruent). it is similar to a part of a circle with shorter diameter. 4) and E(4. so AC should be 10 units long. c: Yes (AA ~).7 units 10-33. ! (2)2 (4) ( 360 ) 45 " 6.1. C 72 Core Connections Geometry . CD ! from P. m = 4 b: 2(x + 4 ! ) = 3x ! 9! . ΔABC ~ ΔLKH b: Not enough information .1. t = 20° 10-21. inscribed angle c: 72°. a: 64° b: 128° c: 64° d: 180° e: 128° f: 52° 10-19. c: 6 2 + 8 2 = 10 units 10-22. A ≈ 795. 7) b: DE = 5 units.28 ft 3 10-36. It must be a rhombus and it could be a square. a: D(0. a: Yes (AA ~). Therefore.Lesson 10. 10-34. CD is longer.51 square units 10-20. 10-35. ΔABC ~ ΔEDC 10-37.3 10-31. a: x ≈ 31. 130°. 6 5 6 c: Possible response: Because all circles are similar.1. MB = 6 feet. b: C ! is at (–5.02 sq.21 sq. The angles. 10-45. Demonstrate that each side is the same length and that two adjacent sides are perpendicular (slopes are opposite reciprocals).05)t 10-59. Therefore we can write a arc length D arc length E proportionality equation. 10-56. The ratio of the volumes must be ( 15 )3 . a: A ≈ 549.9 ft 2 . so the probability is 45 . measure 64°. y = 250 ! 15. –8) and D!! is at (–7.8 ft. 116°. MA = 14 + 17 + 6 = 39 feet.5 in.9°. 6 b: The arc length is 5! . a: The arc length is 30 2 . y ≈ 10. E Selected Answers 73 . a: 1.9 ft 2 . so the ratio is 6 = ! . 1 2 (9)(12) 12 (20)h.9)(12) ! 743. so the ratio 3 = 10-55. cm c: A ≈ 25.05 b: ≈ $130.9) + (5)(6)(12) ! 483. h ! 5. that is.78 sq. and 140°. cm b: A ≈ 74. will be the same for all 30º sectors. 90°.588 c: f (t) = 130588(1. 10-49.7 m c: x ≈ 34. radius D = radius E . C Lesson 10. circle D is similar to circle E.Lesson 10. 10-48. a: 50° b: 50° c: 67° d: 126° e: 54° f: 63° 10-44. a: 4 times b: 360 c: 360! ÷ 5 = 72 10-46. a: It is a square. area of the regular pentagon ! 61. from smallest to largest.4 10-43. total SA ! 2(61.5 10-54. so the volume must be ( 15 )3 ( 500! ) = 4! cm 3 . the radian. 10-58. In fact all sectors with 30º central arc length angles will be similar. so the ratio of radius . so AB = ER and ER = 1485 ! 38.1. volume ! (61.7°. 4). Therefore sector D is similar to sector E (the angles are the same and the radii and arc lengths are proportional by the same ratio of similarity). cm 10-60. b: x ≈ 3.3 ft 3 10-47.4 " 10-57.5 feet " ! 360 ! 2"(3) = 2 . Lesson 10. Erica’s nugget weighs 125 ! 5.4º d: 134.5 c: 67.6 = 700 g . and use the ratios of similarity to calculate the data for Erica’s nugget themselves. it is less likely that there are red marbles. there may be red marbles that she has not selected in her draws. Thus.2. 10-68. the ratio of the volumes is 5 3 = 125 . 1 6 2 b: 5 6 (60) = 50 times c: Answers vary.7 10-71. a: 13 b: 6.1 10-66. c: Since r = 5. b: Since r = 5. a: 1 . c: This is not possible. and all numbers must be less than 5. 20 ! 25 = 500 cm 2 . but no number of trials will ever assure that there are no red ones. but two squares should have an even number. no numbers can be 3. a: True b: False c: False d: True 10-72. a: No. Thus. x ! 25. 2x + 3x + 4x + 5x .8º 10-69. no number of draws will assure this. a: Have Ken’s analyzed for $30. 10-67. which is about 25 ounces. b: No. the ratio of the areas is 25. B 74 Core Connections Geometry . 10-70. or the Triangle Sum Theorem can be used. a: x = y b: y = 2x or x = 1 2 y c: 3y = 5x d: x + y = 180 10-79. B 10-85.Lesson 10. m!C = 0. 