1. Splash Screen

2. Lesson 10-1 Simplifying Radical Expressions
Lesson 10-2 Operations with Radical Expressions
Lesson 10-3 Radical Equations
Lesson 10-4 The Pythagorean Theorem
Lesson 10-5 The Distance Formula
Lesson 10-6 Similar Triangles
Chapter Menu

3. Five-Minute Check (over Chapter 9) Main Ideas and Vocabulary
Key Concept: Product Property of Square Roots
Example 1: Simplify Square Roots
Example 2: Multiply Square Roots
Example 3: Simplify a Square Root with Variables
Key Concept: Quotient Property of Square Roots
Example 4: Rationalizing the Denominator
Example 5: Use Conjugates to Rationalize a Denominator
Concept Summary: Simplest Radical Form
Lesson 1 Menu

4. (over Chapter 9)
Write the equation of the axis of symmetry, and find the coordinates of the vertex of the graph of –3x2 + 5 = y + 12x. Is the vertex a maximum or a minimum?
A. x = 2; (–2, –31); minimum
B. x = –2; (–2, –17); minimum
C. x = 2; (2, –31); maximum
D. x = –2; (–2, 17); maximum A B C D
5Min 1-1

5. (over Chapter 9)
Solve x2 + 4x = 21.
A. 7, 3
B. –7, 3
C. –7, –3
D. 7, –3 A B C D
5Min 1-2

6. (over Chapter 9)
Solve 4x2 + 16x + 7 = 0. A. B. C. D. A B C D
5Min 1-3

7. Which choice shows the graph of ?
(over Chapter 9)
Which choice shows the graph of ?
A B.
C D. A B C D
5Min 1-4

8. (over Chapter 9)
A work of art purchased for $1200 increases in value 5% each year for 5 years. What is its value after 5 years?
A. $
B. $
C. $
D. $ A B C D
5Min 1-5

9. Which equation is equivalent to x2 – 9x = –1?
(over Chapter 9)
Which equation is equivalent to x2 – 9x = –1? A. B. C. D. A B C D
5Min 1-6

10. rationalizing the denominator conjugate
Simplify radical expression using the Product Property of Square Roots.
Simplify radical expression using the Quotient Property of Square Roots.
radical expression
radicand
rationalizing the denominator
conjugate
Lesson 1 MI/Vocab

11. Key Concept 10-1a

12. Prime factorization of 52
Simplify Square Roots
Prime factorization of 52
Product Property of Square Roots
= 2 ● Simplify.
Answer:
Lesson 1 Ex1

13. A. B.
C. 15 D. A B C D
Lesson 1 CYP1

14. Multiply Square Roots Product Property Product Property
= 22 ● Simplify.
Answer: 4
Lesson 1 Ex2

15. A. B. C.
D. 35 A B C D
Lesson 1 CYP2

16. Simplify a Square Root with Variables
Prime factorization
Product Property
Simplify.
Answer:
Lesson 1 Ex3

17. A. B. C. D. A B C D
Lesson 1 CYP3

18. Key Concept 10-1b

19. Rationalizing the Denominator
Product Property of Square Roots
Simplify.
Answer:
Lesson 1 Ex4

20. Rationalizing the DenominatorB.
Product Property of Square Roots
Prime factorization
Lesson 1 Ex4

21. Rationalizing the Denominator
Divide the numerator and denominator by 2.
Answer:
Lesson 1 Ex4

22. A. A. B. C. D. A B C D
Lesson 1 CYP4

23. B. A. B. C. D. A B C D
Lesson 1 CYP4

24. Use Conjugates to Rationalize a Denominatoris
(a – b)(a + b) = a2 – b2
Answer: Simplify.
Lesson 1 Ex5

25. A. B. C. D. A B C D
Lesson 1 CYP5

26. Concept Summary 10-1c

27. End of Lesson 1

28. Five-Minute Check (over Lesson 10-1) Main Ideas
Example 1: Expressions with Like Radicands
Example 2: Expressions with Unlike Radicands
Example 3: Multiply Radical Expressions
Lesson 2 Menu

29. (over Lesson 10-1)
Simplify A. B. C. D. A B C D
5Min 2-1

30. (over Lesson 10-1)
Simplify
A. 288
B. 144 C. D. A B C D
5Min 2-2

31. (over Lesson 10-1)
Simplify A. B. C. D. A B C D
5Min 2-3

32. (over Lesson 10-1)
Simplify A. B. C.
D. 2 A B C D
5Min 2-4

33. (over Lesson 10-1)
The formula for the total surface area of a cube with side s is 6s2. The surface area of a cube is 648 square feet. What is the length of side s? A. B. C. D. A B C D
5Min 2-5

