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```Selected Answers
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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
y
3
0
–1
–4
0
–3
2
–1
1-28. a: 55.5 sq. units
b: 42 sq. units
1-29. a: 5
c: –7
2
b: 12
–5
–8
–2
–5
1
–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
c: ! 63 = ! 12
b:
d:
6
3
0
7
=
2
1
=2
=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
Lesson 1.2.1
1-54. See graph at right.
1-55. a:
27
48
! 56.3%
c: 0
1-56.
y
b:
d:
4
10
1
130 = 13 ! 8%
5 ! 56%
8
y = ! 23 x + 3
–4
5x ! 2 + 2x + 6 = 67 , x = 9, so 5(9) – 2 = 43 miles
1-57. a: x = 3.75
d: x = 3
y = 3x ! 3
b: x = 3
c: x = 0
e: x ≈ 372.25
f: x = –3.4
4
–4
x
y = !4x + 5
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
c: y = –2
4
b: y = 25
Core Connections Geometry
Lesson 1.2.3
1-76. a: Yes
1-77. a: y =
b:
5
2
2
5
c:
x!3
3
5
b: y =
d:
3
2
1
5
e:
4
5
x+7
1-78. C
1-79. a: p = !5
b: w = 20
c: x = 14
d: y = 1.7
1-78. See answers in bold below. y = 3x + 2
x
y
–3
–7
–2
–4
–1
–1
0
2
1
5
2
8
3
11
4
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:
d:
3
10
9
10
e: The outcomes that satisfy part (d) include the outcomes that satisfy part (c), but there
are others on the part (d) list as well.
1-88. a: y =
4
3
x!2
b: The resulting line coincides with the original line; y =
c: The image is parallel; y =
4
3
3
4
x!2
x!7
d: They are parallel, because they all have a slope of
e: y = !
4
3
4
3
.
x + 16
1-89. –14
Selected Answers
5
Lesson 1.2.5
1-94.
1
535
! 0.0019 ; No, this probability is very small.
1-95. a: (9, 3)
b: (3, –3)
1-96. a: 10 square units
1-97. a:
c: (–2, –7)
d: (–52, 1483)
b: 20 square units
c: 208,680 square units
b:
B
A
P
l
c:
d:
C
D
Q
m
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.
6
Core Connections Geometry
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 = 144 square units
1-107. a:
4
52
1
= 13
b:
13
52
=
1
4
c:
1
52
d:
=
39
52
3
4
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:
A
B
l
c:
P
d:
C
D
Q
m
Selected Answers
7
Lesson 1.3.1
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 =
chance.
2
4
, while Robert’s is
3
5
. Therefore, Robert has a greater
b: Sandy (Sandy’s probability = 1 while Robert’s is 0)
c: Sandy’s probability = 43 , while Robert’s is 53 . Therefore, Sandy is more likely to
select a shape with two sides that are parallel.
1-114. a: x = !
9
33
3
= ! 11
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:
c:
7
16
1
16
≈ 44%
b:
≈ 6%
d:
1-125. a: isosceles triangle
d: obtuse scalene triangle
9
16
6
16
≈ 56%
=
3
8
≈ 38%
b: pentagon
c: parallelogram
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
c: x = 3
8
b: x =
3
2
= 1 12
d: No solution
Core Connections Geometry
Lesson 2.1.1
2-8.
a: 33 sq. cm
2-9.
a:
c: 33x 2 ! 50x + 8 sq. units
b: 33x sq. units
1
2
b:
2-10. a: Isosceles triangle
2
6
, parallelogram and square
b: Equilateral triangle
c: Parallelogram
2-11.
KITE
ISOSCELES TRIANGLE
REGULAR HEXAGON
RHOMBUS
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
2-22.
b: (– 4, 5), (1, 5), (– 4, 0), (1, 0)
y = x ! 1 ; No, because 1 ! 3 " 1 .
2-23. a: Vertical; they have equal measure.
Selected Answers
b: They form a “Z.”
9
Lesson 2.1.3
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 if the angle 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
c:
b: 22 girls
2
3
d: 7 boys left, 23 students, so
7
23
.
2-36. a: an isosceles triangle
b: a rectangle
Lesson 2.1.4
2-41. a:
105° 75°
105° 75°
75° 105°
75° 105°
b:
30°
70° 80°
80°
85°
95°
95°
85°
85°
95°
70°
30°
95°
85°
100° 80°
80° 100°
70° 110°
110°
70°
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.
10
Core Connections Geometry
Lesson 2.1.5
2-55.
x = 7º
2-56. a: x = 10 units
c: x = 20!
b: x = 6
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: !
4
3
c: Any equation of the form y = !
3
4
x+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°
b: x = 12° , m!D = 4(12°) + 2° = 50°
2-67. a: 15°
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 =
1
2
x+5
c: Any equation of the form y = !2x + b for all real b-values.
Selected Answers
11
Lesson 2.2.2
2-74. a: 8x 2 ! 26x ! 7
c: 4x 2 ! 47x + 33
b: 10x 2 + 31x ! 14
2
d: !6x + 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:
b:
22
35
22
35
= 62.9%
=
42
?
; she needs to attempt about 67 pancakes.
c: She should add three banana pancakes to make the probability of banana
3
30
.
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
c: 3x 2 ! 5x ! 12
12
b: 2x + 3
c: 13x 2 + 30x
Core Connections Geometry
Lesson 2.2.4
2-94. a: 72 = 49 sq cm
b: 0.5(10)4 = 20 sq. in.
2-95. a: 15x 2 + 21x
c: 0.5(16 + 8)6 = 72 sq ft
b: x 2 + 5x + 6
c: 3x 2 ! 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
3
= 13
b:
20
52
5
= 13
c:
2
52
=
1
26
d: 0
Lesson 2.3.1
2-104. 10 units
2-105. a: (1): (5, 3), (2): (2, –6)
b: p: y = 2x + 8 ; q: y = ! 12 x + 3
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.
106°
74°
74°
106°
b: 20 square units
c: ≈ 23.4 units
106°
74°
74°
106°
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)
y
y
x
x
c: A!!(1,!2) and C !!("2,!5)
d:
1
2
(3)(2 + 7) = 13.5 units
Selected Answers
13
Lesson 2.3.2
2-113. a: (–2, 5)
2-114. a:
b: (1, 5)
7
8
b:
c: (–12, 14)
3
8
c:
d: (2, 2)
5
8
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 = 10 units, B!C ! = 8 units, and A!C ! = 6 units.
c: A = 24 sq. units; P = 24 units
3-6.
a: x = 18
3-7.
a: ≈ 30°, ≈ 40°, ≈ 110°
3-8.
a:
4
5
, y=
b: x = 3
4
5
x + 95
c: x = 6
d: x = 2
b: Obtuse scalene triangle
b: MU =
41 ! 6.40 units
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.
14
Core Connections Geometry
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
20
ratios are possible. One possibility: 15
6 = 8 = 2.5
b: 25 and 10;
25
10
= 2.5 ; yes
3-20. Yes they are parallel because they have the same slope: ! 53 .
3-21. a: 6x 2 ! 8x
3-22.
b: 2x 2 + x ! 15
c: 4x 2 ! 25
d: 2x 3 ! 5x 2 ! 3x
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 =
3
2
= 1.5
3-32. a: y = 3 ! 53 x
b: A = 7.5 sq. units; P = 8 + 34 ! 13.8
c: y = 3 + 53 x
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
3-34. a: alt. int. angles
c: corresponding angles
Selected Answers
and ! 23 , then they are perpendicular.
b: vertical angles
d: supplementary and/or adjacent angles
15
Lesson 3.1.4
3-41. a: f = 9
b: g = 18
c: h =
70
3
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 2x 2 .
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)
b: Perimeters = 28 and 56 units; areas = 52 and 208 sq. units
16
Core Connections Geometry
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:
9
12
, ii:
8
12
, iii:
1
12
y
3-57. See graph at right. Perimeter = 44.9 units; Area = 94 sq. units
3-58. a: ABCD ~ EVOL
x
b: RIGHT ~ RONGW
c: One possible answer: ΔTAC ~ ΔGDO
Selected Answers
17
Lesson 3.2.2
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:
8
12
b:
4
8
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: Not enough information to determine similarity.
c: Rotation, dilated by factor of 2, and translation.
d: Rotation, reflection, and reduced by zoom factor of 0.5 (or translation and reflection
instead of rotation and reflection).
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.
d: Possible
18
Core Connections Geometry
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 =
1
2
x+2
b: A = 4 sq. units; P = 6 + 20 ! 10.47 units
c: y = !2x + 7
3-78. a: x = 51° alternate interior angles and Triangle Sum Theorem
b: x = 43° circle has 360º
c: x = 1 Pythagorean Theorem
double \$200
\$100
keep
3-79. a: See tree diagram at right.
b: Yes
\$300
c:
1
6
,
3
6
d:
1
2
, no—the spinners are independent
e:
2
6
\$600
keep \$300
, because now the possible outcomes are
\$100, \$200, \$1500, \$200, \$400, and \$3000.
3-80. a: n = 32
\$100
double
double
\$1500
\$3000
keep
\$1500
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.
Selected Answers
19
Lesson 3.2.4
3-88. a: Scalene triangle
c: Not possible
b: Isosceles triangle
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: 4.9 2 ! 3.12 = x 2 , x ! 3.79
c: x + (180° ! 51° ! 103°) + 82° = 180° , x = 72!
d: 3x ! 2 = 2x + 9 , x = 11
20
Core Connections Geometry
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:
24
40
= 60%
b:
3-101. a: 12x 2 ! 7x ! 10
18
x
3 , x = 60
= 10
b: 16x 2 ! 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
61 ! 7.81 units
b: LD =
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 3.2.6
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:
13+17
22+8+13+15+17
c:
3
?
=
30
75
= 40%
= 5% ; You need a total of 60 animals in the bin.
3-111. a: y = ! 12 x + 4
c: y =
2
5
x + 57
b: y = 2x ! 1
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
Selected Answers
21
Lesson 4.1.1
4-6.
a: x = 11°
b: x = 45°
4-7.
a: See flowchart at right.
c: x = 30°
b: Yes, because the triangles are similar
(AA ~) and the ratio of the corresponding
side lengths is 1 (because AC = DF).
4-8.
m∠B = m∠E
m∠A = m∠D
ΔABC ~ ΔDEF
AA ~
a: Yes, she used the Pythagorean Theorem.
c: x = 24
b: (x + 1)2 = x 2 + 2x + 1
4-9.
d: x = 68°
d: 56 units
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) =
18
36
; P(10) =
d: The sum of 7. P(7) =
3
36
6
36
; P(15) = 0
=
1
6
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 ! 19
4-18. a: side ratio = 4:1
y
70
b: a = b = 45°
c:
b: perimeter ratio is 4:1
c: 28'
! 52 , y ! 175
4-19. a: Yes, AA ~.
b: No, side ratios not equal
12
64
! 18
98 .
c: Cannot tell, not enough angle values given.