10-87. a: P(on-campus given Engr) = 120 800+120 ! 13.28 1 ft d: 6 square cm e: 1! 14 ! 23 = 12 1 1 (360°) = 30° . then the equation. area = 125º (64! ) " 69. 360 r = !3 . because 8 2 + 15 2 = 17 2 . 0. so it should result 14 (80) = 20 times.94) = 97 b: mAD ! = 125º and the length of AB c: mAB ! = 125º (16! ) " 17. x = 60º 10-84.5(1.0% b: P(on-campus) = 0.2 ft 1 ft red b: 5 ·5 6 6 = 25 36 ! 69.2 ft 0. 12 1. w = !6 or 4 .8 in.2 10-78. a: x = 34 b: x = 4 3 c: x = !5 d: x = 32 5 10-90. a: Both are ! . a: See diagram at right. She is incorrect. 10-89.5 ft 10-82. a: = 40 360 9 ! = 2(97" ) = 194 " . b: Yes. can be solved for x. the Exterior Angle Theorem. Regions B and C have equal weight (which can be confirmed with arc measures). a: See diagram at right.4% blue 5 ! 5 = 25 " 69. a variety of relationships such as Parallel Line Angle Conjectures. 10-81. 540° 10-83. so they should each result (80 – 20) ÷ 2 = 30 times. 10-80. 10-86. A Selected Answers 75 . Note that a radian measure is a ratio and does not have units.5 " . 4 x (2! r) b: If x is the unknown central angle of the circle. Region A is 14 of the circle.2. 10-91. 180 ! 64 ! 26 = 90 .2 360º 360º 10-88. b: 1. y = 71°. which can be tested by substituting both answers into the equation.4% c: 6 6 36 5 (49! ) " 128.4 ft 3 ft ft 25 25 0. a: Yes. x = 109°. Methods vary. The probability of living on campus given that the student is an engineer is much smaller than the probability of living on campus. z = 99°. because of the Triangle Sum Theorem.6 c: No. 0. ! . b: AB c: BC d: BC AC AB AC e: AB f: BC AC AC 10-107.9 b: 18 76 Core Connections Geometry . and in turn. a: If x represents the length of chord AC . 10-102. or radian measure. x ≈ 12.2. Not D a Daily i l y Weekly local 25% 12% 37% Not weekly local 40% 23% 63% 65% 35% 100% b: 25% + 12% + 40% = 77% c: 25+40 77 ! 84. 105. 10-105.75 6. SA = 28 units2. the arc length must be doubled. a: See table below.) F R T b: V = 7 units 3 . c: Since this solid has no “holes. the ratio of arc length to radius.” the surface area can be calculated by adding the areas of each of the views. must also double and thus be ! . a: See views at right. then x 2 = 10 2 + 10 2 ! 2(10)(10) cos 80º . Entries not in bold are given in problem statement. Methods vary.25! " 5. (Or they could be reflections of each other.3 10-101. h = 107. the top and bottom views are the same. Depending on how they are oriented.4% 10-103. a: 236 236+274 = 46. 10-104. Since the angle is doubled.49 in.5 2 )h .75 = ! (2.3% b: P(laptop) ≠ P(laptop given business trip) so they are associated. 10-106. and entries in bold are computed from given information.Lesson 10. SA = 6(100 2 ) = 60. a: 30% b: 42% c: P(Green Fang) ! P(alarm) " P(Green Fang and alarm) .53 " 84. 000 . 5! = 120 10-116. 576. 10-132. so they are weakly associated. Since they are approximately the same. a: 7! = 5040 b: 1 5040 10-128. P(ticket given red) = 507+19493 ! 0.7% yellow 9 18 red 10-118. a: cylinder b: triangle based prism c: cone Selected Answers 77 . a: See table at right. 000 = 140. 000 cm 3 1 10-119. 000 ! 17.2 10-120.000+348 ! 0. a: 12 60 = 20% b: 9 45 = 20% c: Yes 10-117. Area of shaded region ! 93.2 10-127. 158.025 and P(ticket) = 20. they should not charge a higher premium.1 10-114.82 ! 8. a: 24 b: The decision chart tells how many branches there are at each stage. 