34. If x = 98c2 and c > 0, what is A. B. C. D. (over Lesson 10-1) A B C
5Min 2-6

35. Add and subtract radical expressions.
Multiply radical expressions.
Lesson 2 MI/Vocab

36. Expressions with Like Radicands
Distributive Property
Simplify.
Answer:
Lesson 2 Ex1

37. Expressions with Like RadicandsB.
Commutative Property
Distributive Property
Simplify.
Answer:
Lesson 2 Ex1

38. A.
A. 65 B. C. D. A B C D
Lesson 2 CYP1

39. B. A. B. C.
D. 3 A B C D
Lesson 2 CYP1

40. Expressions with Unlike Radicands
Answer:
Lesson 2 Ex2

41. A.
B. 305 C. D. A B C D
Lesson 2 CYP2

42. Multiply Radical Expressions
A. GEOMETRY Find the area of a rectangle in simplest form with a width of and a length of
To find the area of the rectangle, multiply the measures of the length and width.
Lesson 2 Ex3

43. Multiply Radical Expressions
First terms
Outer terms
Inner terms
Last terms
Multiply.
Prime factorization
Simplify.
Lesson 2 Ex3

44. Multiply Radical Expressions
Combine like terms.
Answer:
Lesson 2 Ex3

45. A. B. C. D. A B C D
Lesson 2 CYP3

46. End of Lesson 2

47. Five-Minute Check (over Lesson 10-2) Main Ideas and Vocabulary
Example 1: Real-World Example: Variable in Radical
Example 2: Radical Equation with an Expression
Example 3: Variable on Each Side
Lesson 3 Menu

48. (over Lesson 10-2)
Simplify
A. 8
B. 9 C. D. A B C D
5Min 3-1

49. (over Lesson 10-2)
Simplify A. B. C. D. A B C D
5Min 3-2

50. (over Lesson 10-2)
Find A. B. C. D. A B C D
5Min 3-3

51. (over Lesson 10-2)
Find A. B. C. D. A B C D
5Min 3-4

52. (over Lesson 10-2)
What is the perimeter of a rectangle whose width is meters and whose length is meters?
A. meters
B meters
C. meters
D. meters A B C D
5Min 3-5

53. What is the area of the triangle shown?
(over Lesson 10-2)
What is the area of the triangle shown?
A cm2
B cm2
C cm2
D cm2 A B C D
5Min 3-6

54. Solve radical equations.
Solve radical equations with extraneous solutions.
radical equation
extraneous solution
Lesson 3 MI/Vocab

55. Variable in Radical
FREE-FALL HEIGHT An object is dropped from an unknown height and reaches the ground in 5 seconds. Use the equation to find the height from which the object was dropped.
Original equation
Replace t with 5.
Multiply each side by 4.
Lesson 3 Ex1

56. Check by substituting 400 for h in the original equation.
Variable in Radical
Square each side.
400 = h Simplify.
Check by substituting 400 for h in the original equation.
Answer: 400 ft
Lesson 3 Ex1

57. A. 28 ft
B. 11 ft
C. 49 ft
D. 784 ft A B C D
Lesson 3 CYP1

58. Radical Equation with an Expression
Original equation
Subtract 8 from each side.
Square each side.
x = 52 Add 3 to each side.
Answer: The solution is 52.
Lesson 3 Ex2

59. A. 64
B. 60
C. 4
D. 196 A B C D
Lesson 3 CYP2

60. 0 = y2 + y – 2 Subtract 2 and add y to each side.
Variable on Each Side
Original equation
Square each side.
2 – y = y2 Simplify.
0 = y2 + y – 2 Subtract 2 and add y to each side.
0 = (y + 2)(y – 1) Factor.
y + 2 = 0 or y – 1 = 0 Zero Product Property
y = –2 y = 1 Solve.
Lesson 3 Ex3

61. Variable on Each Side Check X ? ? ? ? X 
Answer: Since –2 does not satisfy the original equation, 1 is the only solution.
Lesson 3 Ex3

62. A. 3
B. –1
C. 0
D. –3 A B C D
Lesson 3 CYP3

63. End of Lesson 3

64. Five-Minute Check (over Lesson 10-3) Main Ideas and Vocabulary
Key Concept: The Pythagorean Theorem
Example 1: Find the Length of the Hypotenuse
Example 2: Find the Length of a Side
Example 3: Standardized Test Example: Pythagorean Triples
Key Concept: Converse of the Pythagorean Theorem
Example 4: Check for Right Triangles
Lesson 4 Menu

