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3. Lesson 11-1 Simplifying Radical Expressions
Lesson 11-2 Operations with Radical Expressions
Lesson 11-3 Radical Equations
Lesson 11-4 The Pythagorean Theorem
Lesson 11-5 The Distance Formula
Lesson 11-6 Similar Triangles
Lesson 11-7 Trigonometric Ratios
Contents

4. Example 1 Simplify Square Roots Example 2 Multiply Square Roots
Example 3 Simplify a Square Root with Variables
Example 4 Rationalizing the Denominator
Example 5 Use Conjugates to Rationalize a Denominator
Lesson 1 Contents

5. Prime factorization of 52
Simplify .
Prime factorization of 52
Product Property of Square Roots
Answer:
Simplify.
Example 1-1a

6. Prime factorization of 72
Simplify .
Prime factorization of 72
Product Property of Square Roots
Simplify.
Answer:
Simplify.
Example 1-1b

7. Simplify. a. b.
Answer:
Answer:
Example 1-1c

8. Find
Product Property
Product Property
Answer:
Simplify.
Example 1-2a

9. Find
Answer:
Example 1-2b

10. The absolute value of ensures a nonnegative result. Answer:
Simplify
Prime factorization
Product Property
Simplify.
The absolute value of ensures a nonnegative result.
Answer:
Example 1-3a

11. Simplify
Answer:
Example 1-3b

12. Product Property of Square Roots
Simplify .
Multiply by .
Product Property of Square Roots
Answer:
Simplify.
Example 1-4a

13. Product Property of Square Roots
Simplify .
Prime factorization
Multiply by
Answer:
Product Property of Square Roots
Example 1-4b

14. Product Property of Square Roots
Simplify .
Multiply by
Product Property of Square Roots
Prime factorization
Example 1-4c

15. Divide the numerator and denominator by 2.
Answer:
Divide the numerator and denominator by 2.
Example 1-4c

16. Simplify. a. b. c.
Answer:
Answer:
Answer:
Example 1-4d

17. Simplify
Simplify.
Answer:
Example 1-5a

18. Simplify
Answer:
Example 1-5b

19. End of Lesson 1

20. Example 1 Expressions with Like Radicands
Example 2 Expressions with Unlike Radicands
Example 3 Multiply Radical Expressions
Lesson 2 Contents

21. Distributive Property
Simplify .
Distributive Property
Answer:
Simplify.
Example 2-1a

22. Distributive Property
Simplify .
Commutative Property
Distributive Property
Answer:
Simplify.
Example 2-1b

23. Simplify a. b.
Answer:
Answer:
Example 2-1c

24. Answer: The simplified form is
Simplify
Answer: The simplified form is
Example 2-2a

25. Simplify
Answer:
Example 2-2b

26. Find the area of a rectangle with a width of and a length of
To find the area of the rectangle multiply the measures of the length and width.
Example 2-3a

27. Answer: The area of the rectangle is square units.
First terms
Outer terms
Inner terms
Last terms
Multiply.
Prime factorization
Simplify.
Combine like terms.
Answer: The area of the rectangle is square units.
Example 2-3a

28. Find the area of a rectangle with a width of and a length of
Answer:
Example 2-3b

29. End of Lesson 2

30. Example 1 Radical Equation with a Variable
Example 2 Radical Equation with an Expression
Example 3 Variable on Each Side
Lesson 3 Contents

31. Use the equation to replace t with 5 seconds.
Free-Fall Height An object is dropped from an unknown height and reaches the ground in 5 seconds. From what height is it dropped?
Use the equation to replace t with 5 seconds.
Original equation
Replace t with 5.
Multiply each side by 4.
Square each side.
Simplify.
Example 3-1a

32. Answer: The object is dropped from 400 feet.
Check
Original equation
and
Answer: The object is dropped from 400 feet.
Example 3-1a

33. Free-Fall Height An object is dropped from an
unknown height and reaches the ground in 7 seconds.
Use the equation to find from what height
it is dropped.
Answer: 784 ft
Example 3-1b

34. Subtract 8 from each side.
Solve
Original equation
Subtract 8 from each side.
Square each side.
Add 3 to each side.
Answer: The solution is 52.
Example 3-2a

35. Solve
Answer: 60
Example 3-2b

36. Subtract 2 and add y to each side.
Solve
Original equation
Square each side.
Simplify.
Subtract 2 and add y to each side.
Factor.
Zero Product Property or
Solve.
Example 3-3a

37. Check
Answer: Since –2 does not satisfy the original equation, 1 is the only solution.
Example 3-3a

38. Solve
Answer: 3
Example 3-3b

39. End of Lesson 3

40. Example 1 Find the Length of the Hypotenuse
Example 2 Find the Length of a Side
Example 3 Pythagorean Triples
Example 4 Check for Right Triangles
Lesson 4 Contents