10-131.28 " 0. they are 507+9 most likely independent.71 in. 1 175760 10-133. 000 cm 2 . c: 1 6 10-115.Lesson 10. 000. 507 No. 0. CIRCLES b: 1 red yellow 9 1 SQUARES c: 9 1+ 1 ! 66. blue V = 100 3 = 1. Edge = 1200/12 = 100 cm.22 . 608. C Lesson 10.64 ! 0. 12 P3 = 12! (12!3)! = 12! 9! = 12 "11"10 = 1320 10-130.025 .3.184.3. 760 b: 1! 15 P5 = 1!15 !14 !13!12 !11 = 360.2% . 9! = 362.3 10-139. a: 16 P6 = 16 !15 !14 !13!12 !11 = 5. you would be dividing by 0. a: 10 P8 = 10! 2! = 1.360 c: d: 15 C 5 = 3003 = 37.3. a: 5! = 5 ! 4 ! 3! 2 !1 = 120 5! = 5!4!3!2!1 = 60 b: 2! 2!1 c: 5! = 5!4!3!2!1 = 30 2!2! 2!1!2!1 d: Because you cannot tell the repeated letters apart.2%. P(coffee and dairy) = P(coffee)  P(dairy) = 63 )( 63 ) ! 22. 814.765.5% 16 C 6 8008 10-144.25% 360. 400 P b: 10 C8 = 108! 8 = 8!2! 10! = 45 c: 6 C1 = 1!5! 6! = 6 10-142. 8 P8 = 8! (8!8)! = 8! . 880 10-140. c: 3! 3 = 2!. 2! 2 = 1!. P(no caffeine and no dairy) = P(no caffeine)  P(no dairy) = ( 42 21 ( 63 )( 63!21 63 ) " 22. a: 20 b: 30 c: 36 10-145. a: 1 b: 8 things taken 8 at a time. 765. 1! 1 = 0! 10-141. Because they are independent. 0! If 0! = 0 . Avoid the common mistake of computing 63!42 P(no caffeine and no dairy) = 1 – P(coffee and dairy) because {coffee and dairy} includes all the drinks that have both coffee and dairy but does not include drinks that have only coffee or only dairy.Lesson 10. 10-143. 8 P8 should equal 8!. 78 Core Connections Geometry .760 = 6. 360 5. there are fewer arrangements when there are repeated letters. 708. Selected Answers 79 . a: There is one way to choose all five. b: There is one way to choose nothing. a: 22 P3 = 9240 b: This is really a permutation lock. c: = 1540 . In this case the common use of the word “combination” conflicts with the mathematical meaning. In order to have the formula give a reasonable result for all situations. d: 22 ! 21! 21 = 9702 10-156. 2!5!8! = 168 arrangements of cars. 500 10-154. 5! 0!5! =1 c: This is a permutation in which some of the items occur multiple times. (12 – 1). a: 162º b: 16 sides c: ! 120.8 cm 2 10-157.3. 5!0! 5! = 1 . (12 – 3) b: n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5) c: n(n – 1)(n – 2)(n – 3)(n – 4)(n – 5) 10-158. it is necessary to define 0! as equal to 1. a: 12. like in an anagram. 3! 7 ! 2 !1! 4 = 168 choices 10-155.Lesson 10. (12 – 2). but this does not make sense for a mechanical lock because it would 22 C 3 imply dialing the numbers in any order to open the lock. 500 C 3 = 20.4 10-153. V = 27. a: See diagram at right. A(face) = (8)(10) = 80. x = 8 . sin b = 15 17 . A(base) = 0.83 or a 17% decrease 15 cm 10-178. 12 C2 . 10-177. Entries not Not bolded are given in the problem OceanView Ocean statement. 80 Core Connections Geometry . 3!2! 10-181. this is (0. 28 hours 10-182.72 0. like an anagram.5 10-173.06 0. Not (0. 12 C1. sin a = 8 17 .24 0. TSA = 2(27. ≈ 0. 12 C 4 . cm.18 0.90 b: P(Senior!Ocean View) = 0. If t = hours. 2 + 14 t = 9 .06) . cm 17 cm d: 90º.5(8)915) ! 60. This is permutation with some of the items to be arranged occurring multiple timers.10 computed. b: h 2 = 17 2 ! 15 2 = x 2 . 10! = 302. SA: A(!) = 12 (8)(4 3) ≈ 27.00 c: They are associated because P(senior) ! P(Ocean View) " P(senior and Ocean View) .20)(0. while bold entries are View Senior (0.10) = 0. 12 C 3 . The first because 3! > 2! 