1-4. a: 40 b: 6 c: 7 d: a: 3 b: 5 c: 6 d: 2

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: x2+5x d: x2+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

1-28. a: 55.5 sq. units b: 42 sq. units

1-29. a: 5 b: 12 c: –7

x 3 –1 0 2 –5 –2 1

Lesson 1.1.4

1-32. a: x= ₂₄9 = 3₈=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+72 1-35. a: 6x + 6

b: 6x+6=78, so x=12 and the rectangle is 15 cm by 24 cm.

c:

(

)

(

)

Lesson 1.1.5

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 cm2 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).

1-54. See graph at right. 1-55. a: 27 48 !56.3% b: 13010 =131 !8% c: 0 d: 5₈!56% 1-56. 5x!2+2x+6=67, x = 9, so 5(9) – 2 = 43 miles 1-57. a: x = 3.75 b: x = 3 c: 24 d: x = 3 e: x≈ 372.25 f:x = –3.4

1-58. The flag would need to be a rectangle. The height of the cylinder would match the height of the rectangle along the pole, and the cylinder’s radius would match the width of the rectangle.

Lesson 1.2.2

1-63. Yes; yes; no

1-64. 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. 19+7x+10x+3=52, so x = 2. Side lengths are 19, 10, and 23.

1-66. a: Area ≈ 16 square units b: Area ≈ 15 square units

1-67. a: y = –4 b: y = 25 c: y = –2 y x y=!4x+5 y=!2 3x+3 y=3x!3 4 –4 –4 4

Lesson 1.2.3

1-78. See answers in bold below. y=3x+2

x –3 –2 –1 0 1 2 3 4

y –7 –4 –1 2 5 8 11 14

Lesson 1.2.4

1-85. a: Yes. It has four sides. mAB=mCD= 1₂ and mBC=mAD=!2, so each pair of consecutive sides is perpendicular and forms 90° angles.

b: A´ (4, 3), B´ (6, –1), C´ (–2, –5), D´ (–4, –1)

1-86. a: x= !4.75 b: x= !94 c: x!1.14 d: a=22

1-87. a: There are 10 combinations: a & b, a & c, a & d, a & e, b & c, b & d, b & e, c & d, c & e, d & e

b: Yes. If the outcomes are equally likely, we can use the theoretical probability computation in the Math Notes box in Lesson 1.2.1.

c: ₁₀3 d: ₁₀9

e: The outcomes that satisfy part (c) includes the outcomes that satisfy (b), but there are others on the (c) list as well.

1-88. a: y= 4₃x!2

b: The resulting line coincides with the original line; y= 4₃x!2

c: The image is parallel; y= 4₃ x!7

d: They are parallel, because they all have a slope of 4₃.

1-94. ₅₃₅1 !0.0019; No, this probability is very small.

1-95. a: (9, 3) b: (3, –3) c: (–2, –7) d: (–52, 1483)

1-96. a: 10 square units b: 20 square units c: 208,680 square units

1-97. a: b:

c: d:

1-98. a: The orientation of the hexagon does not change.

b: The orientation of the hexagon does not change.

c: There are 6 lines of symmetry, through opposite vertices and through the midpoints of opposite sides. C m A l Q D P B

Lesson 1.2.6

1-105. (a) and (b) are perpendicular, while (b) and (c) are parallel.

1-106. a: One possibility: 4(5x + 2) = 48

b:x = 2

c: 12!12=144square units

1-107. a: ₅₂4 =₁₃1 b: 13₅₂= ₄1 c: ₅₂1 d: 39₅₂ = 3₄ 1-108. a: It looks the same as the original.

b: Solution should be any value of 45k where k is an integer.

c: circle 1-109. a: b: c: d: l A m D P B Q C

1-112. Carol: only inside circle #2, Bob: outside both circles, Pedro: only inside circle #1. In order to belong to the intersection of both circles, a person would need to have long hair and study a lot for class.

1-113. a: Sandy’s probability = 2₄, while Robert’s is 3₅. Therefore, Robert has a greater chance.

b: Sandy (Sandy’s probability = 1 while Robert’s is 0)

c: Sandy’s probability = ₄3, while Robert’s is 3₅. Therefore, Sandy is more likely to select a shape with two sides that are parallel.

1-114. a: x=!₃₃9 =!₁₁3 b: x=5 and x=!₂3

c: x=1 d: x=12

13

1-115. a: heart b: square

c: hexagon d: Answer vary.

1-116. a: (–6, –3)

b: The vertices are: (6, 2), (2, 3), (5, 6)

c: (8, –3)

Lesson 1.3.2

1-123. Isosceles: (a) and (c); Scalene: (b)

1-124. a: ₁₆7 ≈ 44% b: ₁₆9 ≈ 56%

c: ₁₆1 ≈ 6% d: ₁₆6 = ₈3 ≈ 38%

1-125. a: isosceles triangle b: pentagon c: parallelogram

d: obtuse scalene triangle e: isosceles right triangle f: trapezoid

1-126. Reflection only: A, B, C, D, E, M, T, U, V, W, Y Rotation only: N, S, Z

Intersection only: H, I, O, X

Outside both regions: F, G, J, K, L, P, Q, R

1-127. a: x=!2 b: x=₂3=11₂

Lesson 2.1.1

2-8. a: 33 sq. cm b: 33x sq. units c: 33x2+50x+8 sq. units

2-9. a: 1₂ b: 2₆, parallelogram and square

2-10. a: Isosceles triangle b: Equilateral triangle c: Parallelogram

2-11.

2-12. Answers vary. The left circle could be “equilateral”, and the right could be “quadrilateral”. Assuming this, you could add an equilateral hexagon to the left, a rhombus to the intersection, and a rectangle to the right circle.

Lesson 2.1.2

2-19. a: Vertical angles, equal measure, 3x+5° =5x!57°, x = 31º

b: Straight angle pair, supplementary, 2x+4x+150° =180°, x = 5º

2-20. 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.

b: Since they are equal, m!B= 1₂(124°)=62°.

c: 71° +x=180°, x=109°

2-21. a: Square b: (– 4, 5), (1, 5), (– 4, 0), (1, 0)

2-22. y=x!1; No, because 1!3"1.

2-23. a: Vertical; they have equal measure. b: They form a “Z.”

2-31. a: (–2, 3) b: (–2, 3); yes

2-32. a: 20 square units

b: 2,600 square units; subtract the x- and y-coordinates to find the length of the two sides.

2-33. a: We do not know the angles measures are equal, because we do not know if BD! "## $EG! "## .

b: The diagram does not have parallel line marks.

2-34. a: x=17.5 (corresponding angles)

b: x=5 (multiple relationships possible)

2-35. a: 12 boys b: 22 girls

c: 2₃ d: 7 boys left, 23 students, so ₂₃7 .