41. Find the length of the hypotenuse of a right triangle if and
Pythagorean Theorem
and
Simplify.
Take the square root of each side.
Use the positive value.
Answer: The length of the hypotenuse is 30 units.
Example 4-1a

42. Find the length of the hypotenuse of a right triangle if and
Answer: 65 units
Example 4-1b

43. Find the length of the missing side.
In the triangle, and units.
Pythagorean Theorem
and
Evaluate squares.
Subtract 81 from each side.
Use a calculator to evaluate .
Use the positive value.
Answer: To the nearest hundredth, the length of the leg is units.
Example 4-2a

44. Find the length of the missing side.
Answer: about units
Example 4-2b

45. Multiple-Choice Test Item What is the area of triangle XYZ?
A 94 units2 B 128 units2 C 294 units2 D 588 units2
Read the Test Item
The area of the triangle is In a right triangle,
the legs form the base and height of the triangle. Use
the measures of the hypotenuse and the base to
find the height of the triangle.
Example 4-3a

46. The height of the triangle is 21 units.
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 units and the leg is units. This triangle is a multiple of a (3, 4, 5) triangle.
The height of the triangle is 21 units.
Example 4-3a

47. Step 2 Find the area of the triangle.
Area of a triangle
and
Simplify.
Answer: The area of the triangle is 294 square units. Choice C is correct.
Example 4-3a

48. Multiple-Choice Test Item What is the area of triangle RST?
A 764 units2 B 480 units2 C 420 units2 D 384 units2
Answer: D
Example 4-3b

49. Answer: Since , the triangle is not a right triangle.
Determine whether the side measures of 7, 12, 15 form a right triangle.
Since the measure of the longest side is 15, let , and Then determine whether
Pythagorean Theorem
and
Multiply.
Add.
Answer: Since , the triangle is not a right triangle.
Example 4-4a

50. Answer: Since the triangle is a right triangle.
Determine whether the side measures of 27, 36, 45 form a right triangle.
Since the measure of the longest side is 45, let and Then determine whether
Pythagorean Theorem
and
Multiply.
Add.
Answer: Since the triangle is a right triangle.
Example 4-4b

51. Determine whether the following side measures form right triangles.
b. 12, 22, 40
Answer: right triangle
Answer: not a right triangle
Example 4-4b

52. End of Lesson 4

53. Example 1 Distance Between Two Points
Example 2 Use the Distance Formula
Example 3 Find a Missing Coordinate
Lesson 5 Contents

54. Find the distance between the points at (1, 2) and (–3, 0).
Distance Formula
and
Simplify.
Evaluate squares and simplify.
Answer: or about 4.47 units
Example 5-1a

55. Find the distance between the points at (5, 4) and (0, –2).
Answer:
Example 5-1b

56. 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?
Draw a model of the situation on a coordinate grid. If the bull’s-eye is at (0, 0), then the location of the first shot is (2, –7). Use the Distance Formula.
Example 5-2a

57. Example 5-2a

58. Answer: The distance is or about 7.28 inches.
Distance Formula
and
Simplify.
or about 7.28 inches
Answer: The distance is or about 7.28 inches.
Example 5-2a

59. 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?
Answer:
Example 5-2b

60. Find the value of a if the distance between the points at (2, –1) and (a, –4) is 5 units.
Distance Formula
Let
and .
Simplify.
Evaluate squares.
Simplify.
Example 5-3a

61. Subtract 25 from each side.
Square each side.
Subtract 25 from each side.
Factor.
Zero Product Property or
Solve.
Answer: The value of a is –2 or 6.
Example 5-3a

62. Find the value of a if the distance between the points at (2, 3) and (a, 2) is
Answer: –4 or 8
Example 5-3b

63. End of Lesson 5

64. Example 1 Determine Whether Two Triangles Are Similar
Example 2 Find Missing Measures
Example 3 Use Similar Triangles to Solve a Problem
Lesson 6 Contents

65. The ratio of sides XY to AB is
Determine whether the pair of triangles is similar. Justify your answer.
The ratio of sides XY to AB is
The ratio of sides YZ to BC is
Example 6-1a

66. The ratio of sides XZ to AC is
Answer: Since the measures of the corresponding sides are proportional, triangle XYZ is similar to triangle ABC.
Example 6-1a

67. Determine whether the pair of triangles is similar. Justify your answer.
Answer: Since the corresponding angles have equal measures, the triangles are similar.
Example 6-1b

68. Find the missing measures if the pair of triangles is similar.
Since the corresponding angles have equal measures, The lengths of the corresponding sides are proportional.
Example 6-2a