4-20.
6 2 ! 32 = 27 , 9 2 ! 32 = 72 . So perimeter is
area is ( 27 + 72 )(3) ÷ 2 ! 20.52 sq. cm.
27 + 72 + 15 ! 28.68 cm. The
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
22
b: Yes
c:
6
12
= 12 ;
8
12
=
2
3
Core Connections Geometry
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; an = 1+ 3(n ! 1) = 3n ! 2
b: neither
c: G; an = 2 ! 2 n"1 = 2 n
d: A; an = 5 + 7(n ! 1) = 7n ! 2
4-30. Answers vary, possible solution: square, equilateral triangle, and equilateral hexagon.
4-31. a:
2
5
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:
d:
1 , because
2
1 ; same
3
only 2 songs are left and only one is sung by Sapphire.
4-32. m∠ABC = 22°, m∠BAC = 68°, sum = 90°; complementary
Selected Answers
23
Lesson 4.1.4
b: p ! 3.215
4-39. a: t = 11.524
c: b ! 148.505
4-40. a: x2 + 182 = 302; x = 24
b: 2x + 20° + 3x + 20° + x + 2x = 360° , x = 40°
c:
5
12
= 3x , x =
36
5
= 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
c: A = 72 sq cm, P = 48 cm
b: A = 696.67 m2, P = 114.67 m
d: A = 130 sq. feet, P = 58 feet
Lesson 4.1.5
4-47. a: Either 3 or
1
3
b: Either 9 or
4-48. a: 3x + 3° + x + 7° = 90° , x = 20°
1
9
b: 9x + 4° = 3x + 14° ; x = 10
6 ! 1.67°
4-49. (a) and (d) are most likely independent.
4-50. b: ≈ 29.44 feet
4-51. a: G; an = 12 ! 12
n"1
=
1n
2
b: A; an = !7.5 ! 2(n ! 1) = !5.5 ! 2n
4-52. 10 2 + (x + 3)2 = 26 2 ; x = 21
24
Core Connections Geometry
Lesson 4.2.1
4-58. a:
10
20
=
1
2
b:
9
19
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.
b: x = ! 12 or x = 3 ; Yes.
a: (! 12 ,!0) and (3,!0)
4-60. a: Slope =
1
2
b: It must be parallel to or coincide with the line graphed at right.
4-61. Francis: y = x + 2, John: y =
4-62. a: x ≈ 2.344
3
4
x + 5 ; 12 seconds
b: x ≈ 0.667
c: x = 1.5 or –5
d: No real solution
4-63. leg ≈ 29.44 cm, hypotenuse ≈ 30.78 cm, so the perimeter ≈ 69.22 cm
Lesson 4.2.2
H
4-69. a: A tree diagram; a third dimension would be needed to represent
the three coins with an area model.
H
b: See tree diagram at right; 8
c: i:
1
8
, ii:
3
8
, iii:
7
8
, iv:
T
3
8
d: They are both the same probability of 50%.
e: The sample space remains the same;
64 , ii: 4 + 4 + 4 = 12 , iii: 61 , iv:
i: 125
125 125 125
125
125
T
H
T
T
H
H
T
T H
T
H
12
125
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.
1
6
; 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
4-74.
b: 10, –50, 250, –1250
x ! 10.39 , y = 12
Selected Answers
25
Lesson 4.2.3
4-81. Both equal
4-82.
y=
1
3
3
8
.
x+9
4-83. a: See diagram at right.
b: Ratio for tan 11° ≈
1
5
, so
170
x
! 15 , and x ! 850 feet.
Alternatively, a calculator could be used and x =
4-84. a: x = 49
11º
b: x = 2
170
tan(11°)
x
170
ft
! 875 feet.
c: x = 16
3
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:
3
36
4-96. a: 5 ways
b:
4
36
b: 6 ways
c:
24
36
c: 11
d:
5
11
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 , x 2 + 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°
7 . Points will vary. y = !0.7x + 82.5
4-100. The slope is ! 10
A few possible solutions: (5, 79), (15, 72), (25, 65), etc.
26
Core Connections Geometry
Lesson 4.2.5
4-110. a:
1
12
b: Intersection
c: No, P(yellow) =
1
6
e: You cannot move
d:
1
6
+ 16 + 13 =
4-111. a: y = 3
2
3
2
3
or you can move
1
3
of the time and 1! 13 = 23 .
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 ) =
b:
4
52
, P(Q) =
4
52
, P(C) =
13
52
16
52
; You can add the probabilities of king and club, but you need to subtract the
4 + 13 ! 1 = 16
number of cards that are both kings and clubs (1). P(K or C) = 52
52 52
52
8 = 2 . There is no overlap in the events so you can just add the
c: P(K or Q) = 52
13
probabilities.
d: P(not a face card) = 1! 12
52 =
40
52
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 ! 11.47 mm, 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)
c: (9, 3)
Selected Answers
27
Lesson 5.1.1
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)! (4 suits)
52
b: 1! 16
52 =
36
52
= 16
52 . If Aces are not included,
12
52
.
40
52 .
. If Aces are included,
c: P(red) + P(face) – P(red and face) =
26
52
6
+ 12
52 ! 52 =
32
52
5-11. area = 74 sq ft, perimeter = 47.66 ft
5-12. a: x = –3
b: m = 10
c: p = !4 or
d: x = 23
2
3
Lesson 5.1.2
x
5-17. a: sin 22° = 17
b: x ! 6.37 , tan 49° = 7x , 6.09
c: cos 60° = 6x , x = 3
5-18. ≈ 26.92 feet
1 )n!1 = 10 3!n
5-19. a: G; an = 100( 10
b: A; an = 0 ! 50(n ! 1) = 50 ! 50n
5-20. Region A is 14 of the circle. Since the spins are independent, the probability of A and A
1
1 (80) = 5
is 14 ! 14 = 16
. In 80 games, we expect A and A to occur 16
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: 6x 2 ! x ! 2
c: !3xy + 3y 2 + 8x ! 8y
b: 6x 3 ! x 2 ! 12x ! 5
d: x 2 ! 9y 2
5-23. ΔABC ~ ΔEFD by SAS ~
28
Core Connections Geometry
Lesson 5.1.3
5-29. a: x = ±5
b: All numbers
c: x = 2
d: No solution
5 , sin A = 12 , or tan A = 12 , A ! 67.4º
5-30. Using cos A = 13
13
5
5-31.
! 11.5 seconds
5-32. a:
1
36
b:
20
36
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°) ! 23 , y ! 23 x + 7 .
5-43. 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
5-44. a: sin ! =
5-45. a:
22
52
b
a
; union
b: tan ! =
b:
3
52
a
b
; intersection
d: an = 117(0.2)n!1 = 585(0.2)n
c: cos ! =
a
b
c: 1! 22
52 =
30
52
18
5-46. a: cos!23° = 18
x or 0.921 = x
b: Since 67° is complementary to 23°, then sin 67° = cos 23° . So sin 67° ! 0.921 .
Selected Answers
29
Lesson 5.2.1
5-52. a: A = 1 sq. m, P = 2 + 2 2 m
b: A =
25 3
2
! 21.65 square ft, P = 15 + 5 3 ! 23.66 ft
5-53. a: y = 111º ;!x = 53º
c: y = 83º ;!x = 53º
b: y = 79º ;!x = 47º
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: an = 500 + 1500(n ! 1) = 1500n ! 1000
b: an = 30(5)n!1 = 6(5)n
5-58. a: Not similar
b: Similar: Rotate ΔGHI, translate, then dilate.
c: Similar: Reflect ΔMNP, translate, then dilate.
d: Similar: Rotate ΔTUV, translate, then dilate.
30
Core Connections Geometry
Lesson 5.2.2
b: 4 yards and 4 2 yards
5-64. a: 16 inches
d: 10 meters and 10 3 meters
c: 24 feet
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 and DC = EC, so
AC
BC
DC = EC and m∠ACB = m∠DCE therefore ΔABC ~ ΔDEC 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° = bc ; tan!52° = ba ; cos!38° = ac ; cos!38° = sin!52°
5-67.
14
27
=
x
40
, x ! 20.74 inches
b: an = !3 + 4(n ! 1) = 4n ! 7
5-68. a: explicit
d: an = 3 ! 13 (n ! 1) = 3 13 ! 13 n
c: a50 = 193
5-69.
135
90
\$3( 135
360 ) + \$5( 360 ) + (!\$6)( 360 ) = \$1.50 . It is not fair because the expected value is not 0.
5-70. a:
1
2
b: 0
c:
3
4
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 = 16
5
c: x = –11 or 3
b: No solution
d: x = 288
5-82. b is correct; if two sides of a triangle are congruent, the angles opposite them must be
equal.
Selected Answers
31
Lesson 5.3.2
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
yogurt
yogurt
5-93. See tree diagram at right (an
area model is not practical).
P(three yogurts) = 12.5%.
yogurt
100% ! 12.5% = 87.5%
chance of not getting three
yogurts.
5-94. a: x =
45
4
red apple
yogurt
green apple
yogurt
green apple
b: x = !10 or x = 10
red apple
c: x = 1.3
yogurt
yogurt
d: No solution
5-95.
green apple
red apple
red apple
= 11.25
green apple
green apple
red apple
(!2,!4)
green
apple
yogurt
green apple
green apple
red apple
red apple
yogurt
green apple
red apple
yogurt
yogurt
green apple
red apple
red
apple
yogurt
green apple
green apple
red apple
red apple
yogurt
green apple
red apple
32
Core Connections Geometry
Lesson 5.3.3
y
42
5-100. a: 29°
b: cos!29° =
5-101. a: (–1, –2)
b: (4, – 4)
c: (3, 4)
b: 3x 2 ! 7x ! 6
c: x = 1 or 7
, y ! 36.73
5-102. tan !1 ( 43 ) " 36.87°
5-103. sin !1
7
8
" 61.0º
5-104. a: 2x 2 + 6x
5-105. a:
1
12
b:
d: y = –3 or 5
1
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.
3
12
2 (10) = 11 " \$1.83
(3) + 127 (!1) + 12
6
The game is not fair because the expected value is not zero.
5-117. 7 years
Selected Answers
33
Lesson 5.3.5
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
x 2 + x 2 = 16 2 , 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) = 15
4 = 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 =
3
4
x+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 = 5.27 cm, 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) = 12 t(n)
c: t(n) = 24 ! 7n
d: t(n) = 5(1.2)n
e: t(4) = 1620
5-137. a: See diagram at right.
Chain
b: x = 103 3 ! 5.77
5-138. a: 5 + 20 + 37 ! 15.55 units
34
b: ! 31.11
x
x
Bush
Shed
c: (–2, 0)
Core Connections Geometry
Lesson 6.1.1
6-4.
a: Alternate interior angles.
b: Vertical angles.
c: ∠u & ∠z,
6-5.