10-175. h = 10 cm c: TSA = 2(60) + 17(10) + 8(10) + 15(10) = 520 sq.10)(0. 400 ways to layer the dip.Lesson 10. n ! 1 .76 1. a: Similar b: Similar 10-183.24) ! (0.71) + 3(80) ≈ 295. a = 28º.43 sq.04 0.71. 700 10-174.13 cu. 12 C5 . a: (n ! 3)(n ! 4)(n ! 5)(n ! 6)(n ! 7) b: n + 2 .24 = 25% senior 0. a: See diagram at right.3.60)(0.90) = 0.71(10) ≈ 277. n ! 2 c: n(n ! 1)(n ! 2) d: (n + 2)(n + 1)(n ! 1) 10-176. cm.06 0. 600 = 60h . 100 C 3 = 161. b = 62º 10-179. 12 C0 = 1586 10-180. a: b: T H T IG H R IG FRONT FRONT R 10-184. n + 1 . b: The missing angle = 180 ! 62 ! 70 = 48 and since the angle opposite side a is bigger. 11-9. 11-21. is 10 units long and the area of the circle is 100! " 314. a: A sphere. so h = 840 ÷ 120 = 7 mm 11-18. a must be larger than b. that is the definition of hexagon.1 11-8. Then the radius. y c: cos ! = xr . c: a = 9 3 ! 15.2 d: 4x ! 2 + 2(8x ! 9) = 180 . x = 37° b: x + 67 = 180 . b: A cylinder with a cone on top and bottom. (10)(12)h = 840 . a: 4 b: 6 c: 4 11-16. so b is larger than a. y 11-19. P = 84 units b: A = 16 square units. B 11-15. then ΔADC is a 30°-60°-90° triangle. a: x 2 + y 2 = r 2 b: sin ! = r . x ! 7. AC . y ! 7. y = 24. a: 2x = 180 ! 106 . a: Yes. P = 28 units 11-13. x = 113! . b: No. n! (n!4)! 11-17. even though we are not told that b is a central angle.Lesson 11.1. x 11-20. a: 7! = 5040 b: 1! 5!!1 = 120 11-10. c: A double cone: two cones attach at the vertices. the sum of the interior angles only depends on the number of sides of the polygon. 11-14. x = 10 11-12. a: A = 144 square units.16 units2 . c: Yes.5 c: sin 73 9 = sin 57 x = sin 50 y .6 units2 and b = 16 units2. If the circle’s center is C and if the midpoint of AB is D. only regular hexagons have a guarantee of having an angle with the measure 120°. C Selected Answers 81 . 5y + 3y ! 16 = 180 . Lateral surface area = (circumference of the base)(height) = 8! "15 = 120! # 377 cm 2 11-11.9 . a: b is larger. 1. 18.5 b: 3 c: 48 square units 11-44. slant height = 32 + 4 2 = 5. 11-30.3 (Day 1) 11-39. PB = PC (radii of a circle must be equal). a: x = 12.30 "10 24 c: ! 1.Lesson 11. a: 8 faces.56 11-29.13 11-31. 11-40. 11-28. B 82 Core Connections Geometry . 626 c: 1! 22 ! 22 = 484 d: 44 ! 22 ! 21 = 1848 11-27. a: x = 12. D Lesson 11. 12 edges. b: A tetrahedron (also called a triangle-based pyramid). a: 10 b: tan ! = 86 . m∠PBA = m∠PCA = 90º (tangents are ⊥ to radii drawn to the point of tangency). SA = 4( 12 ·6 ·5) + 6 2 = 96 units2 3 11-41.48 "10 26 b: 899 C11 ! 7. ±54 11-32.3% 11-42. 626 b: 23 P3 = 10.9 b: x = 2 c: x = 0 c: x = 23 9 ! 2. The areas are all equal because the triangles have the same base and height. a: Yes. ! Therefore AB = AC ( !! s!"!!!parts ). a: 23 P3 = 10. PA = PA (Reflexive Property). y = 7.2 11-25. and 6 vertices b: A square 4 11-43. C = 8! units 11-45. V = 1 (6 2 )(4) = 48 units3.!! " 53. ±6. so ΔPAB ≅ ΔPAC (HL).1. One way is to cut off a corner so the cross-section is a triangle. A = 16! square units. a: 900 C12 ! 5. 12 inches 11-26. a: ! (1. ∠DEA = ∠BEC (vertical angles are congruent). x = 7 c: 2(x + 4°) = 134° .1. region B. m∠ECB = m∠EAD (given). x = 39° 11-52. x = 63° d: x + 9° + 3x + 15° = 180° . c: 2! (1.5)2 (4.3 . ΔAED ≅ ΔCEB (ASA ≅). since area ! 17. Therefore.5)(4.3 b: Volume of the pot is ! (7)2 (10) " 1539.5) " 42. Aimee would need 1539.4 in.8 = 48. E Selected Answers 83 . 