2-36. a: an isosceles triangle

b: a rectangle

Lesson 2.1.4

2-41. a: b:

2-42. The slopes are 1₂ and !₂3. Since the slopes are not opposite reciprocals, the lines must not be perpendicular.

2-43. (3, –1), (7, –1)

2-44. They used different units.

2-45. The lines are parallel, so they do not intersect. Therefore, there is no solution.

30° 110° 110° 30° 70° 70° 70° 70° 80° 80° 80° 80° 100° 100° 85° 95° 75° 85° 95° 85° 75° 75° 75° 105° 105° 105° 105° 95° 95° 85°

Lesson 2.1.5

2-55. x=7º

2-56. a: x=10 units b: x=6 c: x=20! d: x=10!

2-57. a: x=4 and y=18 b: x=!13 and y=6

2-58. a: It should be a triangle with horizontal base of length 4 and vertical base of length 3.

b: ! 4₃

c: Any equation of the form y=!₄3x+b.

2-59. 2

Lesson 2.2.1

2-65. The acute and isosceles triangles.

2-66. Reasoning will vary. a=118°, b = 118°, c = 32°, d=32°

2-67. a: 15° b: x=12°, m!D=4(12°)+2° =50° c: It is equilateral.

2-68. a: A!("6,"3), B!("2,"1), and C!("5,"7) b: B!!(8, 13)

c: A!!!(3, "6)

2-69. a: Yes, because the slopes are opposite reciprocals. b: y= 1₂ x+5

2-74. a: 8x2!26x!7 b: 10x2+31x!14 c: 4x2!47x+33 _d: !6x2+17x!5 2-75. area = 28 square feet

2-76. a: x=8°, right angle is 90° b: x=20°, straight angle is 180°

c: x=20°, sum of angles is 180° d: x=60°, sum of angles is 180°

2-77. Daniel is correct because the definition of a rectangle is a quadrilateral with four right angles. Since a square has four sides and four right angles, it must be a rectangle.

2-78. a: 22₃₅ =62.9%

b: 22₃₅ = 42_? ; she needs to attempt about 67 pancakes.

c: She should add three banana pancakes to make the probability of banana ₃₀3 .

Lesson 2.2.3

2-85. No, it would take 10 months for Sarita to catch up to Berti.

2-86. The unshaded triangle is half the area of the rectangle (0.5(8)(17) = 68 sq. in.), so the shaded area is the other half.

2-87. a: Because when you are not standing up straight, you have changed your height, and you will not get a true measure of your height.

b: Diagram (1) is correct.

c: No, you measure from the very bottom to the very top.

2-88. a: _{If it rains, then Mr. Spelling is unhappy.}

b: If you add two even numbers together, then the result is even.

c: If it is Tuesday, then Marla has a piano lesson.

2-89. a: 8x+13 b: 2x+3 c: 3x2!5x!12 _{c: 13x}2_+30x

Lesson 2.2.4

2-94. a: ₇2 _{= 49 sq cm b: 0.5(10)4 = 20 sq. in. c: 0.5(16 + 8)6 = 72 sq ft}

2-95. a: 15x2+21x b: x2+5x+6 c: 3x2!x!10 _{c: 10x}2_!3x!4 2-96. See graph at right. (–3, 0) and (0, –3)

2-97. a: Isosceles Trapezoid because two sides are parallel and the other two sides are the same length.

b: A!(7,"2), B!(8,"4), C!(2,"4), D!(3,"2) c: 10 square units 2-98. a: 12 52 =133 b: 2052 =135 c: 522 = 261 d: 0

Lesson 2.3.1

c: The slopes indicate that the lines are perpendicular.

d: The solution should be (–2, 4).

2-106. a: Right triangle; slopes are opposite reciprocals.

b: 20 square units

c: ≈ 23.4 units

2-107. See diagram at right.

2-108. See graph at right.

a: It is a trapezoid because it has exactly one pair of parallel sides.

b:A!("2,!"1),!B!("5,!0),C!("5,!2), D!("2,!6) c: A!!(1,!2) and C!!("2,!5) 1 _{+ =} 74° 74° 106° 106° 74° 74° 106° 106° x y x y

2-113. a: (–2, 5) b: (1, 5) c: (–12, 14) d: (2, 2)

2-114. a: 7₈ b: ₈3 c: 5₈

2-115. Height = 12 feet; Using the Pythagorean Theorem, area = 1

2(12)(12+23)=210 sq. feet 2-116. a: x=28.5°, Triangle Angle Sum Theorem

b: x=23°, relationships used varies

c: x=68°, corresponding angles are congruent because the lines are parallel and base angles of an isosceles triangle are congruent.

2-117. 5" and 21"

Lesson 3.1.1

3-5. a: D is not similar. AB = 5, BC = 4, AC = 3

b: A!B!= 100=10units, B!C!=8 units, and A!C!=6 units.

c: A = 24 sq. units; P = 24 units

3-6. a: x=18 b: x=3 c: x=6 d: x=2 3-7. a: ≈ 30°, ≈ 40°, ≈ 110° b: Obtuse scalene triangle

3-8. a: 4₅, y= 4₅ x+9₅ b: MU = 41!6.40units

c: One is a ratio (slope) while the other is a length (distance).

3-9. a: triangle inequality

b: Pythagorean Theorem

c: base angles not equal

3-10. a: If a shape is an equilateral triangle, then it has 120° rotation symmetry.

b: If a shape is a rectangle, then the shape is a parallelogram.

c: If a shape is a trapezoid, then the area of the shape is half the sum of its bases multiplied by its height.

Lesson 3.1.2

3-18. Result should be 12 units tall and 16 units wide.

3-19. a: The 15 corresponds to the 6, while the 20 corresponds to the 8. Multiple equivalent ratios are possible. One possibility: 15₆ = 20₈ =2.5

b: 25 and 10; ₁₀25=2.5; yes

3-20. Yes they are parallel because they have the same slope: !₅3.

3-21. a: 6x2!8x b: 2x2+x!15 c: 4x2!25 d: 2x3!5x2!3x 3-22. x=10°, y=61°

3-23. No, this is not convincing. While the facts are each correct, the conclusion is not based on the facts. As stated in Fact #2, a square is a rectangle because it has four right angles. However, a rhombus does not have to have four right angles, so therefore there is not enough evidence that a rhombus is a rectangle.

Lesson 3.1.3

3-29. a: Zoom factor: 0.5; The sides are only half as long, so the side corresponding to the 16 must become 8, and the side corresponding to the 11 must become 5.5.

b: It is 1:1 because it is congruent.

3-30. P(original) = 18 units and P(new) = 36 units; A(original) = 18 sq. units and A(new) = 72 sq. units. The enlarged perimeter is 2 times greater. The enlarged area is not 2 times greater. Notice that the enlarged area is 4 times greater.