65. Solve A. 2 B. 4 C. 16 D. no solution (over Lesson 10-3) A B C D
5Min 4-1

66. Solve A. 1 B. 2 C. 5 D. no solution (over Lesson 10-3) A B C D
5Min 4-2

67. Solve A. 88 B. 74 C. 16 D. no solution (over Lesson 10-3) A B C D
5Min 4-3

68. Solve A. 1 B. 6 C. 7 D. no solution (over Lesson 10-3) A B C D
5Min 4-4

69. (over Lesson 10-3)
A circular pond has an area of 69.3 square meters. What is the radius of the pond? Round to the nearest tenth of a meter.
A m
B m
C. 4.7 m
D m A B C D
5Min 4-5

70. (over Lesson 10-3)
The square root of the sum of a number and 3 is 15. What is the number?
A. 12
B. 144
C. 222
D. 228 A B C D
5Min 4-6

71. Solve problems by using the Pythagorean Theorem.
Determine whether a triangle is a right triangle.
hypotenuse
legs
Pythagorean triple
converse
Lesson 4 MI/Vocab

72. Key Concept 10-4a

73. Find the Length of the Hypotenuse
Find the length of the hypotenuse of a right triangle if a = 18 and b = 24.
c2 = a2 + b2 Pythagorean Theorem
c2 = a = 18 and b = 24
c2 = 900 Simplify.
Take the square root of each side.
Use the positive value.
Answer: The length of the hypotenuse is 30 units.
Lesson 4 Ex1

74. Find the length of the hypotenuse of a right triangle if a = 25 and b = 60.
A. 45 units
B. 85 units
C. 65 units
D. 925 units A B C D
Lesson 4 CYP1

75. Find the Length of a Side
Find the length of the missing side. If necessary, round to the nearest hundredth.
c2 = a2 + b2 Pythagorean Theorem
162 = 92 + b2 a = 9 and c = 16
256 = 81 + b2 Evaluate squares.
175 = b2 Subtract 81 from each side.
Use the positive value.
Answer: about units
Lesson 4 Ex2

76. Find the length of the missing side.
A. about 12 units
B. about 22 units
C. about units
D. about 5 units A B C D
Lesson 4 CYP2

77. What is the area of triangle XYZ?
Pythagorean Triples
What is the area of triangle XYZ?
A 94 units2 B 128 units2 C 294 units2 D 588 units2
Read the Test Item
Lesson 4 Ex3

78. The height of the triangle is 21 units.
Pythagorean Triples
Solve the Test Item
Step 1 Check to see if the measurements of this triangle are a multiple of a common Pythagorean triple. The hypotenuse is 7 ● 5 units and the leg is 7 ● 4 units. This triangle is a multiple of a (3, 4, 5) triangle.
7 ● 3 = 21
7 ● 4 = 28
7 ● 5 = 35
The height of the triangle is 21 units.
Lesson 4 Ex3

79. Step 2 Find the area of the triangle.
Pythagorean Triples
Step 2 Find the area of the triangle.
Area of a triangle
b = 28 and h = 21
Simplify.
Answer: The area of the triangle is 294 square units. Choice C is correct.
Lesson 4 Ex3

80. What is the area of triangle RST?
A. 764 units2
B. 480 units2
C. 420 units2
D. 384 units2 A B C D
Lesson 4 CYP3

81. Key Concept 10-4b

82. Check for Right Triangles
A. Determine whether the side measures of 7, 12, 15 form a right triangle.
Since the measure of the longest side is 15, let c = 15, a = 7, and b = 12. Then determine whether c2 = a2 + b2.
c2 = a2 + b2 Pythagorean Theorem ?
152 = a = 7, b = 12, and c = 15
225 = Multiply. ?
225 ≠ 193 Add.
Answer: Since c2 ≠ a2 + b2, the triangle is not a right triangle.
Lesson 4 Ex4

83. Check for Right Triangles
B. Determine whether the side measures of 27, 36, 45 form a right triangle.
Since the measure of the longest side is 45, let c = 45, a = 27, and b = 36. Then determine whether c2 = a2 + b2.
c2 = a2 + b2 Pythagorean Theorem
452 = a = 27, b = 36, and c = 45
2025 = Multiply.
2025 = Add.
Answer: Since c2 = a2 + b2, the triangle is a right triangle.
Lesson 4 Ex4

84. A. Determine whether the following side measures form a right triangle: 33, 44, 55.
B. not a right triangle
C. cannot be determined A B C
Lesson 4 CYP4