11-49.46 units2 and perimeter ≈ 24. AE ! CE (definition of midpoint).3 11-48.3 units 11-47. so AD ! CB ! ( !! s!"!!!parts ).1% b: 1! 1000 1000 250 = 750 = 75% c: 1! 1000 1000 d: P(A given under 20) = 152 152+54+44 = 60. x = 20 13 ° b: x 2 + (x + 17)2 = 25 2 .9% 1000 1000 469 = 531 = 53.5) " 31.4 ÷ 31.8% 11-51.4 in. a: 371 1000 + 1000 250 ! 152 = 469 = 46. a: 18x + 174° = 540° .3 (Day 2) 11-46.Lesson 11.8 in. a: 6 P4 = 6 ! 5 ! 4 ! 3 = 360 b: 1! 5 ! 4 ! 3 + 5 !1! 4 ! 3 + !5 ! 4 !1! 3 + 5 ! 4 ! 3!1 = 4 ! 5 P4 = 240 11-50.4 or 49 cans of soup to fill the pot. V = 13 (25! )(12) = 100! m 3 . c: 7 C2 = 21 = 37. b: A circle c: A sphere 11-62. See graph at right.2 . A 84 Core Connections Geometry . Only one right triangle can be built: 3-4-5.1. a: BA = 12 (7)(24) = 84 in.2 b: BA = 25! 2 . rolling a combination of 1.3 . SA = (2)(84) + (12)(7 + 24 + 25) = 840 in. ! c: Straight angle. a: Corresponding angles.Lesson 11. 1. a: Vertically through the vertex of the cone. neither because lines intersect. 11-64. lateral SA = ! (5)(13) = 65! m 2 . 1. ! b: Alternate interior angles. V = (84)(12) = 1008 in. 11-63. total SA = 25! + 65! = 90! " 282. a: !ABC ! FED (AA~) b: !ABC ~ !MKL (SSS~) c: Not similar because the zoom factors for corresponding sides are not equal. 11-65. 1. You cannot use combinations 11-60.4 11-58.7 m 2 11-59. 11-61.5% 8 C3 56 3!2!1 = 6 . 6!6!6 216 because each combination is not equally likely. supplementary d: Alternate exterior angles. For example. 2 on the dice (where order does not matter) is three times more likely than 1. a: 8 C3 = 56 b: There are 6 choices left for the third filling. 7% . !9) Selected Answers 85 .1 cm 3 11-73.3 . SA = 4! (4 2 ) = 64! " 201. a: An icosahedron has 20 faces.4 in. r = 2. 8 C 3 ! 10 C 3 = 6720 11-74. V = 4 3 ! (4)3 = 256 3 ! " 268.3 11-76.06) " 164. a: ( 43 .!10) b: (!1. 11-75. new volume = 188.06 in.6 in.37 in.3 .1.Lesson 11. 72! square units 11-79.4". ≈ 29° 11-78. large cone: V = 13 ! (6)2 (5) " 188.2 .2 . a: 12 C 3 + 12 C 3 = 715 b: If raspberry and custard are known filling. 180 ÷ 12 = 9 cm 2 . so the surface area is (20)(45) = 900 mm 2 . r = 4 cm. c: The area of each face is 1 2 (6)(3 3) = 9 3 ! 15. then there are two fewer fillings to choose C + C from.6) ! 62. 11-77.50 ! 2(12. small cone: 2 5 = 6r . b: Since it has 12 faces. so total SA = 4(15.5 11-72.4)2 (2) " 12.1 cm 2 . V = 13 ! (2.50 in. so 10 271510 1 = 715 55 ! 7. 182. 2 ! 52!51 4!12 " 3.132. b: cos 89.5 times greater. 253. a: 52 C2 = 1326 16 C 2 = 120 ! 9.Lesson 11.1 mm 360 11-86. a: Earth’s circumference: 8000! = 25. Sample responses: The rings in a tree.132. a: 1 (9 2 )(12) = 324 cm 3 b: 12 cm3 3 11-87. the ripples created when a stone is tossed in a pond.85° = 238. 11-90.0% or using permutations 12 P2 = 12!11 " 5.1 11-84.0% 52 2 C 16 2 c: 52 C2 d: 12 C 2 = 66 ! 5.4 miles !! 11-85.7 ! 9. a: 108 177 ! 61% b: 44 88 = 50% 11-89. 000 ÷ 25.900 x .7 miles. Answers vary.6 . Therefore the distance to the moon is 230.6% or using permutations.0% or using permutations 16 P2 = 16!15 b: 1326 1326 P 52!51 " 9. the rings of a dartboard. etc.6% 1326 1326 11-88.6) ! 9. so mAB 32 (102. C ! 102.2. D 86 Core Connections Geometry . x ! 91.0% 52 C 2 1326 52 P2 52!51 e: 4 C1 ! 12 C1 = 48 " 3. 3 inches 11-97.977 2 " 2. 