3-31. a: x= 42

5 =8.4 b: m=22 c: t=12.5 d: x=23=1.5

3-32. a: y=3!3₅x

b: A=7.5 sq. units; P=8+ 34 !13.8 c: y=3+5₃x

3-33. a: If the lines have the same slope, then they are parallel.

b: If a line is vertical, then the slope is undefined.

c: If lines have slopes 2

3 and !23, then they are perpendicular. 3-31. a: alt. int. angles b: vertical angles

3-41. a: f =9 b: g=18 c: h= 70₃

3-42. a: 180°!38°!63° =79° and 180°!38°!79° =63°; corresponding angles are equal.

b: Upon inspection, all unmarked angles are the same since the difference with 180° will be the same.

3-43. a: Frank:0.25x+1.95=y; Alice:0.40x+1.5=y b: They will be 3 years old.

3-44. a: If a rectangle has base x and height 2x, then the area is 2x2.

b: If a rectangle has base x and height 3y, then the perimeter is 2x+6y.

c: If a rectangle has base of 2 feet and a height of 3 feet, then the area is 864 sq. inches.

3-45. In theory, 3 < x < 13 but some of these lengths are not practical.

3-46. a: The coordinates of the image are A(–6, –4), B(10, –4), C(10, 6), D(2, 12), E(–6, 6)

Lesson 3.2.1

3-53. a: Yes, since all trees are green and the oak is a tree.

b: No, only trees must be green according to the statement.

c: No, the second statement reverses the first.

3-54. a: Yes, AA ~. Dilate from right vertex.

b: Yes, AA ~ since all angles are 60°.

c: Yes, zoom factor of 2.5; translate so that one pair of corresponding vertices coincide, rotate so that rays coincide, and dilate.

d: No, since corresponding angles are not equal. Note that you cannot apply zoom factor to angles.

3-55. a: One strategy: Translate one so that the centers coincide. Then dilate so that the radius is the same as the other circle.

b: Equilateral triangles, which from part (b) of 3-54 were similar because they have equal angle measures. Squares or other regular polygons are also always similar.

3-56. a: There are 12 combinations. One way to systematically list them all is to list a bus number (such as 41) and then match it with each possible activity. This can be repeated for each of the possible bus numbers.

b: i: 129 , ii: 128 , iii: 121

3-57. See graph at right. Perimeter = 44.9 units; Area = 94 sq. units

3-58. a: ABCD~EVOL b: RIGHT ~RONGW

c: One possible answer: ΔTAC ~ ΔGDO

x y

3-65. a:x = 20 mm b: w = 91 mm

3-66. a: Impossible: can be rejected using Triangle Inequality or Pythagorean Theorem.

b: Possible

c: Impossible: rejected because the sum of the angles is 179°.

3-67. a: 128 b: 48

3-68. This reasoning is incorrect. Rewrite “it is raining” in the lower left oval, and “Andrea’s flowers must be closed up” in the right oval.

3-69. a: Reflection, rotation, and translation

b: Rotation and translation

c: Rotation, dilated by factor of 2, and translation.

3-70. a: Possible

b: Not possible because the sum of the measures of an obtuse and right angle is more than 180°.

c: Not possible because a triangle with sides of equal length obviously cannot have sides of different lengths.

Lesson 3.2.3

3-76. a: (5, –2) b: (–4, 2)

c: (3, 3); It is the center of the figure, or the midpoint of each diagonal.

3-77. a: y= 1₂ x+2

b: A=4 sq. units; P=6+ 20 !10.47 units

c: y=!2x+2

3-78. a: x=51° alternate interior angles and Triangle Sum Theorem

b: x=43° circle has 360º

c: x=1 Pythagorean Theorem

3-79. a: See tree diagram at right.

b: Yes c: 1₆, 3₆

d: 1₂, no—the spinners are independent

e: 2₆, because now the possible outcomes are $100, $200, $1500, $200, $400, and $3000.

3-80. a: n=32 b: m!14.91

3-81. Missing side length of first rectangle must be 4 m because the perimeter is 26 m. Missing side length of second rectangle must be 9" because the area is 36 sq. in. Since angles are equal and ratios of corresponding side lengths are equal, therefore, the rectangles are similar. In fact, they are congruent because r = 1.

$100 $300 $1500 double keep double keep double keep $200 $100 $600 $300 $3000 $1500

3-88. a: Scalene triangle b: Isosceles triangle

c: Not possible d: Equilateral triangle

3-89. 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: When solving a system of equations that has no solution, the equations combine to create an impossible equality, such as 3 = 0. Another special case occurs when the resulting equality is always true, such as 2 = 2. This is the result when the two lines coincide, creating infinite points of intersection.

3-90. a: Not similar, interior angles are different.

b: Must be similar by AA ~.

c: Similar, all side lengths have the same ratio.

3-91. Perimeter = 10+10+4+3+4+3+4=38 units, height of triangle = 8 units, area = 60 square units.

3-92. This reasoning is correct.

3-93. a: 3(4x!12)=180°, x=18 b: 4.92!3.12=x2, x!3.79

c: x+(180°!51°!103°)+82° =180°, x=72! d: 3x!2=2x+9, x=11

Lesson 3.2.5

3-99. 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.

3-100. a: 24₄₀ =60% b: 18_x =₁₀3 ,x=60

3-101. a: 12x2!7x!10 b: 16x2!8x+1

b: x=!5₉ c:x = 3

3-102. !y=48!because of vertical angles; !z=48! because of reflection of !y or because of angle of incidence = angle of reflection with !x.

3-103. a: ! 5₆

b: LD = 61!7.81units

c: Calculate ∆x and ∆y by determining the difference in the corresponding coordinates.

3-104. Original: A = 135 sq. units, P = 48 units; New: A = 15 sq. units, P = 16 units

Lesson 3.2.4

3-108. x=137°,!y=76°

3-109. h = 5 feet; perimeter ≈ 24.2 feet

3-110. a: 28_? = 2₅; There are 70 animals in the bin.

b: _{22+8+13+15+17}13+17 = 3075=40%

c: 3_? =5%; You need a total of 60 animals in the bin.

3-111. a: y=! 1

2x+4 b: y=2x!1

c: y= 2₅x+7₅ d: C=15+7(t!1)=8+7t

3-112. ≈ 13.2 miles

3-113. Possible response: Rotate WXYZ clockwise, translate it to the left, and dilate it by a factor of 0.4. y=7.5,!z=9.6

4-6. a: x=11° b: x=45° c: x=30° d: x=68° 4-7. a: See flowchart at right.

b: Yes, because the triangles are similar (AA ~) and the ratio of the corresponding side lengths is 1 (because AC = DF).