85. B. Determine whether the following side measures form a right triangle: 15, 12, 24.
B. not a right triangle
C. cannot be determined A B C
Lesson 4 CYP4

86. End of Lesson 4

87. Five-Minute Check (over Lesson 10-4) Main Ideas and Vocabulary
Key Concept: The Distance Formula
Example 1: Distance Between Two Points
Example 2: Real-World Example
Example 3: Find a Missing Coordinate
Lesson 5 Menu

88. (over Lesson 10-4)
In the figure show, find the length of the missing side. If necessary, round to the nearest hundredth. A B C
D. 84 A B C D
5Min 5-1

89. (over Lesson 10-4)
In the figure shown, find the length of the missing side. If necessary, round to the nearest hundredth. A B
C. 14
D. 9.9 A B C D
5Min 5-2

90. (over Lesson 10-4)
Find c if c is the measure of the hypotenuse of a right triangle, a = 5, and b = 9. If necessary, round to the nearest hundredth. A B
C. 7.48
D. 6.71 A B C D
5Min 5-3

91. (over Lesson 10-4)
Find b if c is the measure of the hypotenuse of a right triangle, a = 6, and c = If necessary, round to the nearest hundredth. A B C
D. 9.11 A B C D
5Min 5-4

92. (over Lesson 10-4)
A triangular plot of land has sides of 52 feet, 48 feet, and 22 feet. Is the plot of land a right triangle? Explain.
A. yes; 522 >
B. yes; 522 =
C. no;
D. no; 522 > A B C D
5Min 5-5

93. (over Lesson 10-4)
What is the height of an equilateral triangle whose sides measure 10 inches?
A inches
B. 5 inches
C inches
D. 10 inches A B C D
5Min 5-6

94. Find the distance between two points on the coordinate plane.
Find a point that is a given distance from a second point on a plane.
Distance Formula
Lesson 5 MI/Vocab

95. Key Concept 10-5a

96. Distance Between Two Points
Find the distance between the points at (1, 2) and (–3, 0).
Distance Formula
(x1, y1) = (1, 2) and (x2, y2) = (–3, 0)
Simplify.
Evaluate squares and simplify.
Answer:
Lesson 5 Ex1

97. Find the distance between the points at (5, 4) and (0, –2).
A. 29 units
B. 61 units
C units
D. 10 units A B C D
Lesson 5 CYP1

98. BIATHLON Julianne is sighting her rifle for an upcoming biathlon competition. Her first shot is 2 inches to the right and 7 inches below the bull’s- eye. What is the distance between the bull’s-eye and where her first shot hit the target?
Model the situation. If the bull’s-eye is at (0, 0), then the location of the first shot is (2, –7). Use the Distance Formula.
Lesson 5 Ex2

99. Distance Formula (x1, y1) = (0, 0) and (x2, y2) = (2, –7) Simplify.
Answer:
Lesson 5 Ex2

100. HORSESHOES Marcy is pitching a horseshoe in her local park
HORSESHOES Marcy is pitching a horseshoe in her local park. Her first pitch is 9 inches to the left and 3 inches below the pin. What is the distance between the horseshoe and the pin?
A. 9 in.
B. 3 in.
C. 12 in.
D in. A B C D
Lesson 5 CYP2

101. Find a Missing Coordinate
Find the value of a if the distance between the points at (2, –1) and (a, –4) is 5 units.
Distance Formula
Let d = 5, x2 = a, x1 = 2, y2 = –4, and y1 = –1.
Simplify.
Evaluate squares.
Simplify.
Lesson 5 Ex3

102. Find a Missing Coordinate
25 = a2 – 4a + 13 Square each side.
0 = a2 – 4a – 12 Subtract 25 from each side.
0 = (a – 6)(a + 2) Factor.
a – 6 = 0 or a + 2 = 0 Zero Product Property
a = a = –2 Solve. The value of a is –2 or 6.
Answer: –2 or 6
Lesson 5 Ex3

103. Find the value of a if the distance between the points at (2, 3) and (a, 2) is units.
A. –4 or 8
B. 4 or –8
C. –4 or –8
D. 4 or 8 A B C D
Lesson 5 CYP3

104. End of Lesson 5

105. Five-Minute Check (over Lesson 10-5) Main Ideas and Vocabulary
Key Concept: Similar Triangles
Example 1: Determine Whether Two Triangles Are Similar
Example 2: Find Missing Measures
Example 3: Real-World Example
Lesson 6 Menu

106. (over Lesson 10-5)
Find the distance between the points (9, 2) and (3, 10). Express answers in simplest radical form and as decimal approximations rounded to the nearest hundredth, if necessary.
A. 10
B. 6 C. D. A B C D
5Min 6-1