∠s & ∠x,
∠v & ∠w,
and ∠t & ∠y
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 = 127
3 ! 42.33°
6-7.
a: 8 cm
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:
3
4
b: ≈ 14.97 ft
or 75%
c: 1 or 100%
c: ≈ 15.2 in.
b:
3
20
or 15%
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.
6-18. a: (0.8)(0.8) = 0.64 = 64%
b: (0.8)(0.2) = 0.16 = 16%
6-19. A
Selected Answers
35
Lesson 6.1.3
6-23. a: Not similar because there are not three pairs of corresponding angles that are
congruent.
b: Similar (AA ~)
6-24. a: y =
6-25. b:
14
22
5
2
=
x!8
10
DE
b: y =
3
2
x +1
, 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) =
of A and B (the overlap) was 0.
64
212
176
+ 112
212 ! 0 = 212 " 83.0% ; the probability
56
b: P(A or B) = P(A) + P(B) – P(A and B) ! 75% = 114
212 + 212 ! x " x # 5.1%
36
Core Connections Geometry
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.
4
10
=
5
x+5
, 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:
1
4
c:
1
9
+ 16 + 16 + 14 =
25
36
! 69%
parents
1
3
niece
6-40. C
1
6
boyfriend
1
2
Selected Answers
parents
niece
boyfriend
1
3
1
6
1
2
1
9
1
18
1
6
1
18
1
36
1
12
1
6
1
12
1
4
37
Lesson 6.1.5
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: c 2 = 9 + 36 ! 2 · 3 ·6 cos 60° , c = 3 3 ! 5.196 units;
For right-hand triangle: c 2 = 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 = 13
4
6-51. a:
c:
6-52.
38
b: y = –2
3
8
3
8
sin 40° =
b:
d:
h
600
1
8
1
8
c: 4 23 "
d: x = 4
; sum must be equal to one.
, h ≈ 385.67 feet
Core Connections Geometry
Lesson 6.2.1
6-55. a: x ! 45.56
b: x ! 10.63
c: x ! 265.48
6-56. 9 square units; First find AC = 5 and then calculate
and calculate 12 (2 + 4)(3) .
1
2
d: x = 5
(5)(3.6) , or use BC as the base
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
always true.
1
6
, then a 6-sided dice was rolled. Not
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
c: x = 23° and y = 43°
6-75.
b: x = 55°
d: x = 5.5 and y = 45.2
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 ≅)
c: Not necessarily congruent.
b: Congruent (AAS ≅)
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
16
8 , x ! 9.8 , corresponding parts of similar figures have equivalent ratios
= 13
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
Lesson 6.2.5
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 =
6
5
1
3
x!3
b: y = !
c: y = x
1
4
x + 4.5
d: y = 2
6-96. a: x ! 8.1
b: Not enough information.
6-97. a: x = 15° , Triangle Angle Sum
c: t = 9° and w = 131° , parallel lines
6-98. a: Lose \$1.50
c: x ! 10.67
b: k = 5 , Isosceles triangle
d: x ! 49.94 , Triangle Angle Sum
b: Lose \$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 = 20 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
=
7-11. a: (6, –13)
42
8 3
3
, y=
16
3
= 163 3
b: Not possible, these curves do not intersect.
Core Connections Geometry
Lesson 7.1.2
7-15. Using the Pythagorean Theorem, AB = 8 and JH = 5. Then, since
ΔABC ~ ΔHGJ because of SSS ~.
3
6
=
4
8
5 ,
= 10
7-16. 7 cm
7-17. Line L: y = ! 16 x + 6 ; line M: y =
2
3
x + 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 nonconvex). 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.
c 2 and a 2 + b 2
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.
c: AC :!y =
1
2
x + 12 , BD :!y = !x + 5 ; No
7-33. a: No solution, lines are parallel.
Selected Answers
b: 63.4 °; They are equal.
d: (3, 2)
b: (0, 3) and (4, 11)
43
Lesson 7.1.3 (Day 2)
7-34. Side length =
50 !·! 2 = 100 = 10 units.
50 units; diagonal is
7-35. a: It is a rhombus. It has four sides of length 5 units.
b: HJ : y = !2x + 8 and GI : y =
1
2
x+3
c: They are perpendicular.
d: (6, –1)
e: 20 square units
7-36. 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. 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: k 2 = 15 2 + 25 2 ! 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).
7-39. ≈ 35.24 m
44
Core Connections Geometry
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 = 15 2 ! 12 2 = 9 mm.
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.
ΔFED ~ ΔBUG
AA ~
SSS ~
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 =
Selected Answers
40 ! 6.32 , BC =
34 ! 5.83 , therefore C is closer to B.
45
Lesson 7.2.1
7-54. a: x = 8.5°
b: x = 11
7-55. a: 360° ÷ 72° = 5 sides
c: x = 14°
b: 360° ÷ 9 = 40°
7-56. ≈ 36.4 feet from the point on the street closest to the art museum.
7-57. a: an = 20 + 20n = 40 + 20(n ! 1)
b: an = 6( 12 )n = 3( 12 )n!1
7-58. a: (0.7)(0.7) = 0.49 = 49%
b: (0.3)(0.7) = 0.21 = 21%
7-59. a: Similar (SSS ~)
b: Congruent (ASA ≅ or AAS ≅)
c: Similar, because if the Pythagorean Theorem is used to solve for each unknown side,
then 3 pairs of corresponding sides have a common ratio; thus, the triangles are similar
(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.
46
Core Connections Geometry
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)
E is a midpoint
7-69.
Given
∠AEB ≅ ∠CED
≅
≅
Vertical angles
are congruent.
Definition of
midpoint.
Definition of
midpoint.
ΔAEB ≅ ΔCED
SAS ≅
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.
Selected Answers
47
Lesson 7.2.3
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 ≅
c: tan 18° =
4
AD
b: 18°, 4
, AD ≈ 12.3 units
d: ≈ 49.2 square units
7-79. a: Parallelogram because the opposite sides are parallel.
!##"
!##"
b: AC : y = 43 x ; BD : y = ! 23 x + 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°
c:
sin 77°
x
=
sin 72° ,
8
b: 4x ! 3° + 3x + 1° = 180° , 26°
d: 5x + 6° = 2x + 21° , x = 5°
x ≈ 8.2
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
c: x ≈ 15.3', Law of Cosines
b: x ≈ 7.86 mi, Law of Sines
7-86. a: Congruent (SAS ≅) and x = 2
b: Congruent (HL ≅) and x = 32
7-87.
A = 24 square units
7-88. a:
b:
4
1
20 = 5
4
5 ; 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
sandbox is unchanged.
7-89.
48
4
20
=
1
5
because the ratio of the shaded region to the whole
a50 = !130
Core Connections Geometry
Lesson 7.2.5
7-96. a: See diagram at right.
A
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)
18
14
T
21
P
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 ( !! s!"!!!parts ).
Thus, BC // AD (if alt. int. angles are congruent, then the lines cut by the transversal are
congruent).
!
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!! = 52
38 , ! " 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 ≅
Selected Answers
49
Lesson 7.2.5
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.
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.
6. a = d
Substitution
7-114. This problem is similar to the Interior Design problem (7-21).
Her sink should be located 3 23 feet from the right front edge of the counter. This will
make the perimeter ≈ 25.6 feet, which will meet industry standards.
50
Core Connections Geometry
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.
7-124. No. Her conclusion in Statement #3 depends on Statement #4, and thus must follow it.
7-125. a: Must be a quadrilateral with all four sides of equal length.
b: Must be a quadrilateral with one pair of opposite sides that are parallel.
Selected Answers
51
Lesson 7.3.2
7-131. a: Yes, each side has the same length ( 29 units). See graph at right.
!##"
!##"
b: BD is y = x ; AC is y = 5 ! x
c: The slopes are 1 and –1. Therefore the diagonals are perpendicular.
7-132. Multiple answers are possible. Any order is valid as long as Statement #1 is first,
Statement #6 is last, and Statement #4 follows both Statements #2 and #3. Statements #2,
#3, and #5 are independent of each other and can be in any order as long as #2 and #3
follow Statement #1.
7-133. a: 6
b: 3
c: – 6.5
7-134. a: Yes, by SAS ~.
b: ∠FGH ≅ ∠FIJ, ∠FHG ≅ ∠FJI
c: Yes, because corresponding angles are congruent. (Triangle Midsegment Theorem)
d: 2(4x ! 3) = 3x + 14 , so x = 4 and GH = 4(4) ! 3 = 13 units.
7-135. a: It is a right triangle because the slopes of AB and AC ( 43 and !
opposite reciprocals.
4
3
, respectively) are
b: B! is at (–2, 0). 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).
7-136. a: ≈ 23.83 feet
c: x ≈ 66.42
b: x ≈ 7 yds
d: ≈ 334.57 feet
7-137. a: Must be: trapezoid.
Could be: isosceles trapezoid, parallelogram, rhombus, rectangle, and square.
b: Must be: parallelogram. Could be: rhombus, rectangle, and square.
52
Core Connections Geometry
Lesson 7.3.3
7-140. a: (8, 8)
b: (6.5, 6)
7-141. a: X and Y
b: Y and Z
7-142. a: Must be: none. Could be: right trapezoid, rectangle, square.
b: Must be: none. Could be: kite, rhombus, square.
7-143. a: If a polygon is a parallelogram, then its area equals its base times its height.
b: If the area of a polygon is the base times its height, then the polygon is a
parallelogram. This is always true.
c: “If a polygon is a triangle, then its area equals one half its base times its height.”
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, then the polygon is a triangle.
This is always true.
7-144. a: 360° ÷ 18° = 20 sides
b: It can measure 90° (which forms a square). It cannot be 180° (because this polygon
would only have 2 sides) or 13° (because 13 does not divide evenly into 360°).
7-145. The expected value is 14 (\$3) + 43 (\$1) = \$1.50 per spin, so each player should pay \$1.50
so that there is no net gain or loss over many games.
7-146. a: They are perpendicular because their slopes are opposite reciprocals.
b: It could be a kite, rhombus, or square because the diagonals are perpendicular and at
least one diagonal is bisected.
Selected Answers
53
Lesson 8.1.1
8-6.
a: 110°
b: 70°
c: 48°
d: 108°
8-7.
a: no
b: yes
c: no
d: yes
e: no
f: yes
g: yes
h: no
8-8.
b: The measure of an exterior angle of a triangle equals the sum of the measures of its
remote interior angles.
c: a + b + c = 180º (the sum of the interior angles of a triangle is 180°), x + c = 180º
(straight angle); therefore, a + b + c = x + c (substitution) and a + b = x (subtracting c
from both sides).