4-8. a: Yes, she used the Pythagorean Theorem.

b: (x+1)2=x2+2x+1 c: x=24 d: 56 units 4-9. x=9,!y=4,!z=6 2₃ 4-10. a: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12. b: Yes. c: P(even)= 18₃₆; P(10)= ₃₆3 ; P(15)=0 d: The sum of 7. P(7)= 6 36 =16

4-11. If h represents the number of hours and t represents the temperature, then t=77+3h and

t=92!2h; h=3 hours and the temperature will be 86°F.

Lesson 4.1.2

4-17. a: ! =11°, x

95! 15, x!18.46 b: a=b=45° c: y

70! 52,y!175 4-18. a:side ratio = 4:1 b: perimeter ratio is 4:1 c: _28'

4-19. a:Yes, AA ~.

b: No, side ratios not equal 12 64 !1898.

c: _{Cannot tell, not enough angle values given.}

4-20. 62!32 = 27_, 92!32 = 72_{. So perimeter is} 27+ 72+15!28.68 _{cm. The}

area is ( 27+ 72 )(3)÷2!20.52 _{sq. cm.}

4-21. Since the slope ratio for 11° ≈ 0.2, AB ≈ 50 feet. The slope ratio for 68° ≈ 2.5, so BC ≈ 4 feet. Thus, AB is actually longer.

4-22. a: 12 b: Yes c: ₁₂6 = 1₂; ₁₂8 = 2₃

AA ~ m∠A = m∠D m∠B = m∠E

Lesson 4.1.3

4-27. They both could be. It depends on which angle is used as the slope angle.

4-28. a: Yes, since the slope ratio is greater than 1, the angle must be greater than 45°.

b: Isiah is correct. Since the angle is less than 45°, the slope ratio must be less than 1.

c: Since the angle is greater than 45°, x must be less than 9.

4-29. a: A; a_n=1+3(n!1)=3n!2 b: neither

c: G; a_n=2!2n"1₌₂n

d: A; a_n=5+7(n!1)=7n!2

4-30. Answers vary, possible solution: square, equilateral triangle, and equilateral hexagon.

4-31. a: 2₅

b: Yes. If the first song is a country song, then there is only 1 country song left to play out of 4 songs. Therefore, the chance that the second song is a country song is 1₄.

c: 1₂, because only 2 songs are left and only one is sung by Sapphire.

d: 1₃; same

4-39. a:t = 11.524 b: p!3.215 c: b!148.505 4-40. a:x2 + 182 = 302 ; x = 24 b: 2x+20° +3x+20° +x+2x=360°, x=40° c: ₁₂5 = 3_x, x= 36₅ =7.2

4-41. They are congruent. Possible response: Reflect ΔADS across a vertical line, then translate it.

4-42. 24 possible ways: ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, BACD, BADC, BCAD, BCDA, BDAC, BDCA, CABD, CADB, CBAD, CBDA, CDAB, CDBA, DABC, DACB, DBAC, DBCA, DCAB, DCBA

4-43. Her father’s eyes were ≈ 69.126 inches high.

4-44. a: A = 144 cm2 , P = 52 cm b: A = 696.67 m2 , P = 114.67 m c: A = 72 sq cm, P = 48 cm d: A = 130 sq. feet, P = 58 feet

Lesson 4.1.4

4-50. b:≈ 29.44 feet

4-51. a: G; a_n= 1₂!1₂n"1=1₂n b: A; a_n=!7.5!2(n!1)=5.5!2n

Lesson 4.2.1

4-58. a: 10₂₀= 1₂ b: ₁₉9

c: No, they are not independent. The probability the second contestant is a girl depends on whether the first contestant was a girl or not.

4-59. See graph at right.

a: (! 1₂,!0) and (3,!0) b: x=!1₂ or x=3; Yes.

4-60. a: Slope = 1₂

b: It must be parallel to or coincide with the line graphed at right.

4-61. Francis: y = x + 2, John: y= ₄3x+5; 12 seconds

4-62. a:x ≈ 2.344 b: x≈ 0.667 c:x = 1.5 or –5 d: No solution

4-63. leg ≈ 29.44 units, hypotenuse ≈ 30.78 cm, so the perimeter ≈ 69.22 cm

Lesson 4.2.2

4-69. a: A tree diagram; a third dimension would be needed to represent the three coins with an area model.

b: See tree diagram at right; 8

c:i: 1₈ , ii: ₈3, iii: 7₈, iv: 3₈

d: They are both the same probability of 50%.

e: The sample space remains the same; i: 64

125 , ii: 1254 +1254 +1254 =12512 , iii: 12561 , iv: 12512

4-70. Yes, they are similar due to AA ~ because m∠B = m∠E and m∠C = m∠C (triangles share an angle).

4-71. 1₆; If the die is “fair,” each roll of the die is an independent event.

4-72. a: It implies that because Brian is always late on Tuesday, then today must be Tuesday.

b: The “Brian is always late on Tuesdays” and “Today is Tuesday” ovals should be next to each other, both with arrows pointing to “Brian will be late today.”

4-73. a: 3, 15, 75, 375 b: 10, –50, 250, –1250 4-74. x!10.39, y=12 H H H H T T T T T T T H H H

4-81. Both equal 3₈.

4-82. y=1₃x+9

4-83. a: See diagram at right.

b: Ratio for tan 11° ≈ 1₅, so 170_x ! 1₅, and x!850 _feet.

Alternatively, a calculator could be used and x= _tan(11170_°)!875 feet.

4-84. a: x=49 b: x=2 _c: x=16₃ d: x=!5 or 1

4-85. No. Triangle Inequality property prevents this because 7+10<20 and 20!10>7.

4-86. $450

Lesson 4.2.4

4-95. a: ₃₆3 b: ₃₆4 c: 24₃₆

4-96. a: 5 ways b: 6 ways c: 11 d: ₁₁5

4-97. a: x=13, Pythagorean Theorem

b: x=80°, Alternate interior angles and the Triangle Angle Sum

4-98. (x+2)(x+5)=40, x2+7x!30=0, so x=!10 or 3. Since x cannot be negative, x = 3. Therefore, the dimensions of the rectangle are 5 and 8 units.

4-99. a: Less than 45° b: Equal to 45° c: More than 45° 4-100. The slope is !₁₀7 . Points will vary. y=!0.7x+82.5

A few possible solutions: (5, 79), (15, 72), (25, 65), etc.

170 ft 11º

Lesson 4.2.5

4-110. a: ₁₂1 b: Intersection

c: No, P(yellow)=1₆ d: 2₃

e: You cannot move 1₆+1₆+1₃= 2₃ or you can move 1₃ of the time and 1!1₃= 2₃.