69. Corresponding sides of similar triangles are proportional.
and
Find the cross products.
Divide each side by 18.
Example 6-2a

70. Corresponding sides of similar triangles are proportional.
and
Find the cross products.
Divide each side by 18.
Answer: The missing measures are 27 and 12.
Example 6-2a

71. Find the missing measures if the pair of triangles is similar.
Corresponding sides of similar triangles are proportional.
and
Example 6-2b

72. Find the cross products.
Divide each side by 4.
Answer: The missing measure is 7.5.
Example 6-2b

73. Find the missing measures if each pair of triangles is similar. a.
Answer: The missing measures are 18 and 42.
Example 6-2c

74. Find the missing measures if each pair of triangles is similar. b.
Answer: The missing measure is 5.25.
Example 6-2c

75. Shadows Richard is standing next to the General Sherman Giant Sequoia three 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.
Example 6-3a

76. Let the height of the tree.
Simplify.
Let the height of the tree.
Richard’s shadow
Tree’s shadow
Richard’s height
Tree’s height
Cross products
Answer: The tree is about 84 meters tall.
Example 6-3a

77. Answer: The length of Trudie’s shadow is about 0.98 meter.
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?
Answer: The length of Trudie’s shadow is about meter.
Example 6-3b

78. End of Lesson 6

79. Example 1 Sine, Cosine, and Tangent
Example 2 Find the Sine of an Angle
Example 3 Find the Measure of an Angle
Example 4 Solve a Triangle
Example 5 Angle of Elevation
Lesson 7 Contents

80. Find the sine, cosine, and tangent of each acute angle of
Find the sine, cosine, and tangent of each acute angle of Round to the nearest ten-thousandth.
Write each ratio and substitute the measures. Use a calculator to find each value.
Example 7-1a

81. Answers:
Answer:
Example 7-1a

82. Answers:
Answer:
Example 7-1a

83. Find the sine, cosine, and tangent of each acute angle of
Find the sine, cosine, and tangent of each acute angle of Round to the nearest ten-thousandth.
Answer:
Example 7-1b

84. 65 Find cos 65° to the nearest ten thousandth. Keystrokes COS ENTER
Answer: Rounded to the nearest ten thousandth,
Example 7-2a

85. Find tan 32° to the nearest ten thousandth.
Answer:
Example 7-2b

86. Find the measure of to the nearest degree.
Since the lengths of the adjacent leg and the hypotenuse are known, use the cosine ratio.
and
Now use [COS–1] on a calculator to find the measure of
the angle whose cosine ratio is
Example 7-3a

87. [COS–1]
ENTER 12
Keystrokes 20
2nd
Answer: To the nearest degree, the measure of is 53°.
Example 7-3a

88. Find the measure of to the nearest degree.
Answer: 29°
Example 7-3b

89. Find all of the missing measures in
You need to find the measures of and
Step 1 Find the measure of The sum of the measures of the angles in a triangle is 180.
The measure of is 28°.
Example 7-4a

90. Find the cross products.
Step 2 Find the value of y, which is the measure of the hypotenuse. Since you know the measure of the side opposite use the sine ratio.
Definition of sine
Evaluate sin 62°.
Find the cross products.
is about 17.0 centimeters long.
Example 7-4a

91. Find the cross products.
Step 3 Find the value of x, which is the measure of the side adjacent Use the tangent ratio.
Definition of sine
Evaluate tan 62°.
Find the cross products.
is about 8.0 centimeters long.
Answer: So, the missing measures are 28, 8 cm, and 17 cm.
Example 7-4a

92. Find all of the missing measures in
Answer: The missing measures are 47, 11 m, and 16 m.
Example 7-4b

93. Indirect Measurement In the diagram, Barone is flying his model airplane 400 feet above him. An angle of depression is formed by a horizontal line of sight and a line of sight below it. Find the angles of depression at points A and B to the nearest degree.
Explore In the diagram two right triangles are formed. You know the height of the airplane and the horizontal distance it has traveled.
Plan Let A represent the first angle of depression. Let B represent the second angle of depression.
Example 7-5a

94. Solve Write two equations involving the tangent ratio.
and
Answer: The angle of depression at point A is 45° and the angle of depression at point B is 37°.
Example 7-5a

95. Examine. Examine the solution by finding the horizontal
Examine Examine the solution by finding the horizontal distance the airplane has flown at points A and B.
The solution checks.
Example 7-5a

96. Indirect Measurement In the diagram, Kylie is flying a kite 350 feet above her. An angle of depression is formed by a horizontal line of sight and a line of sight below it. Find the angle of depression at points X and Y to the nearest degree.
Answer: The angle of depression at point X is 38° and the angle of depression at Y is 32°.
Example 7-5b

97. End of Lesson 7

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