8-9.
x = 72° and y = 54°
8-10. 360° ÷ 15° = 24
8-11. a: congruent (SAS ≅), x = 79°
c: congruent (AAS ≅), x ≈ 5.9 units
b: Cannot be determined.
d: congruent (SAS ≅), x ≈ 60.9°
8-12. a: True
b: False (counterexample is a quadrilateral without parallel sides.)
c: True
d: True
e: False (counterexample is a parallelogram that is not a rhombus)
54
Core Connections Geometry
Lesson 8.1.2
8-17. a: Isosceles right triangle, because AC = BC and AC ! BC .
b: 45°; methods vary
8-18. a = 87°, b = 83°, c = 96°, d = 94°; 360°
8-19. A = 40.5 square miles, P ≈ 27.7 miles
8-20. (–5, 1), (–3, 7), and (–6, 2)
8-21. a: an = !45 + 15n = !30 + 15(n ! 1)
1 n
1 n!1
b: an = 27( 3 ) = 9( 3 )
8-22. See answers in bold below.
Statements
Reasons
1. BC // EF , AB // DE , and AF = DC
1. Given
2. m∠BCF = m∠EFC and
m∠EDF = m∠CAB
2. If two lines cut by a transversal
are parallel, then alternate
interior angles are equal.
3. FC = FC
3. Reflexive Property
4. AF + FC + CD = FC
4. Additive Property of Equality
(adding the same amount to both
sides of an equation keeps the
equation true)
5. AC = DF
5. Segment addition
6. ΔABC ≅ ΔDEF
6. ASA ≅
7. BC ! EF
7. !! s!"!!!parts
!
8-23. B
Selected Answers
55
Lesson 8.1.3
8-29. a: A = 36 sq. ft, P = 28 ft
b: A = 600 sq. cm, P ≈ 108.3 cm
8-30. QP = RS and PR = SQ (given), QR = QR (Reflexive Property), so ΔPQR ≅ ΔSRQ (SSS ≅)
and ∠P ≅ ∠S (≅ Δs → ≅ parts).
8-31. a: Isosceles triangles
b: The central vertex must be 360° ÷ 10 = 36°. The other two angles must be equal since
the triangle is isosceles. Therefore, (180° – 36°) ÷ 2 = 72°.
c: 10!·!14.5 = 145 square inches
8-32. a: (6.5, 5)
b:
3
8
c: Using the strategy developed in Lesson 7.3.2, !x = 14 " 2 = 12 and !y = 10 " 2 = 8 .
Then the x-coordinate is 2 + 83 (12) = 6.5 and the y-coordinate is 2 + 83 (8) = 5 .
8-33. a: x + x + 82° = 180° , x = 49°
b: 2(71°) + x = 180° , x = 38°
8-34. a: The region can be rearranged into a rectangle with dimensions 14 and 7 units.
b: (14)(7) = 98 square units
8-35. B
56
Core Connections Geometry
Lesson 8.1.4
8-40. The reflections are all congruent triangles with equal area. Therefore, the total area is
(6)(11.42) = 68.52 square inches.
8-41. a: 1.04
b: f (t) = 135000(1.04)t
8-42. a: Non-convex
b: Convex
8-43. a: 64 units2
b: ≈ 27.0 units2
8-44. a: 3
b: 15
c: ≈ \$199,833
c: Convex
d: Non-convex
c: 8 3 ! 13.9 units2
c: 4
d: 9
8-45. All circles are similar, or use similarity transformations to justify the similarity.
8-46. a: A = 192 cm2, P = 70 cm
b: The length of each side is 5 times the corresponding side in the floor plan.
A = 4,800 cm2 and P = 350 cm.
c: The ratio is
5
1
= 5; the ratio of the perimeters equals the zoom factor.
d: The ratio of the areas is
factor (52).
Selected Answers
25
1
= 25 . The ratio of the areas equals the square of the zoom
57
Lesson 8.1.5
8-53. a: The interior and exterior angles must be supplementary. 180° – 20° = 160°.
b: Use 360° ÷ 20° = 18 sides or solve the equation
180º(n!2)
n
= 160° to find n = 18.
8-54. a: The perimeter of both triangles is ≈ 48.25 units.
b: ≈ 32.9° and ≈ 57.1°
8-55. Since the diagonals of a parallelogram bisect each other, they must intersect at the
midpoint of BD . Thus, they intersect at (6, 21).
8-56. A = 100 3 ! 173.2 mm2
8-57. ≈ 103.8 meters
8-58. a: w = ±
17
5
! ±!1.84
b: w ≈ 2.17 and –1.57
c: No real solution
8-59. E
8-60. a: 60°
b: 82°
8-61. a: Equilateral triangle
c: Nonagon
c: 14°
d: 117°
b: Rectangle
d: Rhombus or kite
8-62. The x-coordinate must lie on the perpendicular bisector of segment AB. Thus, since the
midpoint M of segment AB is (6, 0), the x-coordinate of point C must be 6. ΔAMC is a
right triangle and the hypotenuse must have a length of 12 units for ΔABC to be
equilateral. Therefore, MC = 12 2 ! 6 2 = 6 3 because of the Pythagorean Theorem. So
the y-coordinate of point C could be 6 3 or – 6 3 .
8-63. a: Yes; since BC = BC (Reflexive Property), AB ! DC (given), and ∠ABC ≅ ∠DCB
(given), then ΔABC ≅ ΔDCB (SAS ≅). Therefore, AC = DB (≅ Δs → ≅ parts).
b: No; the relationships in the figure are true, as long as the two angles remain congruent.
See the diagram for problem 8-30 for a similar diagram.
8-64. a: (–2.5, 0) and (3, 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.
8-65. B
8-66. D
58
Core Connections Geometry
Lesson 8.2.1
8-71. a: A = 34 units2; P ≈ 25.7 units
b: A = 306 units2; P ≈ 77 units
c: ratio of the perimeters = 3; ratio of the areas = 9
8-72. 80 inches or ≈ 6.67 feet
8-73. The area of the hexagon ≈ 23.4 ft2. Adding the rectangles makes the total area ≈ 41.4 ft2.
8-74.
4
21
! 19.05% ; k = 0, 6, 10, 12 are factorable.
8-75. a: Reasoning will vary. For example, it is most likely you will earn more extra credit if
the class spins the spinner with the options of 5 and 10 points.
b: Reasoning will vary, but now the first spinner is definitely more attractive.
8-76.
4x 2 = 2x 2 + 17x ! 30 , x = 2.5 or 6; yes, there are two possible answers.
8-77. B
Selected Answers
59
Lesson 8.2.2
8-83. a:
3
4
b: rp
8-84. a: ≈ 403.1 cm2
8-85.
a150 = !74
c: ar 2
b: ≈ 100.8 cm2
1
2
8-86. a: 5 + 1 = 6, so two sides will collapse on the third side.
b: Answers vary. One solution is 2, 5, and 6.
8-87.
//
Given in
diagram
∠ADB ≅ ∠CBD
∠A ≅ ∠C
Given in
diagram
≅
If lines are //,
then alt. int.
angles are ≅.
Shared side
(or Reflexive
Property)
ΔADB ≅ ΔCBD
AAS ≅
≅
≅ Δs → ≅ parts
8-88. a: AAS ≅, ΔABC ≅ ΔDCB
b: ASA ≅, ΔABC ≅ ΔEDC
8-89. D
60
Core Connections Geometry
Lesson 8.3.1
8-93. Area of the entire pentagon ≈ 172.05 square units, so the shaded area
≈ 53 (172.05) ! 103.23 square units.
8-94. a: x = 14 3 , 30°-60°-90° pattern.
b: x ≈ 5.78, Law of Sines.
c: No solution, hypotenuse must be longest side.
d: 24 units, triangle area formula.
8-95. 168°
8-96.
BC ! DC and ∠A ≅ ∠E (given) and ∠BCA ≅ ∠DCE (vertical angles are ≅).
So ΔABC ≅ ΔEDC (AAS ≅) and AB ! ED ( ! s " ! parts ).
!
8-97. a: x + x + 125º + 125º + 90º = 540º, x = 100°
b: 6x + 18º = 2x + 30º, x = 3°
8-98. a: (1.5, 5)
b: (3, 7)
c: y =
4
3
x+3
d:
2 2 + 1.5 2 = 6.25 = 2.5 units
8-99. B
Lesson 8.3.2
8-105. The area of the hexagon is 24 3 square units, so the side length of the square is
24 3 ! 6.45 units.
8-106. a: A = 42 square units, P ≈ 30.5 units
8-107. a: 20
b: A = 4
2
3
square units, P ≈ 10.2 units
b: ≈ 126.3 units2
8-108. a: x = 26; if lines are parallel and cut by a transversal, then alt. int. angles are equal.
b: x = 33, n = 59°; If lines are parallel and cut by a transversal, then same-side exterior
angles are supplementary.
8-109. a: 8
b: ≈ 18
c: (1, 33), (3, 11), (11, 3), or (33, 1)
8-110. a: cos!58.5° =
x
7
b: x ≈ 3.657
8-111. D
Selected Answers
61
Lesson 8.3.3 Day 1
8-116. a: C = 28 π units, A = 196 π units2
b: C = 10 π units, A = 25 π units2
c: d = 100 units, r = 50 units, area = 2500 π units2
d: A = C2/4π units2
8-117. See bold answers in table below.
Statements
Reasons
1. AB ! DE and DE is a diameter of !C .
1. Given
2. ∠AFC and ∠BFC are right angles.
2. Definition of Perpendicular
3. FC = FC
4. AC = BC
3. Reflexive Property
4. Definition of a Circle (radii must be equal)
5. ΔAFC ≅ ΔBFC
5. HL ≅
6. AF ! FB
6. !! s!"!!!parts
!
8-119. a: (3)
b: (1)
c: (4)
8-120. a: –3
b: –4
c: 2.8 and –2.8
d: (2)
d: 2 and –2
8-121. a = 97° , b = 15 , c = 68° , d = 68°
8-122. D
62
Core Connections Geometry
Lesson 8.3.3 Day 2
8-123. a: (55)(60) + 900! " 6127.4 square feet
b: 110 + 60 π ≈ 298.5 feet, 298.5 · 8 = \$2387.96 or approximately \$2,388
c: Area is four times as large ≈ 24,509.6 square feet; perimeter is twice as
long ≈ 597 units.
8-124. a: supplementary angles sum to 180º; x = 26º
b: alternate exterior angles are congruent; x = 5º
c: Triangle Angle Sum Theorem; x = 15º
d: exterior angle equals sum of remote interior angles; x = 35º
8-125. a: CD = 22 , BC = 7 , and ED = 6 ; the perimeter is 22 + 14 + 12 = 48 cm
b: 19(4) = 76 cm2
8-126. a: 16 3 ! 27.71 square units
b: 36 square units; more
c: 24 3 ! 41.57 sq. units; its area is greater than both the square and the equilateral Δ.
d: A circle
8-127. a:
1
8
b:
5
6
144
8-128. a: 2πr = 24; r = 12
! ; A = ! " 45.84 square cm
b: 2πr = 18π; r = 9; A = 81π ≈ 254.47 square cm
8-129. E
Selected Answers
63
Lesson 9.1.1
9-7.
a: Solutions vary. Possible solution shown at right.
b: Answers vary. Assuming there are no hidden cubes, V = 11 units3
4
25
9-8.
a:
9-9.