4-111. a: y=3 b: y=9

4-112. It assumes that everyone who likes bananas is a monkey.

4-113. ≈ 1469.27 feet

4-114. 6 "<x<14 "

4-115. Methods vary: θ = 68º (could be found using corresponding and supplementary angles),

α = 85º (could be found using corresponding angles since lines are parallel.

4-116. a: P(K)=₅₂4 , P(Q)=₅₂4 , P(C)=13₅₂

b: 16₅₂; You can add the probabilities of king and club, but you need to subtract the number of cards that are both kings and clubs (1). P(K or C)= 4

52+1352!521 =1652

c: P(K or Q) = ₅₂8 =₁₃2 . There is no overlap in the events so you can just add the probabilities.

d: P(not a face card) = 1!12 52= 4052 4-117. ≈ 26 years

4-118. a: Yes, ΔABD ~ ΔEBC by AA ~.

b: Yes. Since DB = 9 units (by the Pythagorean Thm), the common ratio is 1.

4-119. LE = MS and LI = ES = MI

4-120. AB!11.47mm, A!97.47 sq. mm

4-121. a:A′(–3, –3), B′(9, –3), C′(–3, –6)

b: A″(–3, 3), B″(–3, –9), C″(–6, 3)

5-7. x≈ 7.50 and y≈ 8.04 units; Use either sine or cosine to get the first leg, then any one of the trig ratios or the Pythagorean Theorem to get the other.

5-8. a: False (a rhombus and square are counterexamples)

b: True

c: False (it does not mention that the lines must be parallel)

5-9. B

5-10. a: (4 cards less than 5)₅₂ !(4 suits)=16₅₂. If Aces are not included, 12₅₂.

b: 1!16₅₂ = 36₅₂. If Aces are included, 40₅₂.

c: P(red) + P(face) – P(red and face) = 26₅₂+12₅₂!₅₂6 = ₅₂32 5-11. area = 74 sq ft, perimeter = 47.66 ft

5-12. a: x = –3 b: m=10 c: p=!4or 2₃ d: x=23

Lesson 5.1.2

5-17. a: sin 22° =₁₇x b: x!6.37, tan 49° = 7_x, 6.09 c: cos 60° = x₆, x=3 5-18. ≈ 26.92 feet

5-19. a: G; a_n=100(₁₀1)n!1=103!n b: A; a_n=0!50(n!1)=50!50n

5-20. Region A is 1₄ of the circle. Since the spinners are independent, the probability of A and A is 1₄!1₄ =₁₆1 . In 80 games, we expect A and A to occur ₁₆1 (80)=5 times.

5-21. a: False (a 30°- 60°- 90° triangle is a counterexample)

b: False (this is only true for rectangles and parallelograms)

c: True

5-22. a: 6x2!x!2 b: 6x3!x2!12x!5 c: !3xy+3y2+8x!8y d: x2!9y2

Lesson 5.1.3

5-29. a: x= ±5 b: All numbers c: x=2 _{d: No solution} 5-30. Using cosA=₁₃5 , sinA=12₁₃, or tanA=12₅ ,A!67.4º

5-31.

!

5-32. a: ₃₆1 b: 20₃₆

5-33. Area ≈ 294.55 sq m, perimeter ≈ 78.21 m

5-34. a: It uses circular logic.

b: Reverse the arrow between“Marcy likes chocolate” and “Marcy likes Whizzbangs.” Also, remove the arrow connecting “Marcy likes chocolate” and “Whizzbangs are 100% chocolate.”

5-35. 9.38 minutes

Lesson 5.1.4

5-41. All of the triangles are similar. They are all equilateral triangles.

5-42. Since tan(33.7°)!2

3, y!23x+7.

5-43. a: a₁=108, a_n+1=a_n+12 b: a₁= 2₅,a_n+1=2!a_n

c: a_n =3741!39(n!1)=3780!39n d: a_n =117(0.2)n!1=585(0.2)n 5-44. a: sin! =b_a b: tan! =a_b c: cos! = a_b

5-45. a: 22₅₂; union b: ₅₂3 ; intersection c: 1!22₅₂ = 30₅₂ 5-46. a: cos!23° =18_x or 0.921=18_x

5-52. a:A = 1 sq. m, P=2+2 2 m

b: A= 25 3₂ !21.65 square ft, P=15+5 3!23.66 _ft 5-53. a: y=111º ;!x=53º b: y=79º ;!x=47º

c: y=83º ;!x=53º d: y=3;!x=3 2 units

5-54. a: 4 2 units; Use the Pythagorean Theorem or that it is a 45°- 45°- 90° triangle.

b: It is a trapezoid; 24 square units

5-55. 10.1% by using the Addition Rule.

5-56. a: Answers vary; sample responses: x < 3, x is even, etc.

b: The length of each leg is 6 units.

5-57. a: a_n =500+1500(n!1)=1500n!1000 _b: a_n =30(5)n!1=6(5)n

Lesson 5.2.2

5-64. a: 16 inches b: 4 yards and 4 2 yards

c: 24 feet d: 10 meters and 10 3 meters

5-65. a:m∠A = 35º, m∠B = 35º, m∠ACB = 110º, m∠D = 35º, m∠E = 35º, m∠DCE = 110º

b: Answers vary. Once all the angles have been found, state that two pairs of

corresponding angles have equal measure, such as m∠A = m∠D and m∠B = m∠E to reach the conclusion that ΔABC ~ ΔDEC by AA ~ or AC = BC, DC = EC, and

m∠ACB = m∠DCE by SAS ~.

c: They are both correct. Since both triangles are isosceles, we cannot tell if one is the reflection or the rotation of the other (after dilation).

5-66. cos!52° = b_c , tan!52° =_ba, cos!38° = a_c, sin!38°

5-67. 14₂₇ = ₄₀x , x!20.74 inches

5-68. a: explicit b: a_n =!3+4(n!1)=4n!7

c: a₅₀=193 d: a_n =3!1₃(n!1)=31₃!1₃n

5-69. $3(135₃₆₀)+$5(135₃₆₀)+(!$6)(₃₆₀90)=$1.50. It is not fair because the expected value is not 0.

5-70. a: 1₂ b: 0 c: ₄3 d: 1

Lesson 5.3.1

5-77. ≈ 61°

5-78. a: Impossible because a leg is longer than the hypotenuse.

b: Impossible because the sum of the angles is more than 180°.

5-79. William is correct.

5-80. a: A!(–3, –6), B!(–5, –4), C!(0, –4) b: A!!(3, 3), B!!(1, 1), C!!(1, 6)

5-81. a: x=16₅ b: No solution c:x = –11 or 3 d: x = 288

5-89. They must have equal length. Since a side opposite a larger angle must be longer than a side opposite a smaller angle, sides opposite equal angles must be the same length.