4 C1!6 C1!2 C1 = 12 55 " 21.8% 11-98.2. SA = 1007. and 12 P3 = 12 !11!10 ways to arrange twelve pens.777 units.7 square units 11-100. 414. The surface area of the moon ! 4" (1080)2 ! 14. SA = 1100 ! 3 + 240 " 1391.Lesson 11. !ART ! !PIT by AA~ 11-102.2 11-96. SA = 144 in. so. V = 324 ! 12 = 312 cm 3 11-101. 657.7 which makes it larger than Africa and smaller than Asia.34 units2 b: V = 1000! cm 3 .977 in. A = 12 (20)(30. or permutations and the Fundamental Principle of 12 C3 Counting could be used: there are 3 P3 = 6 ways to arrange the colors.92 cm 2 c: V = 60 in. Central angle = 36°. a: V = 2100 units 3 .. distance from center to midpoint of side = 30.3 . .8% .777)(10) ! 3077. D Selected Answers 87 . 6 ! 12!11!10 = 21. 2 11-103.5 2 ! 0. x ≈ 2. radius of slice = 3 ! " 0. 6!4!2 11-99. 3 units2.3 11-110. SA = (6)(16)(160 = 1536 units2 11-116. slant height = 109 ! 10. V (cylinder) = ! (38)2 (71) " 322. V (prism) = (34)(84)(99) = 282. a: ≈ 436. Since P(academic)  P(Arts) = P(academic and Arts) the events are independent. a: x = 270 b: x = 132° . a = 9 23 11-118.3 units2 11-114. 744 units3.Lesson 11. The solution is shown with dashed lines in the diagram above. But some may notice that the vertical angles (72°) are each the average of the two arcs they intercept. This will be revisited in problem 11-120.7 c: 3(x + 2) = 6x . x = 103° 11-121. so the cylinder has more volume. That means that if the sun were placed next to the Earth. a: x = 117°. Base area = 36 units2.4 b: r = 5 sin 25° ! 11.295. V (16)(16)(16) = 4096 units3. 088.44 units. y ! 15.2.8 . 11-113. b = 28°. D 88 Core Connections Geometry . Another possible method is P(Arts given academic) = P(arts). x = 2 11-111.3 ! 161. C 11-117. c = 56° b: Some students may just report that c is smaller. 11-119. There is no association between winning an academic award and a Fine Arts award. its center would be farther away than the moon! c: 1. 6 C3 + 2 !6 C2 + 6C1 = 20 + 2(15) + 6 = 56 11-120.000 miles b: The sun’s radius is almost double the distance between the Earth and the moon. lateral SA = 12 109 ! 125. a: a = 44°. 11-123. or that P(academic given Arts) = P(academic). 11-112. y = 88 ! 9. z = 310° c: 9(9 + a) = 8(21) .029 11-115. 21° + x = 2(62°) . f (x) = 4( 23 ) x 11-122. total SA ! 36 + 125.6 units3. 005)t c: ≈ $11.9° .1. a: x 2 + 8 2 = (x + 2)2 . a: V = 13 ! (32 )(10) = 30! " 94.3 units 12-12.2 units3 b: One method: BA = (21)(18) – (12)(12) = 234 units2. a: 124° b: 25! units2 c: ≈ 12. x = 15 b: tan !1 ( 15 8 ) " 28.1° " 61. 180° ! 90° ! 28. Think of this as an anagram ( 46! 12-9. by SAS ≅ Selected Answers 89 . V = (234)(10) = 2340 units3 !2! = 15 ). C. 12-8.Lesson 12.005 b: f (t) = 8500(1.1 12-6 a: x 2 + y 2 = 9 b: 7 12-7. Sample tools: Trigonometry and the Triangle Sum Theorem 12-11.1° . a: 1.465 12-10. 17 b: sin 102° 7 = sin 62° x . a: sin 27° = 18 x . 2x + 9° = 4x ! 2° . !3). EF seems to be the average of AB and CD. x = 5. x ≈ 56. 116° + (3x + 8) + 32° + (2x ! 1) = 360° . a: The slant height of the cone is ≈ 9. 12-27. 3 so total the volume is ≈ 1244. x ≈ 8. d = 72º 12-23. a: Vertical angles are equal. 3) .22 m.78 m2. while AB = 6 and CD = 2 c: Sample response: The midsegment of a trapezoid is parallel to the bases and has a length that is the average of the lengths of the bases.07 + 263. 12-22. x ≈ 6.55°. LA(cone) ≈ 6π(9.89 m3. so 7x ! 