Since the perimeter is 100, each side is 10. The central angle is 360 ÷ 10 = 36°.
The right triangle has acute angles 18° and 72°. Area = 769.4 units2.
9-10.
(36 ! 9" ) ÷ 2 # 3.86 units2
9-11. a: (4, –2)
b: 196:1
3 2 0
2 0 1
1 1 1
b: (–6, – 4)
c: 9:1
c: ( 12 ,!12)
9-12. a: An area model is not possible because there are more than two events. A tree diagram
follows part (b).
b: From the following tree diagram,
2 + 2 + 1 + 1 = 1 , or,
P(long pants) = 12
12 12 12
2
# successes
total # possible
=
6 long pants
12 students
= 12 .
shoes
clothing
gender
long pants
tennis
other
tennis
shorts
male
other
0
dress or skirt
tennis
long pants
other
tennis
female
shorts
dress or skirt
other
tennis
other
c: Using M or F for gender, and L, S, or D for clothing, and T or O for shoes, the union is
{MLT, MLO, MST, FLT, FLO, FST, and FDS}. The intersection is {MLT, FLT}
9-13. A
64
Core Connections Geometry
Lesson 9.1.2
9-20. a: ≈ 986.16 square mm
b: 400% is 4 times as large. Therefore, its area increases by a factor of 42;
986.16 · 16 ≈ 15,778.61 square mm.
b: 42 units2
9-21. a: See solutions at right.
F
9-22. a: 151°
R
T
b: Yes; a regular 36-gon has interior angles of 170°.
c: 27 · 180° = 4860°
9-23. a: 2x + 4x ! 3 + 7x ! 6 + 3x + 12 + x + 10 = 540 , x = 31°
b: 4x + 20° + 5x ! 2° = 180° , x = 18°
9-24. ≈ 0.271 or 72.9% decrease
9-25. a: 2 2 units
b: (–1, 6), (x, y) ! (x, "y)
c: (8, 5)
d: (2, –3)
9-26. 240 cm3
9-27. D
Lesson 9.1.3
9-33. 24 square units; As a midsegment, DE must be half the length of BC. If the ratio of
lengths is 0.5, then the ratio of areas is 0.52 = 0.25.
9-34. Base Area = 509.23 cm2; Height = 5 cm; SA = 1438.44 cm2
9-35. Yes, by AAS ≅.
11 + 4 ! P(long and lost) , resulting in a probability of
9-36. By the Addition Rule, 0.07 = 200
200
1 % that the food took too long and the rider got lost.
2
9-37. a: 6 or –6
b: No solution because absolute value cannot be negative.
c: x = 3 or –17
9-38.
! (6)2 (14.5) = 522! in.3;
Selected Answers
522! !in.3 !·! 1!gallon
1
231!in.3
" 7.1!gal
65
Lesson 9.1.4
9-45. If she needs the balloon to double in width, then the volume will increase by a factor of 8.
That means the balloon requires 24 breaths to blow it up. Since she has already used 3
breaths, she needs 21 more to fill the balloon.
9-46. a: SA = 180π ≈ 565.5 in.2; V = 324π ≈ 1017.9 in.3
b: 324π · 27 = 27,482.65 in.3
9-47. Circumference of each circle = 10π; total distance = 20π ≈ 62.8 feet
9-48. a: x ≈ 10.3
b: No solution because the hypotenuse must be the longest side of a right triangle.
c: The length of the base of the composite triangle must be 6 3 .
The smaller triangle has a base length 6 3 ! 5 " 5.39 ; x ≈ 8.07
9-49. a: See diagram at right.
b: V = 16 cubic units; SA = 52 square units
F
R
T
9-50. The line should be solid, and the shading should be to the right of the line.
9-51. See the area model below. A tree diagram would have worked as well.
3
4
7
45 + 45 = 45 ! 15.6%
red oak
granite
1
5
tile
4
5
3
45
3
9
white oak
4
9
maple
2
9
4
45
9-52. D
66
Core Connections Geometry
Lesson 9.1.5
9-56. b: 2
c: 24 and 96; ratio = 4; It is the square of the linear scale factor.
d: 6 and 48; ratio = 8 = 23. It is the cube of the linear scale factor.
9-57. a: Height of the tank = 6 3 ! 10.4 in., so V = 7!·!13!·!6 3 = 546 3 ! 945.7 inches 3
b:
c:
25 fish
945.7 in 3
25 fish
945.7 in 3
!
0.026 fish
1 in 3
or about 0.026 fish per cubic inch.
inch 12 inch 12 inch
! 12
1 foot ! 1 foot ! 1 foot "
45.68 fish
1 ft 3
9-58. a: False (isosceles trapezoid)
b: True
c: True
9-59.
d: False (parallelogram)
2(12x + 7) = 30x ! 4 , so x = 3
9-60. a: x = –2
b: x = 5,!! 12
9-61. a: 0.85
b: f (t) = 27000(0.85)t
c: x = 2
d: x = –2, 3
c: \$11,980
9-62. a: It is possible.
b: Not possible. Same-side interior angles should add up to 180°.
c: Not possible. One pair of alternate interior angles are equal, but the other is not for the
same pair of lines cut by a transversal; or, the vertical angles are not equal.
9-63. D
Selected Answers
67
Lesson 9.2.1
9-68. a: A rhombus
b: They are perpendicular and intersect each other at their midpoints. Or, the diagonals
bisect the angles of the quadrilateral.
c: 32 3 ≈ 55.4 units2
d: 32 3 ≈ 55.4 units2
9-69. a: See diagram at right.
b: 3000 ft3
20'
3
c: 4.8 ft
d: 0.2 butterflies per cubic foot because
9-70.
625butterflies
3000ft 3
.
! 0.21 butterflies
3
15'
10'
ft
4x ! 5° = 2x + 9° ; x = 7°
9-71. a: Not congruent since side ratio ≠ 1
b: ΔABC ≅ ΔDBC (HL≅)
c: ΔABC ≅ ΔMLK (SSS≅)
9-72.
y = 6x ! 2
9-73. See tree diagram below.
9-74. B
Bread
Protein
salami
white
turkey
ham
salami
grain
turkey
ham
Condiment
mayo
plain
mayo
plain
mayo
plain
mayo
plain
mayo
plain
mayo
plain
68
Core Connections Geometry
Lesson 9.2.2
9-79. b: It got longer.
c: Although the change is minute, the line segment got a bit longer.
d: It is possible, although not using Karen’s incremental strategy. One way: Construct a
hypotenuse for a right isosceles triangle with leg lengths 1 cm.
9-80. Possible response: Construct a circle of any radius. Then choose a point on the circle and
mark off two radius lengths in each direction along the circle circumference. Connect the
original point to each of the other two points.
9-81. Area of whole circle = 16! m2; Area for dog to roam =
9-82. Height = 9 3 m, so A =
9-83.
(20+14)(9 3)
2
240
360
(16! ) =
32!
3
" 33.51 m2
= 153 3 ! 265 m2
36! " 113.10 ft 3
(7!2)!·!180°
" 128.6° , so the
9-84. Equilateral triangle: 360° ÷ 3 sides = 120°; regular heptagon:
7
measure of the interior angle of a regular heptagon has greater measure.
9-85. See graph at right.
x-intercept: (4, 0), y-intercept: (0, 6)
y
9-86. C
x
Selected Answers
69
Lesson 9.2.3
6 0 0
3 0 1
2 6 6
9-91. a: See diagram below right.
b: 24 cubic units
c: The volume of the new solid must be 18 the original, so the
reduced volume must be 3 cubic units.
mat plan
!
9-92. ΔABC ≅ ΔEDF (given), so ∠B ≅ ∠D ( !! s!"!!!parts ), so DG ! BG (if base angles of
a triangle are congruent, then the two sides opposite those angles are also congruent) and
ΔDBG is isosceles (definition of an isosceles triangle).
9-93. a: ≈ 74.2 m2
b: ≈ 36.2 meters
c:
6 goats
74.2 m 2
! 0.1
goats
m2
, or approximately
1
10
goat per square meter.
9-94. Using equations: n + q = 14 and 0.05n + 0.25q = 2.90 , so n = 3 and q = 11.
9-95. a: It has 8 sides.
b: ! 30177.67 ft 3
DA ! CB and m∠DAB ≅ m∠CBA (given), AB ! BA (reflexive property), so
Then ∠C ≅ ∠D ( !! s!"!!!parts ).
9-96.
!DAB !!!CBA (SAS≅).
9-97.
f (x) = 7.68(2.5) x
!
9-98. B
70
Core Connections Geometry
Lesson 9.2.4
9-103. a: 14! " 43.98 cm 3
b: BA ≈ 19.3 ft2, so V ≈ (19.3)(7) ≈ 135.2 ft3
9-104. 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). She should then construct two more circles with
radius k, centered at R and S. The fourth vertex lies where these two circles intersect.
9-105. 1000 · 180° = 180,000°
y
9-106. ≈ 5626.5 square miles
b: f (x) = 10(2.3) x
9-107. a: See graph at right.
c: y = 42, 000(0.75)5 , 9967
x
d: 60 = 25(b)10 ,!b = 1.09, 9% increase
9-108. a:
b:
9-109. C
Lesson 10.1.1
10-6. a: V = (2)(5)(6) ! "(0.5 2 )(6) # 55.3 cm 3
b: Answers vary. One possibility: It could represent a pencil sharpener.
10-7. a: 70°
b: 50°
c: 2x
10-8. She is constructing an angle bisector.
10-9. Height = 6 3 , area = 15(6 3) + 16 (144!) = 90 3 + 24! " 231.3 in.2 ;
perimeter = 15 + 12 + 12 + 12 + 15 + 16 (24!) = 54 + 4! " 66 in.
10-10. a: 17, !17,17, !17,17 ; geometric
c: 81, 81, 81, 81, 81 ; arithmetic and geometric
b: 32,11, 12 , !4 43 , !7 83 ; neither
10-11. They intersect only once at (3, 5).
10-12. D
Selected Answers
71
Lesson 10.1.2
10-18. a: 64°
b: 128°
c: 64°
d: 180°
e: 128°
f: 52°
10-19. Central angle = 3.6°; A ≈ 795.51 square units
b: 2(x + 4 ! ) = 3x ! 9! , x = 17°
10-20. a: 5m + 1 = 3m + 9, m = 4
d: 18t = 360! , t = 20°
c: ( p ! 2)2 + 6 2 = p 2 , p = 10
10-21. a: D(0, 4) and E(4, 7)
b: DE = 5 units, so AC should be 10 units long.
6 2 + 8 2 = 10 units
c:
10-22. b: 108°, inscribed angle
c: 72°, central angle
10-23. a: 9 cubic units
b: 10 cubic units
d: 216°, yes
c: Possible response: There are no “floating” blocks and that there are no hidden blocks
behind the ones visible.
d: Either a left or back view would reveal any hidden blocks.