5-90. ≈ 10.6 mm

5-93. 72 square units

5-92. 12.6

5-93. See tree diagram at right (an area model is not practical). P(three yogurts) = 12.5%.

100%!12.5%=87.5%

chance of not getting three yogurts. 5-94. a: x= 45₄ =11.25 b: x = !10 or x = 10 c: x=1.3 d: No solution 5-95. (!2,!4) yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple yogurt green apple red apple

Lesson 5.3.3

Lesson 5.3.4

5-111. a: The diagram should be a triangle with sides marked 116 ft. and 224 ft. and the angle between them marked 58°.

b:≈ 190 feet, Law of Cosines

5-112. a: Corresponding angles have equal measure.

b: The ratio of corresponding sides is constant, so corresponding sides are proportional.

5-113. y=(tan 25°)x+4 or y!0.466x+4

5-114. It must be longer than 5 and shorter than 23 units.

5-115. 31 terms

5-116. ₁₂3 (3)+₁₂7 (!1)+₁₂2 (10)=11₆ "$1.83

The game is not fair because the expected value is not zero.

5-126. The third side is 12.2 units long. The angle opposite the side of length 10 is

approximately 35.45°, while the angle opposite the side of length 17 is approximately 99.55°.

5-127. x!11.3 _{units; Methods include using the Pythagorean Theorem to set up the equation} x2+x2=162, using the 45°- 45°- 90° triangle shortcut to divide 16 by 2, or to use sine or cosine to solve using a trigonometric ratio.

5-128. No, because to be a rectangle, the parallelogram needs to have 4 right angles. Counterexample: A parallelogram without 4 right angles.

5-129. a: P ≈ 40.32 mm, A = 72 sq mm b: P = 30 feet, A = 36 square feet

5-130. A(2, 4), B(6, 2), C(4, 5)

5-131. The expected value per throw is 1₄(2)+1₄(3)+1₂(5)=15₄ =3.75, so her expected

winnings over 3 games are 3(3.75) = 11.25; yes, she should win enough tickets to get the panda bear.

5-132. y= ₄3x+4

5-133. a:m∠ABE = 80º, m∠EBC = 60º, m∠BCE = 40º, m∠ECD = 80º, m∠DEC = 40º,

m∠CEB = 80º, m∠BEA = 60º

b: 360°

5-134. a: ≈ 8.64 cm

b: PS=SR=5.27cm, so the perimeter is ≈ 25.5 cm

5-135. Area ≈ 21.86 sq. units, perimeter ≈ 24.59 units

5-136. a: Explicit t(n)=!2+3n;Recursive t(0)=!2,!t(n+1)=t(n)+3 b: Explicit t(n)=6(1₂)n; Recursive t(0)=6,!t(n+1)= 1₂t(n)

c: t(n)=24!7n

d: t(n)=5(1.2)n e: t(4)=1620

5-137. a: See diagram at right. b: x=10 3₃ !5.77

5-138. a: 5+ 20+ 37!15.55 units b: !31.11 c: (–2, 0)

Chain

Shed Bush

Lesson 6.1.1

6-4. a: Alternate interior angles.

b: Vertical angles.

c:∠u & ∠z, ∠s & ∠x, ∠v & ∠w, and ∠t & ∠y

6-5. a: They are similar by SAS ~.

b: Yes, because they are similar and the corresponding sides have a ratio of 1.

6-6. 3x+1° +52° =180°, x=127₃ !42.33°

6-7. a: 8 cm b:≈ 14.97 ft c:≈ 15.2 in.

6-8. 1a and 1b: statements ii and iv, 2: The cupcakes are burned, 3: The fans will not buy the cupcakes because they are burned, 4: The team will not have enough money for

uniforms.

6-9. a: ₄3 or 75% b: ₂₀3 or 15%

c: 1 or 100% c: (b) is an intersection, and (c) is a union.

6-10. A

Lesson 6.1.2

6-14. a = 97º, b = 15º, c = 68º, d = 68º

6-15. a:≈ 3.75, tangent

b: 7 2!9.9_{, Pythagorean Theorem or 45°- 45° - 90° ratios} c:≈ 9.54, Law of Cosines

6-16. a: 25 units b: 56 sq. units and 350 sq. units

6-17. a:A′(–2, –7), B′(–5, –8), C′(–3, –1)

b:A″(2, 7), B″(5, 8), C″(3, 1)

c: Reflection across the y-axis.

6-23. a: Not similar because there are not three pairs of corresponding angles that are congruent.

b: Similar (AA ~)

6-24. a: y= 5₂ x!8 b: y= ₂3x+1 6-25. b: 1422= DE10 , DE≈ 15.71

6-26. a: Yes because of AAS ≅ or ASA ≅; ΔDEF ≅ΔLJK.

b: One possible answer, a reflection across line segment JK and then a translation of

ΔDEF to line up point J and point E, followed by a rotation.

c:KL ≈ 4.3 units

6-27. c = 10 by substitution.

6-28. a: P(A or B) = P(A) + P(B) – P(A and B) = ₂₁₂64 +112₂₁₂!0= 176₂₁₂ "83.0%; the probability of A and B (the overlap) was 0.

Lesson 6.1.4

6-34. Reasoning can vary. See sample responses below.

a:a = 123°, when lines are //, corr. ∠s are =,b = 123°, when lines are //, alt. int. ∠s are =, c = 57°, suppl. ∠s

b: all = 98°, suppl. ∠s, then when lines are //, alt. int. ∠s = and corres. or vert. ∠s =

c:g = h = 75°, when lines are //, alt. int. or corres. ∠s =, then vert. ∠s =

6-35. a: Similar (SSS ~) b: Similar (AA ~)

6-36. a: x=!4 and y=0 b: No solution; the lines are parallel.

6-37. ₁₀4 = _x+5₅, x = 7.5

6-38. Let B represent the measure of angle B. Then (3B + 5º) + B + (B – 20º) = 180º, so

m∠A = 122º, m∠B = 39º, and m∠C = 19º.

6-39. a: See possible area model at right.

b: 1₄ c: 1₉+1₆+1₆+ 1₄= 25₃₆ !69% 6-40. C parents 1 3 niece 1 6 boyfriend 1 2 boyfriend 1 2 niece 1 6 parents 1 3 1_{9 18}1 1₆ 1 18 1 36 1 12 1 12 14 1 6

6-46. Justifications and order may vary: a=53°, given; b=55°, straight angle (with ∠g); c=72°, triangle angle sum; d=53°, when lines are parallel, alternate interior angles are equal; e=55°, when lines are parallel, alternate interior angles are equal; f =127°, straight angle (with ∠a), so they are supplementary.