107. (over Lesson 10-5)
Find the distance between the points (–2, –4) and (3, 8). Express answers in simplest radical form and as decimal approximations rounded to the nearest hundredth, if necessary.
A. 13
B. 12 C. D. A B C D
5Min 6-2

108. (over Lesson 10-5)
Find the distance between the points (–5, 0) and (1, –7). Express answers in simplest radical form and as decimal approximations rounded to the nearest hundredth, if necessary. A. B. C. D. A B C D
5Min 6-3

109. (over Lesson 10-5)
Find the possible values of a if (9, –12) and (2, a) are 25 units apart.
A. –12 or 36
B. –13 or 37
C. –36 or 12
D. –37 or 13 A B C D
5Min 6-4

110. (over Lesson 10-5)
Find the length of the shorter diagonal of parallelogram ABCD shown in the figure.
A units
B units
C units
D units A B C D
5Min 6-5

111. (over Lesson 10-5)
Point P is located at (2, 3). Which point is a distance of 2 units away from point P?
A. (3, 0)
B. (2, 2)
C. (2, 0)
D. (0, 3) A B C D
5Min 6-6

112. Determine whether two triangles are similar.
Find the unknown measures of sides of two similar triangles.
similar triangles
Lesson 6 MI/Vocab

113. Animation: Similar Triangles
Key Concept 10-6a

114. Determine Whether Two Triangles Are Similar
Determine whether the pair of triangles is similar. Justify your answer.
Lesson 6 Ex1

115. Determine Whether Two Triangles Are Similar
Answer: The corresponding sides of the triangles are proportional, so the triangles are similar.
Lesson 6 Ex1

116. Determine whether the pair of triangles is similar.
A. The triangles are similar.
B. The triangles are not similar.
C. cannot be determined A B C
Lesson 6 CYP1

117. A. Find the missing measures if the pair of triangles is similar.
Find Missing Measures
A. Find the missing measures if the pair of triangles is similar.
Corresponding sides of similar triangles are proportional.
Lesson 6 Ex2

118. 216 = 18y Find the cross products. 12 = y Divide each side by 18.
Find Missing Measures
CE = 8, GI = y, ED = 18, and GH = 27
216 = 18y Find the cross products.
12 = y Divide each side by 18.
Corresponding sides of similar triangles are proportional.
CD = 18, GH = 27, ED = 18, and IH = x
18x = 486 Find the cross products.
x = 27 Divide each side by 18.
Answer: The missing measures are 27 and 12.
Lesson 6 Ex2

119. B. Find the missing measure if the pair of triangles is similar.
Find Missing Measures
B. Find the missing measure if the pair of triangles is similar.
Corresponding sides of similar triangles are proportional.
XY = 4, XZ = 10, XW = 3, and XV = a.
4a = 30 Find the cross products.
Lesson 6 Ex2

120. Answer: The missing measure is 7.5.
Find Missing Measures
a = 7.5 Divide each side by 4.
Answer: The missing measure is 7.5.
Lesson 6 Ex2

121. A. Find the missing measures if the pair of triangles is similar.
A. 14 and 28
B. 6 and 42
C. 18 and 28
D. 18 and 42 A B C D
Lesson 6 CYP2

122. B. Find the missing measure if
C. 16
D. 14 A B C D
Lesson 6 CYP2

123. SHADOWS Richard is standing next to the General Sherman Giant Sequoia tree in Sequoia National Park. The shadow of the tree is 22.5 meters, and Richard’s shadow is 53.6 centimeters. If Richard’s height is 2 meters, how tall is the tree?
Since the length of the shadow of the tree and Richard’s height are given in meters, convert the length of Richard’s shadow to meters.
Lesson 6 Ex3

124. Let x = the height of the tree.
1 m = 100 cm
= m Simplify.
Let x = the height of the tree.
Richard’s shadow
Tree’s shadow
Richard’s height
Tree’s height
0.536x = 45 Cross products
x ≈ 83.96
Answer: The tree is about 84 meters tall.
Lesson 6 Ex3

125. TOURISM Trudie is standing next to the Eiffel Tower in France
TOURISM Trudie is standing next to the Eiffel Tower in France. The height of the Eiffel Tower is 317 meters and casts a shadow of 155 meters. If Trudie’s height is 2 meters, how long is her shadow?
A. 2 m
B m
C. 3.2 m
D m A B C D
Lesson 6 CYP3

126. End of Lesson 6

127. Image Bank Math Tools Similar Triangles
Exploring Radical Expressions and Equations
CR Menu

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