10-24. D
Lesson 10.1.3
10-31. a: 3
b: 6
c: 2
d: 1
e: 4
f: 5
! is a dilation of AB
! from P.
10-32. a: They are similar. CD
!
b: They have the same measure because they have the same central angle. However, CD
is longer; it is similar to a part of a circle with shorter diameter.
c:
60
360
(28! ) " 14.7 units
10-33. OY ! KY ! EY ! PY (all radii are congruent), !PYO " !EKY (arc measures are equal),
so !POY " !EKY (SAS ! ). Therefore, PO ! EK because ! "s #! parts.
10-34. It must be a rhombus and it could be a square.
10-35. ! (2)2 (4)
45 " 6.28 ft 3
( 360
)
10-36. a: Yes (AA ~), ΔABC ~ ΔLKH
b: Not enough information .
c: Yes (AA ~), ΔABC ~ ΔEDC
10-37. C
72
Core Connections Geometry
Lesson 10.1.4
10-43. a: 50°
b: 50°
c: 67°
10-44. a: x ≈ 31.9°, y ≈ 10.5 in.
d: 126°
e: 54°
f: 63°
b: x ≈ 3.7 m
c: x ≈ 34.7°, y = 250 ! 15.8 ft.
10-45. a: 4 times
c: 360! ÷ 5 = 72
b: 360
10-46. area of the regular pentagon ! 61.9 ft 2 ; total SA ! 2(61.9) + (5)(6)(12) ! 483.9 ft 2 ;
volume ! (61.9)(12) ! 743.3 ft 3
10-47. a: It is a square. Demonstrate that each side is the same length and that two adjacent
sides are perpendicular (slopes are opposite reciprocals).
b: C ! is at (–5, –8) and D!! is at (–7, 4).
10-48. The angles, from smallest to largest, measure 64°, 90°, 116°, 130°, and 140°, so the
probability is 45 .
10-49. C
Lesson 10.1.5
10-54. MA = 14 + 17 + 6 = 39 feet, MB = 6 feet, so AB = ER and ER = 1485 ! 38.5 feet
10-55. a: The arc length is
b: The arc length is
"
30
2
360 ! 2"(3) = 2 , so the ratio 3
5! , so the ratio is 6 = ! .
6
5
6
=
!
6
.
c: Possible response: Because all circles are similar, circle D is similar to circle E.
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). Therefore we can write a
arc length D
arc length E
proportionality equation, radius D = radius E . In fact all sectors with 30º central
arc length
angles will be similar, so the ratio of radius , that is, the radian, will be the same for
all 30º sectors.
10-56.
1
2
(9)(12) 12 (20)h, h ! 5.4 "
10-57. The ratio of the volumes must be
( 15 )3 , so the volume must be ( 15 )3 ( 500! ) = 4! cm 3 .
10-58. a: 1.05
b: ≈ \$130,588
c: f (t) = 130588(1.05)t
10-59. a: A ≈ 549.78 sq. cm
b: A ≈ 74.21 sq. cm
c: A ≈ 25.02 sq. cm
10-60. E
Selected Answers
73
Lesson 10.2.1
10-66. a: No, there may be red marbles that she has not selected in her draws.
b: No, it is less likely that there are red marbles, but no number of trials will ever assure
that there are no red ones.
c: This is not possible, no number of draws will assure this.
10-67. a:
b:
1
6
5
6
;
1
2
(60) = 50 times
c: Answers vary, but two squares should have an even number, no numbers can be 3, and
all numbers must be less than 5.
10-68. a: 13
b: 6.5
c: 67.4º
d: 134.8º
10-69. a: Have Ken’s analyzed for \$30, and use the ratios of similarity to calculate the data for
Erica’s nugget themselves.
b: Since r = 5, the ratio of the areas is 25. Thus, 20 ! 25 = 500 cm 2 .
c: Since r = 5, the ratio of the volumes is 5 3 = 125 . Thus, Erica’s nugget weighs
125 ! 5.6 = 700 g , which is about 25 ounces.
10-70. 2x + 3x + 4x + 5x , x ! 25.7
10-71. a: True
b: False
c: False
d: True
10-72. B
74
Core Connections Geometry
Lesson 10.2.2
b: y = 2x or x =
10-78. a: x = y
1
2
c: 3y = 5x
y
d: x + y = 180
10-79. a: Yes, because of the Triangle Sum Theorem, 180 ! 64 ! 26 = 90 .
0.2 ft
10-81. a: See diagram at right.
b:
5 ·5
6 6
=
25
36
0.2 ft
0.
25
ft
b: 1.4 ft 3
0.
25
10-80. a: See diagram at right.
ft
b: Yes, because 8 2 + 15 2 = 17 2 .
1 ft
red
! 69.4%
5 ! 5 = 25 " 69.4%
6 6
36
5 (49! ) " 128.28
d: 6
square cm
1
1 (360°) = 30°
e: 1! 14 ! 23 = 12
, 12
c:
blue
1 ft
1.5 ft
10-82. 540°
10-83. a: Both are
!
4
. Note that a radian measure is a ratio and does not have units.
b: If x is the unknown central angle of the circle, then the equation,
solved for x. x = 60º
x (2! r)
360
r
= !3 , can be
10-84. B
10-85. a: P(on-campus given Engr) =
120
800+120
! 13.0%
b: P(on-campus) = 0.6
c: No. The probability of living on campus given that the student is an engineer is much
smaller than the probability of living on campus.
10-86. Region A is 14 of the circle, so it should result 14 (80) = 20 times. Regions B and C have
equal weight (which can be confirmed with arc measures), so they should each result
(80 – 20) ÷ 2 = 30 times.
= 40
! = 2(97" ) = 194 " ; m!C = 0.5(1.94) = 97
b: mAD
! = 125º and the length of AB
! = 125º (16! ) " 17.5 " ; area =
c: mAB
360º
10-87. a:
360
9
125º
360º
(64! ) " 69.8 in.2
10-88. Methods vary, a variety of relationships such as Parallel Line Angle Conjectures, the
Exterior Angle Theorem, or the Triangle Sum Theorem can be used. x = 109°, y = 71°,
z = 99°.
10-89. a: x = 34
b: x =
4
3
c: x = !5
d: x =
32
5
10-90. She is incorrect, which can be tested by substituting both answers into the equation;
w = !6 or 4 .
10-91. A
Selected Answers
75
Lesson 10.2.3
10-101.
a:
236
236+274
= 46.3%
b: P(laptop) ≠ P(laptop given business trip) so they are associated.
10-102.
a: See table below. Entries not in bold are given in problem statement, and entries in
bold are computed from given information.
Not
D
a
i
l
y
Daily
Weekly local
25%
12%
37%
Not weekly local
40%
23%
63%
65%
35%
100%
b: 25% + 12% + 40% = 77%
c:
25+40
77
! 84.4%
10-103.
105.75 = ! (2.5 2 )h , h = 107.75
6.25! " 5.49 in.
10-104.
! ; Since the angle is doubled, the arc length must be doubled, and in turn, the ratio
of arc length to radius, or radian measure, must also double and thus be ! .
10-105.
a: See views at right. Depending on how they are oriented, the
top and bottom views are the same. (Or they could be
reflections of each other.)
F
R
T
b: V = 7 units 3 , SA = 28 units2; Methods vary.
c: Since this solid has no “holes,” the surface area can be calculated by adding the
areas of each of the views.
10-106.
b:
e:
10-107.
AB
AC
AB
AC
c:
f:
BC
AB
BC
AC
d:
BC
AC
a: If x represents the length of chord AC , then x 2 = 10 2 + 10 2 ! 2(10)(10) cos 80º ;
x ≈ 12.9
b: 18
76
Core Connections Geometry
Lesson 10.3.1
10-114.
a: 24
b: The decision chart tells how many branches there are at each stage.
1
6
10-115.
5! = 120
10-116.
a:
10-117.
a: See table at right.
b:
c:
10-118.
12
60
= 20%
1
9 1
9
1+ 1
9 18
b:
9
45
= 20%
c: Yes
CIRCLES
red
! 66.7%
Edge = 1200/12 = 100 cm,
SA = 6(100 2 ) = 60, 000 cm 2 ;
V = 100 3 = 1, 000, 000 cm 3
SQUARES
c:
yellow
yellow
red
blue
1
10-119.
Area of shaded region ! 93.53 " 84.82 ! 8.71 in.2
10-120.
C
Lesson 10.3.2
a: 7! = 5040
10-128.
12 P3
10-130.
507
! 0.025
No, they should not charge a higher premium. P(ticket given red) = 507+19493
507+9
and P(ticket) = 20,000+348 ! 0.025 . Since they are approximately the same, they are
most likely independent.
10-131.
a: 30%
=
12!
(12!3)!
b:
1
5040
10-127.
= 12!
9! = 12 "11"10 = 1320
b: 42%
c: P(Green Fang) ! P(alarm) " P(Green Fang and alarm) , 0.64 ! 0.28 " 0.22 , so they
are weakly associated.
10-132.
158,184, 000 ! 17, 576, 000 = 140, 608, 000 ;
10-133.
a: cylinder
Selected Answers
1
175760
b: triangle based prism
c: cone
77
Lesson 10.3.3
10-139.
9! = 362, 880
10-140.
a: 1
b: 8 things taken 8 at a time.
would be dividing by 0.
10-141.
= 2!,
c:
3!
3
a:
10 P8
2!
2
= 1!,
1!
1
8 P8
should equal 8!.
8 P8
=
8!
(8!8)!
=
8! .
0!
If 0! = 0 , you
= 0!
= 10!
2! = 1, 814, 400
P
10! = 45
b: 10 C8 = 108! 8 = 8!2!
6! = 6
c: 6 C1 = 1!5!
10-142.
Because they are independent, P(coffee and dairy) = P(coffee)  P(dairy) =
21
( 42
63 )( 63 ) ! 22.2% . P(no caffeine and no dairy) = P(no caffeine)  P(no dairy) =
63!42
( 63 )( 63!21
63 ) " 22.2%. Avoid the common mistake of computing
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.
10-143.
a:
16 P6
= 16 !15 !14 !13!12 !11 = 5, 765, 760
b: 1! 15 P5 = 1!15 !14 !13!12 !11 = 360, 360
c:
d:
360,360
5,765,760 = 6.25%
15 C 5 = 3003 = 37.5%
8008
16 C 6
10-144.
a: 20
10-145.
a: 5! = 5 ! 4 ! 3! 2 !1 = 120
b:
c:
b: 30
c: 36
5! = 5!4!3!2!1 = 60
2!
2!1
5! = 5!4!3!2!1 = 30
2!2!
2!1!2!1
d: Because you cannot tell the repeated letters apart, there are fewer arrangements
when there are repeated letters.
78
Core Connections Geometry
Lesson 10.3.4
= 20, 708, 500
10-153.
500 C 3
10-154.