6-47. a: For left-hand triangle: c2_{=9+36!2 · 3 ·6 cos 60°, c=3 3!5.196 units;}

For right-hand triangle: c2 =36+27!2 ·6 · 3 3 cos 30°, c=3 units; They are congruent.

b: Yes; by SSS ≅ or SAS ≅.

6-48. a: Converse: If the ground is wet, then it is raining. Not always true.

b: Converse: If a polygon is a rectangle, then it is a square. Not always true.

c: Converse: If a polygon has four 90° angles, then it is a rectangle. Not always true.

d: Converse: If a polygon is a triangle, then it has three angles. Always true.

e: Converse: If vertical angles are congruent, then two lines intersect. Always true.

6-49. x-intercept: (4, 0), y-intercept: (0, 6)

6-50. a: y=13₄ b: y = –2 c: 42₃" d: x=8₃ 6-51. a: 3

8 b: 18

c: ₈3 d: 1₈; sum must be equal to one.

6-52. sin 40° = h

Lesson 6.2.1

6-55. a: x!45.56 _{b: x!10.63 c: x!266.49 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)

6-62. 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 lines 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 sin 56₅ ° !sin 72₈ °, 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 1₆ , then a 6-sided dice was rolled. Not always true.

c: Converse: If a triangle is a right triangle, then it has a 90° angle. Always true.

6-65. 19₄ 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)

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: ₂₀x = x₂₄+2, x = 10

6-74. a: x=4 b: x=55°

c: x=23° and y=43° d: x=5.5 andy=45.2 6-75. P(A or B) = P(A) + P(B) – P(A and B) = 4%+1

2%!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: ₁₆x =₁₃8 , x!9.8, corresponding parts of similar figures have equivalent ratios

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.

6-94. a: 5x + 3 = 4x + 9 because if lines are parallel, then alternate interior angles are equal,

x = 6º.

b:q = t because if lines are parallel, then corresponding angles are equal; c + t = 180º because if lines are parallel, then same side interior angles are supplementary; 66°

c: 180° – 88° = 92°; g + q = 180° because when lines are parallel, same-side interior angles are supplementary.

6-95. a: y= 6₅ x!3 b: y=! 1₄ x+4.5 c: y= 1

3x d: y = 2

6-96. a: x!8.1 b: Not enough information. c: x!10.67 6-97. a: x=15°, Triangle Angle Sum b: k=5, Isosceles triangle

c: t=9° and w=131°, parallel lines d: x!49.94°, Triangle Angle Sum

6-98. a: $1.50 b: $12

6-99. B

Lesson 7.1.1

7-6. a: They are congruent by ASA ≅ or AAS ≅.

b:AC ≈ 9.4 units and DF = units

7-7. Relationships used will vary, but may include alternate interior angles, Triangle Angle Sum Theorem, etc.; a = 26°, b = 65°, c = 26°, d = 117°

7-8. width = 60 mm, area = 660 mm2 7-9. A quadrilateral. 7-10. a:x = 18, y = 9 3 b: x = 24 2 , y = 24 c:x = 8 3 = 8 3 3 , y = 16₃ = 16 3 3

Lesson 7.1.2

7-15. Using the Pythagorean Theorem, AB = 8 and JH = 5. Then, since ₆3= 4₈ =₁₀5 ,

ΔABC ~ ΔHGJ because of SSS ~.

7-16. 7 cm

7-17. Line L: y=!1₆ x+6; line M: y= 2₃x+1; point of intersection: (6, 5)

7-18. a: 3m=5m!28, m = 14° b: 3x+38° +7x!8° =180°, x = 15°

c: 2(n+4)=3n!1, n = 9 units d: 2(3x+12)=11x!1, x = 5 units

7-19. Rotating about the midpoint of a base forms a hexagon (one convex and one

non-convex). Rotating the trapezoid about the midpoint of either of the non-base sides forms a parallelogram.

7-20. a: 10 units b: (–1, 4)

c: 5 units, it must be half of AB because C is the midpoint of AB.

Lesson 7.1.3 (Day 1)

7-28. a: The 90° angle is reflected, so m∠XYZ =90º. Then m∠YZY′=180º.

b: They must be congruent because rigid transformations (such as reflection) do not alter shape or size of an object.

c: XY !XY", XZ!XZ, YZ! "Y Z, ∠Y ≅∠Y′, ∠YXZ ≅∠Y′XZ, and ∠YZX ≅∠Y′ZX

7-29. M(0, 7)

7-30. c2 and a2+b2

7-31. a: The triangles are similar by SSS ~.

b: The triangles are similar by AA ~.

c: Not enough information is provided.

d: The triangles are congruent by AAS ≅ or ASA ≅.

7-32. a: It is a parallelogram; opposite sides are parallel. b: 63.4 °; They are equal.

c: AC:!y= 1₂x+1₂, BD:!y=!x+5; No d: (3, 2)

7-34. Side length = 50 units; diagonal is 50!·! 2 = 100=10 units.

7-35. a: It is a rhombus. It has four sides of length 5 units.

b: HJ:y=!2x+8 and GI:y= 1₂ x+3 c: They are perpendicular.

d: (6, –1)

e: 20 square units

7-36. a: P(scalene) = 1₄

b: P(isosceles) = 2₄= 1₂

c: P(side of the triangle is 6 cm) =2₄ = 1₂

7-37. a: 6n!3° =n+17°, n = 4°

b: 7x!19° +3x+14° =180°, so x = 18.5°. Then 5y!2° =7(18.5)!19°, so y = 22.5°

c: 5w+36° +3w=180°, w = 18°

d: k2 =152+252!2(15)(25) cos120°, k = 35

7-38. The graph is a parabola with roots (–3, 0) and (1, 0), and y-intercept at (0, –3).

Lesson 7.1.4

7-43. a: 360° ÷ 36° = 10 sides b: Regular decagon

7-44. If the diagonals intersect at E, then BE = 12 mm, since the diagonals are perpendicular bisectors. Then ΔABE is a right triangle and AE = 152!122 =9mm.

Thus, AC = 18 mm.

7-45. Yes, she is correct. Show that the lengths on both sides of the midpoint are equal and that (2, 4) lies on the line that connects (–3, 5) and (7, 3).

7-46. a: See flowchart at right.

b: Not similar because corresponding sides do not have the same ratio.

c: See below.

7-47. (a) and (c) are correct because if the triangles are congruent, then corresponding parts are congruent. Since alternate interior angles are congruent, then AB// DE.

7-48. AB = 40!6.32, BC = 34 !5.83, therefore C is closer to B.

AA ~

ΔFED ~ ΔBUG SSS ~

7-54. a:x = 8.5° b: x = 11 c:x = 14°

7-55. a: 360° ÷ 72° = 5 sides b: 360° ÷ 9 = 40°

7-56. ≈ 36.4 feet from the point on the street closest to the art museum.