3° + 4x + 12° = 180° and x = 15.1. x = 41° c: When lines are parallel.96 m3. b: Yes. so total surface area is ≈ 173. !3). EF = 4 .Lesson 12. 3) and F(7. DC = 3 . AB = 9 .32 c: tan x = 64 . r = 5 b: (!1.78 + 414. 12-24. same-side exterior angles are supplementary.47 m2 b: V(cylinder) = 36π(11) = 396π ≈ 1244.31° 12-26.2 (Day 1) 12-21. r = 4 12-25. EF = 6 . D 90 Core Connections Geometry .07 m3 and V(cone) = 1 (36! )(7) = 84! " 263.5 b: The sum of the angles of a quadrilateral is 360°. a: (2. a: 36° b: b = c = 108°.69 = 588.89 = 1507. and LA(cylinder) = 12π(11) = 132π ≈ 414. a: E(1.22) ≈ 173.69 m2. 25 m2. Also. a: Since the hypotenuse is 1.!! 3 8 c: k = 1!or!9 d: p = !4. a: (x ! 4)2 + (y ! 2)2 = 9 b: Pick any two points. cos ! = 1x . BA = (10)(10) = 100 ft2. b: It must be 1 because of the Pythagorean Theorem. c: Yes.51 m2 b: Slant height = 12 ft. A Selected Answers 91 . total SA ≈ 2(77. such as those that are parallel to the axes on the circle and then find the distance between the points using the Pythagorean Theorem.2 (Day 2) 12-28. total SA = 240 + 100 = 340 ft2 12-29. a: BA ≈ 77. k = 12 b: k = 7 12-34. LA = (8)(4)(16) = 512 m2. LA = 4( 12 )(10)(12) = 240 ft2. this appears to be true for all angles.5!or!1 y 12-30. 12-31. 12-32. a: k 2 = (8)(18) = 144 . a: u = !1 b: x = 5.1.Lesson 12. sin ! = 1 .25) + 512 ≈ 666. a: 2 ! 5 ! 5 = 50 b: 2 ! 4 ! 3 = 24 12-33. and y = sin ! . so x = cos ! . a: Yes. 2 V= 4 3 ! ( 20 ! ) " 1080. This can be shown using the fact that all of the points on the circle are 3 units away from the origin and then finding the distance from the 2 origin to the point (1. rectangle.3 12-40. Typical cross-sections: regular hexagon. the graph only includes the points outside the circle. then r = 20 ! ≈ 6. a: x = 6 b: x = 4 or – 4 c: x = 4 d: x = 30 e: x = 3 or –3 f: x = 3 or –5 12-42. 10! 5!5! = 252 . b: x = –2 or 2 c: Possible answers will satisfy the equation x 2 + y 2 = 9 .8 ft 3 3 12-41.1. 12-45. circle. 12-44. a: The point is not on the circle. B 92 Core Connections Geometry . 12-43. SA ≈ 4! ( 20 ! ) " 509. the graph includes the circle and all of the points inside the circle. Answers vary.Lesson 12. 12-46. b: No. Since 2πr = 40 feet. 5 ) with the Pythagorean Theorem. It is not a combination because order matters. This is an anagram of five Hs and five Ts.4 feet. The circle itself would be dashed.3 square feet. 12 + 5 ! 32 . etc. a: FG = 3. a: C: (0.5 cm3 12-51.1. r = 75 ! 8. base radius = 14 in.8% c: 12 C3 12-56.5 b: C: (0.2% 12 C3 220 5 C3 + 4 C3 + 3 C3 = 10+4+1 220 ! 6. r = 4. 0). y = 120° b: a ! 36. r = 1 d: C: (2. B Selected Answers 93 ..5 cm. V = 820( 12 )3 = 102.8% 12 C3 220 b: 5 C2 !4 C1 = 40 " 18. b = 4 c: z = 12 .4 12-52. 0). r = 19 ! 4.7 c: C: (3. BC = 14 cm b: 16(3) – 4(3) = 36 cm2 12-54. 0).9° . See solution graph at right.2 12-55.Lesson 12. 1). a: x = 33° . a: 4 C3 = 4 = 1.4 12-50. w = 5 d: x = 55° 12-53. V = 13 (196! )(18) = 1176! ≈ 3695 in.3 SA = ! rl = ! (14)( 520 ) " 1003 in. Lesson 12.2.1 12-61. 65°; One method: The base angles of ΔPSR must add up to 40° so that the sum of all three angles is 180°. Then add the 40° and 35° of ∠QPS and ∠QRS, respectively, and the sum of the base angles of ΔPQR must be 115°. Thus, m∠Q must be 180° – 115° = 65°. 12-62. A ! 1, 459, 379.5 square feet 12-63. a: No; the triangle is equilateral, so all angles must be 60°. b: Yes; 5 2 + 12 2 = 132 . 