3! 7 ! 2 !1! 4 = 168 choices
10-155.
a:
22 P3
= 9240
b: This is really a permutation lock. In this case the common use of the word
“combination” conflicts with the mathematical meaning.
c:
= 1540 , but this does not make sense for a mechanical lock because it would
imply dialing the numbers in any order to open the lock.
22 C 3
d: 22 ! 21! 21 = 9702
10-156.
a: 162º
b: 16 sides
10-157.
a: 12; (12 – 1), (12 – 2), (12 – 3)
c: ! 120.8 cm 2
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.
5! = 1 . In order to have the formula give a
a: There is one way to choose all five. 5!0!
reasonable result for all situations, it is necessary to define 0! as equal to 1.
b: There is one way to choose nothing.
5!
0!5!
=1
c: This is a permutation in which some of the items occur multiple times, like in an
8! = 168 arrangements of cars.
anagram. 2!5!
Selected Answers
79
Lesson 10.3.5
= 161, 700
10-173.
100 C 3
10-174.
The first because 3! > 2!
10-175.
a: (n ! 3)(n ! 4)(n ! 5)(n ! 6)(n ! 7)
b: n + 2 , n + 1 , n ! 1 , n ! 2
c: n(n ! 1)(n ! 2)
d: (n + 2)(n + 1)(n ! 1)
10-176.
a: See diagram at right. Entries not
bolded are given in the problem
statement, while bold entries are
computed.
b: P(Senior!Ocean View) =
0.06
0.24
(0.60)(0.10) = 0.06
Not
Ocean
View
0.04
0.10
(0.20)(0.90) = 0.18
0.72
0.90
0.24
0.76
1.00
OceanView
Senior
Not
senior
= 25%
c: They are associated because
P(senior) ! P(Ocean View) " P(senior and Ocean View) ,
this is (0.10)(0.24) ! (0.06) .
10-177.
≈ 0.83 or a 17% decrease
15 cm
10-178.
a: See diagram at right.
b: h 2 = 17 2 ! 15 2 = x 2 , x = 8 , A(base) = 0.5(8)915) ! 60,
600 = 60h , h = 10 cm
c: TSA = 2(60) + 17(10) + 8(10) + 15(10) = 520 sq. cm
d: 90º; sin a =
8
17
, a = 28º; sin b =
15
17
17 cm
; b = 62º
= 1586
10-179.
12 C5 , 12 C 4 , 12 C 3 , 12 C2 , 12 C1, 12 C0
10-180.
This is permutation with some of the items to be arranged occurring multiple timers,
10! = 302, 400 ways to layer the dip.
like an anagram. 3!2!
10-181.
If t = hours, 2 + 14 t = 9 ; 28 hours
10-182.
a: Similar
10-183.
a:
b:
FRONT
10-184.
80
b: Similar
R
IG
H
T
FRONT
R
IG
H
T
SA: A(!) = 12 (8)(4 3) ≈ 27.71; A(face) = (8)(10) = 80;
TSA = 2(27.71) + 3(80) ≈ 295.43 sq. cm. V = 27.71(10) ≈ 277.13 cu. cm.
Core Connections Geometry
Lesson 11.1.1
11-8. a: Yes; the sum of the interior angles only depends on the number of sides of the polygon.
b: No; only regular hexagons have a guarantee of having an angle with the measure 120°.
c: Yes; that is the definition of hexagon.
11-9. a: 7! = 5040
b: 1! 5!!1 = 120
11-10. Lateral surface area = (circumference of the base)(height) = 8! "15 = 120! # 377 cm 2
11-11. a: 2x = 180 ! 106 , x = 37°
b: x + 67 = 180 , x = 113! , 5y + 3y ! 16 = 180 , y = 24.5
c:
sin 73
9
=
sin 57
x
=
sin 50
y
, x ! 7.9 , y ! 7.2
d: 4x ! 2 + 2(8x ! 9) = 180 , x = 10
11-12. a: A = 144 square units, P = 84 units
b: A = 16 square units, P = 28 units
11-13. a: A sphere.
b: A cylinder with a cone on top and bottom.
c: A double cone: two cones attach at the vertices.
11-14. B
11-15. a: 4
11-16.
b: 6
c: 4
n!
(n!4)!
11-17. (10)(12)h = 840 , so h = 840 ÷ 120 = 7 mm
11-18. a: b is larger, even though we are not told that b is a central angle.
b: The missing angle = 180 ! 62 ! 70 = 48 and since the angle opposite side a is
bigger, a must be larger than b.
c: a = 9 3 ! 15.6 units2 and b = 16 units2, so b is larger than a.
11-19. a: x 2 + y 2 = r 2
y
b: sin ! = r ; y
c: cos ! = xr ; x
11-20. If the circle’s center is C and if the midpoint of AB is D, then ΔADC is a 30°-60°-90°
triangle. Then the radius, AC , is 10 units long and the area of the circle is
100! " 314.16 units2 .
11-21. C
Selected Answers
81
Lesson 11.1.2
11-25. 12 inches
11-26. a:
23 P3
= 10, 626
b:
c: 1! 22 ! 22 = 484
23 P3
= 10, 626
d: 44 ! 22 ! 21 = 1848
11-27. The areas are all equal because the triangles have the same base and height.
11-28. a: x = 12.9
b: x = 2
c: x = 0
c: x =
23
9
! 2.56
11-29. PA = PA (Reflexive Property), m∠PBA = m∠PCA = 90º (tangents are ⊥ to radii drawn to
the point of tangency), PB = PC (radii of a circle must be equal), so ΔPAB ≅ ΔPAC (HL).
Therefore AB = AC ( !! s!"!!!parts ).
!
b: tan ! = 86 ,!! " 53.13
11-30. a: 10
11-31. ±6, 18, ±54
11-32. D
Lesson 11.1.3 (Day 1)
11-39. a: Yes. One way is to cut off a corner so the cross-section is a triangle.
b: A tetrahedron (also called a triangle-based pyramid).
11-40. V =
11-41. a:
1 (6 2 )(4) =
3
900 C12
48 units3; slant height =
! 5.48 "10 26
b:
899 C11
11-42. a: 8 faces, 12 edges, and 6 vertices
11-43. a: x = 12, y = 7.5
b:
4
3
32 + 4 2 = 5; SA = 4( 12 ·6 ·5) + 6 2 = 96 units2
! 7.30 "10 24
c: ! 1.3%
b: A square
c: 48 square units
11-44. A = 16! square units; C = 8! units
11-45. B
82
Core Connections Geometry
Lesson 11.1.3 (Day 2)
11-46. region B, since area ! 17.46 units2 and perimeter ≈ 24.3 units
11-47. a: ! (1.5)2 (4.5) " 31.8 in.3
b: Volume of the pot is ! (7)2 (10) " 1539.4 in.3 . Therefore, Aimee would need
1539.4 ÷ 31.8 = 48.4 or 49 cans of soup to fill the pot.
c: 2! (1.5)(4.5) " 42.4 in.3
11-48. m∠ECB = m∠EAD (given), AE ! CE (definition of midpoint), ∠DEA = ∠BEC
(vertical angles are congruent), ΔAED ≅ ΔCEB (ASA ≅), so AD ! CB
( !! s!"!!!parts ).
!
11-49. 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. a:
371
1000
250 ! 152 = 469 = 46.9%
+ 1000
1000
1000
469 = 531 = 53.1%
b: 1! 1000
1000
250 = 750 = 75%
c: 1! 1000
1000
d: P(A given under 20) =
152
152+54+44
11-51. a: 18x + 174° = 540° , x = 20 13 °
c: 2(x + 4°) = 134° , x = 63°
= 60.8%
b: x 2 + (x + 17)2 = 25 2 , x = 7
d: x + 9° + 3x + 15° = 180° , x = 39°
11-52. E
Selected Answers
83
Lesson 11.1.4
11-58. a: BA = 12 (7)(24) = 84 in.2 , V = (84)(12) = 1008 in.3 ,
SA = (2)(84) + (12)(7 + 24 + 25) = 840 in.2
b: BA = 25! 2 , V = 13 (25! )(12) = 100! m 3 , lateral SA = ! (5)(13) = 65! m 2 ,
total SA = 25! + 65! = 90! " 282.7 m 2
11-59. a: 8 C3 = 56
b: There are 6 choices left for the third filling.
c:
7 C2
8 C3
=
21
56
= 37.5%
3!2!1 = 6 . You cannot use combinations
11-60. Only one right triangle can be built: 3-4-5; 6!6!6
216
because each combination is not equally likely. For example, rolling a combination
of 1, 1, 2 on the dice (where order does not matter) is three times more likely than
1, 1, 1.
11-61. a: Vertically through the vertex of the cone.
b: A circle
c: A sphere
11-62. a: !ABC ! FED (AA~)
b: !ABC ~ !MKL (SSS~)
c: Not similar because the zoom factors for corresponding sides are not equal.
11-63. See graph at right.
11-64. a: Corresponding angles, !
b: Alternate interior angles, !
c: Straight angle, supplementary
d: Alternate exterior angles, neither because lines intersect.
11-65. A
84
Core Connections Geometry
Lesson 11.1.5
11-72. r = 4 cm, SA = 4! (4 2 ) = 64! " 201.1 cm 2 , V =
11-73.
8 C 3 ! 10 C 3
11-74. a:
4
3
! (4)3 =
256
3
! " 268.1 cm 3
= 6720
12 C 3 + 12 C 3
= 715
b: If raspberry and custard are known filling, then there are two fewer fillings to choose
C + C
55 ! 7.7% .
from, so 10 271510 1 = 715
11-75. small cone:
2
5
= 6r , r = 2.4"; V = 13 ! (2.4)2 (2) " 12.06 in.3 ;
large cone: V = 13 ! (6)2 (5) " 188.50 in.3 ;
new volume = 188.50 ! 2(12.06) " 164.37 in.3
11-76. a: An icosahedron has 20 faces, so the surface area is (20)(45) = 900 mm 2 .
b: Since it has 12 faces, 180 ÷ 12 = 9 cm 2 .
c: The area of each face is
1
2
(6)(3 3) = 9 3 ! 15.6 in.2 , so total SA = 4(15.6) ! 62.4 in.2 .
11-77. ≈ 29°
11-78. 72! square units
11-79. a: ( 43 ,!10)
Selected Answers
b: (!1, !9)
85
Lesson 11.2.1
11-84. a: Earth’s circumference: 8000! = 25,132.7 miles. Therefore the distance to the moon
is 230, 000 ÷ 25,132.7 ! 9.5 times greater.
b: cos 89.85° =
238,900
x
!!