7-57. a: a_n =20+20n=40+20(n!1) b: a_n =6(1₂)n=3(1₂)n!1 7-58. a: (0.7)(0.7) = 0.49 = 49% b: (0.3)(0.7) = 0.31 = 21%

7-59. a: Similar (SSS ~)

b: Congruent (ASA ≅ or AAS ≅)

c: Congruent, because if the Pythagorean Theorem is used to solve for each unknown side, then 3 pairs of corresponding sides are congruent; thus, the triangles are congruent by SSS ≅).

d: Similar (AA ~) but not congruent since the two sides of length 12 are not corresponding.

7-60. Possible response: Rotate the second triangle 180° and then translate it to match the sides with the first triangle.

Lesson 7.2.2

7-65. 4x!1=x+8, x=3; 5y+2=22, y=4

7-66. a: 0.8 b: 1200(0.8)3_{=$614.40 c: 1200(0.8)}!2_=$1875 7-67. a: !a=36°,!r=54°,!m=54°, !y=72°,!z=108°

b: Possible response: !y and !z are supplementary interior same side angles.

7-68. a: It is a parallelogram, because MN// PQ and NP// MQ.

b: (1, –5)

7-69.

7-70. a: 50%; The sum must be 100%.

b: central angle for red = 0.4(360°) = 144°, white = 0.1(360°) = 36°, blue = 0.5(360°) = 180°

c: Yes; there could be more than three sections to the spinner, but the ratio of the areas for each color must match the ratios for the spinner in part (b).

7-71.

Vertical angles are congruent.

Definition of

midpoint. Definition of midpoint.

SAS ≅ Given E is a midpoint ≅ ≅ ∠AEB ≅∠CED ΔAEB ≅ ΔCED

7-75. a: Congruent (SSS ≅) b: Not enough information

c: Congruent (ASA ≅) d: Congruent (HL ≅)

7-76. See answers in problem 7-75.

7-77. a: 83° b: 92°

7-78. a: Yes; HL ≅ b: 18°, 4

c: tan 18° = 4

AD, AD≈ 12.3 units d:≈ 49.2 square units 7-79. a: Parallelogram because the opposite sides are parallel.

b: AC! "## : y= 3₄x; BD! "## : y=!₂3x+9

7-80. a: 68 ≈ 8.2, since 64 = 8, then 68 must be a little greater.

b: (1) 2.2, (2) 9.2, (3) 7.1, (4) 4.7

7-81. a: 2x+52° =180°, 64° b: 4x!3° +3x+1° =180°, 26°

c: sin 77_x °= sin 72₈ °, x≈ 8.2 d: 5x+6° =2x+21°, x = 5°

Lesson 7.2.4

7-83. 36 3!62.4 square units

7-84. No. Using the Pythagorean Theorem and the Law of Cosines, the perimeter of the triangle is ≈ 26.3 feet.

7-85. a:x≈ 10.73 cm, tangent b:x≈ 7.86 mi, Law of Sines

c:x≈ 15.3', Law of Cosines

7-86. a: Congruent (SAS ≅) and x = 2 b: Congruent (HL ≅) and x = 32

7-87. A=24 square units

7-88. a: ₂₀4 = 1₅

b: 4₅; Since the sum of the probabilities of finding the ring and not finding the ring is 1, you can subtract 1!1₅ = 4₅ .

c: No, his probability is still ₂₀4 = 1₅ because the ratio of the shaded region to the whole sandbox is unchanged.

Lesson 7.2.5

7-96. a: See diagram at right.

b: Since corresponding parts of congruent triangles are congruent, 2y+7=21 and y = 7.

7-97. y-intercept: (0, 6), x-intercept: (4, 0)

7-98. m∠a = 132º, m∠b = 108º, and m∠c = 120º, m∠a + m∠b + m∠c = 360º

7-99. 15%

7-100. AB!CD and AB// CD (given), so ∠BAC ≅∠DCA (alt. int. angles). AC!CA

(Reflexive Property) so ΔABC ≅ΔCDA (SAS ≅). ∠BCA ≅∠DAC (!!

!

7-101. Because alternate interior angles are congruent, the angle of depression equals the angle formed by the line of sight and the ground. Then tan!! = 52₃₈, !"53.85°.

7-102. a:ΔADC; AAS ≅ or ASA ≅ b: ΔSQR; HL ≅

c: No solution, only angles are congruent.

d:ΔTZY; SAS ≅ and vertical angles

e:ΔGFE; alternate interior angles equal and ASA ≅

f: ΔDEF; SSS ≅ T A P 14 18 21

7-108. ZY !WX, ∠YZX≅∠WXZ, ∠ZYW≅∠YWX (alt. int. !’s), ΔXWM≅ΔZYM, ASA !,

YM !WM and XM !ZM, ! " # !parts.

7-109. Typical responses: right angles, congruent diagonals, 2 pairs of opposite sides that are congruent, all sides congruent, congruent adjacent sides, diagonals that bisect each other, congruent angles, etc.

7-110. a: The triangles should be ≅ by SSS ≅ but 80º ≠ 50º.

b: The triangles should be ≅ by SAS ≅ but 80º ≠ 90º and 40º ≠ 50º.

c: The triangles should be ≅ by SAS ≅ but 10 ≠ 12.

d: Triangle is isosceles but the base angles are not equal.

e: The large triangle is isosceles but base angles are not equal.

f: The triangles should be ≅ by SAS ≅ but sides 13 ≠ 14.

7-111. The triangles described in (a), (b), and (d) are isosceles.

7-112. a: 12 b: 15 c: 15.5

7-113. See reasons in bold below.

7-114. This problem is similar to the Interior Design problem (7-21).

Her sink should be located 32₃ feet from the right front edge of the counter. This will make the perimeter ≈ 25.6 feet, which will meet industry standards.

Statements Reasons

1. AD// EH and BF//CG Given

2. a=b If lines are parallel, alternate interior angle measures are equal.

3. b=c If lines are parallel, corresponding angle measures are equal.

4. a=c Substitution

5. c=d Vertical angle measures are equal.

Lesson 7.3.1

7-119. a: (4.5, 3) b: (–3, 1.5) c: (1.5, –2)

7-120. a:ΔSHR ~ ΔSAK because ΔSHR can be dilated by a factor of 2.

b: 2HR = AK, 2SH = SA, SH = HA c: 6 units

7-121. a: 4 b: 1 c: –8

7-122. a:ΔCED; vertical angles are equal, ASA ≅

b:ΔEFG; SAS ≅

c:ΔHJK; HI + IJ = LK + KJ, ∠J≅∠J, SAS ≅.

d: Not ≅, all corresponding pairs of angles equal is not sufficient.

7-123. See answers in problem 7-122.

7-1