12-64. See graph at right. x-intercepts: (4, 0) and (–4, 0), y-intercepts: (0, –2) and (0, 8) 4 ! 3) = 11 . 12-65. a: This has one solution, because 4( 23 b: This has no real solution because x 2 must be positive or zero. c: This has two solutions because x = ± 6 . d: This has no solution because the absolute value must be positive or zero. 8 12-66. See graph at right. 6 4 12-67. C 2 -10 -5 5 10 -2 -4 -6 -8 94 Core Connections Geometry Lesson 12.2.2 12-71. a: a = 120° , b = 108° , so a is greater. b: Not enough information is given since it is not known if the lines are parallel. c: Third side is approximately 8.9 units, so b is opposite the greater side and must be greater than a. d: a is three more than b, so a must be greater. e: a = 7 tan 23° ! 2.97 and b = 2 ÷ cos 49° ! 3.05 , so b is greater than a. 12-72. x = 8 and y = 12.5 12-73. See graph at right. a: y-intercept: (0, 6); x-intercepts: (3, 0) and (–1, 0) b: (1, 8) c: f (100) = !19, 594 and f (!15) = !504 12-74. 49π sq. units 12-75. 2(5!)(3!) = 1440 12-76. SA = 76 units2; V = 40 units3 12-77. B Selected Answers 95 Lesson 12.2.3 12-85. Side length = 4, so height of triangle is 2 3 . Thus, the y-coordinate of point C could be 2 ± 2 3 ; (5,!! 5.46) or (5,!! "1.46) . 12-86. length of diagonal = 4sin54º ≈ 6.4721, or 32 ! 32 cos108 " 6.4721 (using the Law of Cosines) height of shaded triangle = 6.47212 ! 2 2 " 6.155 units, area of triangle = ! 12 (4)(6.155) ! 12.310 sq. units 12-87. The graph should include a circle with radius 5, center (0, 0), and a line with slope 1 and y-intercept (0, 1). (3, 4) and (–4, –3). 12-88. Jamila’s product does not equal zero. She cannot assume that if the product of two quantities is 8, then one of the quantities must be 8; Correct solution: x = !6!or!3 12-89. 14 12-90. a: 3 C1!10 C3 = 360 " 50.3% , or 4 P1 3 P1 ! 10 P3 = 4 ! 13!12!11!10 3!10!9!8 = 50.3% 13 C4 715 13 P4 b: 11 = 11 ! 1.5% or 4 P4 11 = 24 ! 17160 11 " 1.5% 13 C4 715 13 P4 12-91. C 12-92. Base length ≈ 9.713 units, perimeter ≈ 30.97 units, area ≈ 62.84 square units 12-93. a: m∠PCQ = 46º, tan 46° = 5 CP , CP = CQ ≈ 4.83, sin 46° = 5 CR , CR ≈ 6.95, so x ≈ 6.95 – 4.83 ≈ 2.12 b: x 2 = 7 2 + 7 2 ! 2(7)(7) cos102° , x ! 10.88 12-94. a: It is a prism with dimensions 2 × 3 × 4 units. b: SA = 52 units2; V = 24 units3 c: SA = 52(32 ) = 468 units2; V = (24)(33 ) = 648 units3 12-95. a: y = ! 65 x + 4 b: y = 1 2 x!2 8 6 12-96. See graph at right. 4 2 12-97. A -10 -5 5 10 -2 -4 -6 -8 96 Core Connections Geometry 2. 12-104. 0). r = 15 y 12-106. x = 5 seconds 12-103.282)(7) ! 50. the triangles in (a) and (c) are congruent because corresponding sides have equal length. b: 11 cubic units c: The volume of the new solid must be Front Right Top 5 3 = 125 times the original. units b: Side length ≈ 7. so they are all similar to each other. 12-105.90° triangles.60°.Lesson 12. a: C: (!5. x –4 –3 –2 –1 0 1 2 3 4 x y 6 5 4 3 2 3 4 5 6 12-107. units.282 units. a: See views at right. See answers in table below and graph at right. b: !16(3)2 + 64(3) + 80 = 128 feet. so the increased volume must be 11(5 3 ) = 1375 cubic units. r = 10 b: C: (3. !16( 12 )2 + 64( 12 ) + 80 = 108 feet c: !16x 2 + 64x + 80 = 0 . 12-102. In addition. a: It was 80 feet above ground because y = 80 when x = 0 . They all are 30°. a: 60 sq.4 12-101. so area ≈ (7.97 sq. B Selected Answers 97 .1).