11-85. C ! 102.6 , so mAB
, x ! 91, 253,182.4 miles
32
360
(102.6) ! 9.1 mm
11-86. a:
1 (9 2 )(12) =
3
11-87. a:
52 C2 = 1326
16 C 2 = 120 ! 9.0% or using permutations 16 P2
1326
1326
P
52 2
C
16 2
52 C2
12 C 2 = 66 ! 5.0% or using permutations 12 P2
1326
52 C 2
52 P2
4 C1 ! 12 C1 = 48 " 3.6% or using permutations,
1326
1326
b:
c:
d:
e:
11-88. a:
108
177
! 61%
324 cm 3
b: 12 cm3
b:
44
88
= 16!15
52!51 " 9.0%
=
12!11
52!51
" 5.0%
4!12 " 3.6%
2 ! 52!51
= 50%
11-89. Answers vary. Sample responses: The rings in a tree, the ripples created when a stone is
tossed in a pond, the rings of a dartboard, etc.
11-90. D
86
Core Connections Geometry
Lesson 11.2.2
3
!
" 0.977 in., x ≈ 2.5 2 ! 0.977 2 " 2.3 inches
11-96.
radius of slice =
11-97.
The surface area of the moon ! 4" (1080)2 ! 14, 657, 414.7 which makes it larger
than Africa and smaller than Asia.
4 C1!6 C1!2 C1
12 C3
11-98.
, or permutations and the Fundamental Principle of
= 12
55 " 21.8%
Counting could be used: there are 3 P3 = 6 ways to arrange the colors, and
6!4!2
12 P3 = 12 !11!10 ways to arrange twelve pens, so, 6 ! 12!11!10 = 21.8% .
11-99.
Central angle = 36°, distance from center to midpoint of side = 30.777 units,
A = 12 (20)(30.777)(10) ! 3077.7 square units
11-100.
V = 324 ! 12 = 312 cm 3
11-101.
!ART ! !PIT by AA~
11-102.
a: V = 2100 units 3 , SA = 1007.34 units2
!
2
b: V = 1000! cm 3 , SA = 1100
3 + 240 " 1391.92 cm
2
c: V = 60 in.3 , SA = 144 in.
11-103.
D
Selected Answers
87
Lesson 11.2.3
11-110.
a: x = 270
11-111.
a: a = 44°, b = 28°, c = 56°
b: x = 132° , y ! 15.7
c: 3(x + 2) = 6x , x = 2
b: Some students may just report that c is smaller. 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.
11-112.
Since P(academic)  P(Arts) = P(academic and Arts) the events are independent.
There is no association between winning an academic award and a Fine Arts
award. Another possible method is P(Arts given academic) = P(arts), or that
P(academic given Arts) = P(academic).
11-113.
Base area = 36 units2, slant height = 109 ! 10.44 units,
lateral SA = 12 109 ! 125.3 units2, total SA ! 36 + 125.3 ! 161.3 units2
11-114.
a: ≈ 436,000 miles
b: The sun’s radius is almost double the distance between the Earth and the moon.
That means that if the sun were placed next to the Earth, its center would be
farther away than the moon!
c: 1,295,029
11-115.
V (16)(16)(16) = 4096 units3; SA = (6)(16)(160 = 1536 units2
11-116.
C
11-117.
a: x = 117°, y = 88 ! 9.4
b: r =
5
sin 25°
! 11.8 , z = 310°
c: 9(9 + a) = 8(21) , a = 9 23
11-118.
V (prism) = (34)(84)(99) = 282, 744 units3;
V (cylinder) = ! (38)2 (71) " 322, 088.6 units3, so the cylinder has more volume.
11-119.
6 C3
11-120.
21° + x = 2(62°) , x = 103°
11-121.
+ 2 !6 C2 + 6C1 = 20 + 2(15) + 6 = 56
f (x) = 4( 23 ) x
11-122.
The solution is shown with dashed lines in the diagram above.
11-123.
D
88
Core Connections Geometry
Lesson 12.1.1
12-6
a: x 2 + y 2 = 9
b: 7
12-7. a: V = 13 ! (32 )(10) = 30! " 94.2 units3
b: One method: BA = (21)(18) – (12)(12) = 234 units2, V = (234)(10) = 2340 units3
12-8. Think of this as an anagram ( 46!
!2! = 15 ).
12-9. a: 1.005
b: f (t) = 8500(1.005)t
c: ≈ \$11,465
12-10. a: x 2 + 8 2 = (x + 2)2 ; x = 15
8 ) " 28.1° , 180° ! 90° ! 28.1° " 61.9° ; Sample tools: Trigonometry and the
b: tan !1 ( 15
Triangle Sum Theorem
12-11. a: 124°
b: 25! units2
c: ≈ 12.3 units
12-12. C, by SAS ≅
Selected Answers
89
Lesson 12.1.2 (Day 1)
12-21. a: The slant height of the cone is ≈ 9.22 m, LA(cone) ≈ 6π(9.22) ≈ 173.78 m2, and
LA(cylinder) = 12π(11) = 132π ≈ 414.69 m2, so
total surface area is ≈ 173.78 + 414.69 = 588.47 m2
b: V(cylinder) = 36π(11) = 396π ≈ 1244.07 m3 and V(cone) =
so total the volume is ≈ 1244.07 + 263.89 = 1507.96 m3.
12-22. a: 36°
1 (36! )(7) =
3
84! " 263.89 m3,
b: b = c = 108°, d = 72º
12-23. a: E(1, 3) and F(7, 3) ; AB = 9 , DC = 3 , EF = 6 ; EF seems to be the average of AB
and CD.
b: Yes; EF = 4 , 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.
b: (!1, !3), r = 4
12-24. a: (2, !3), r = 5
x , x ≈ 8.17
12-25. a: sin 27° = 18
b:
sin 102°
7
=
sin 62°
x
, x ≈ 6.32
c: tan x = 64 , x ≈ 56.31°
12-26. a: Vertical angles are equal, 2x + 9° = 4x ! 2° , x = 5.5
b: The sum of the angles of a quadrilateral is 360°, 116° + (3x + 8) + 32° + (2x ! 1) = 360° ,
x = 41°
c: When lines are parallel, same-side exterior angles are supplementary, so
7x ! 3° + 4x + 12° = 180° and x = 15.55°.
12-27. D
90
Core Connections Geometry
Lesson 12.1.2 (Day 2)
12-28. a: BA ≈ 77.25 m2, LA = (8)(4)(16) = 512 m2, total SA ≈ 2(77.25) + 512 ≈ 666.51 m2
b: Slant height = 12 ft, LA = 4( 12 )(10)(12) = 240 ft2, BA = (10)(10) = 100 ft2,
total SA = 240 + 100 = 340 ft2
12-29. a: u = !1
c: k = 1!or!9
8
b: x = 5,!! 3
d: p = !4.5!or!1
y
12-30. a: Since the hypotenuse is 1, sin ! = 1 , and y = sin ! . Also, cos ! = 1x , so x = cos ! .
b: It must be 1 because of the Pythagorean Theorem.
c: Yes, this appears to be true for all angles.
12-31. a: (x ! 4)2 + (y ! 2)2 = 9
b: Pick any two points, such as those that are parallel to the axes on the circle and then
find the distance between the points using the Pythagorean Theorem.
12-32. a: 2 ! 5 ! 5 = 50
b: 2 ! 4 ! 3 = 24
12-33. a: k 2 = (8)(18) = 144 , k = 12
b: k = 7
12-34. A
Selected Answers
91
Lesson 12.1.3
12-40. Since 2πr = 40 feet, then r =
V=
4
3
20
!
2
≈ 6.4 feet; SA ≈ 4! ( 20
! ) " 509.3 square feet;
3
3
! ( 20
! ) " 1080.8 ft
12-41. a: x = 6
d: x = 30
b: x = 4 or – 4
c: x = 4
e: x = 3 or –3
f: x = 3 or –5
12-42. Answers vary. Typical cross-sections: regular hexagon, circle, rectangle, etc.
12-43. a: Yes; the graph includes the circle and all of the points inside the circle.
b: No; the graph only includes the points outside the circle. The circle itself would be
dashed.
12-44. This is an anagram of five Hs and five Ts.
order matters.
10!
5!5!
= 252 . It is not a combination because
12-45. a: The point is not on the circle. 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, 5 ) with the Pythagorean Theorem, 12 + 5 ! 32 .
b: x = –2 or 2
c: Possible answers will satisfy the equation x 2 + y 2 = 9 .
12-46. B
92
Core Connections Geometry
Lesson 12.1.4
12-50. V = 820( 12 )3 = 102.5 cm3
12-51. See solution graph at right.
a: C: (0, 0); r = 4.5
b: C: (0, 0); r = 75 ! 8.7
c: C: (3, 0); r = 1
d: C: (2, 1); r = 19 ! 4.4
12-52. a: x = 33° , y = 120°
c: z = 12 , w = 5
b: a ! 36.9° , b = 4
d: x = 55°
12-53. a: FG = 3.5 cm, BC = 14 cm
b: 16(3) – 4(3) = 36 cm2
12-54. base radius = 14 in.; V = 13 (196! )(18) = 1176! ≈ 3695 in.3
SA = ! rl = ! (14)( 520 ) " 1003 in.2
12-55. a:
4 C3
12 C3
=
4
220
= 1.8%
b:
5 C2 !4 C1
12 C3
c:
5 C3 + 4 C3 + 3 C3
12 C3
=
40
220
" 18.2%
= 10+4+1
220 ! 6.8%
12-56. B
Selected Answers
93
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)
12-65. a: This has one solution, because 4( 23
4 ! 3) = 11 .
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:
b:
3 C1!10 C3
13 C4
11
13 C4
=
=
360
715
11
715
" 50.3% , or 4 P1
! 1.5% or 4 P4
11
13 P4
3 P1 ! 10 P3
13 P4
3!10!9!8 = 50.3%
= 4 ! 13!12!11!10
11 " 1.5%
= 24 ! 17160
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° =
so x ≈ 6.95 – 4.83 ≈ 2.12
5
CP
, CP = CQ ≈ 4.83, sin 46° =
5
CR
, CR ≈ 6.95,
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
Lesson 12.2.4
12-101.
a: 60 sq. units
b: Side length ≈ 7.282 units, so area ≈ (7.282)(7) ! 50.97 sq. units.
12-102.
a: It was 80 feet above ground because y = 80 when x = 0 .
b: !16(3)2 + 64(3) + 80 = 128 feet; !16( 12 )2 + 64( 12 ) + 80 = 108 feet
c: !16x 2 + 64x + 80 = 0 ; x = 5 seconds
12-103.
They all are 30°- 60°- 90° triangles, so they are all similar to each other. In addition,
the triangles in (a) and (c) are congruent because corresponding sides have equal
length.
12-104.
a: See views at right.
b: 11 cubic units
c: The volume of the new solid must be
5 3 = 125 times the original, so the increased
volume must be 11(5 3 ) = 1375 cubic units.
12-105.
a: C: (!5, 0), r = 10
12-106.
See answers in table below and graph at right.
x
y
–4
6
12-107.
B
Selected Answers
–3
5
–2
4
Front
Right
Top
b: C: (3,1), r = 15
–1
3
0
2
1
3
2
4
y
3
5
4
6
x
97
```
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