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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

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

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

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

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

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

Find Answer: Example 1-2b

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

Simplify Answer: Example 1-3b

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

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

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

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

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

Simplify Simplify. Answer: Example 1-5a

Simplify Answer: Example 1-5b

End of Lesson 1

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

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

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

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

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

Simplify Answer: Example 2-2b

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

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

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

End of Lesson 2

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

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

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

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

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

Solve Answer: 60 Example 3-2b

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

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

Solve Answer: 3 Example 3-3b

End of Lesson 3

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

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

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

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 13.23 units. Example 4-2a

Find the length of the missing side. Answer: about 16.25 units Example 4-2b

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

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

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

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

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

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

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

End of Lesson 4

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

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

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

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

Example 5-2a

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

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

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

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

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

End of Lesson 5

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

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

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

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

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

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

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

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

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

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

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

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

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

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 0.98 meter. Example 6-3b

End of Lesson 6

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

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

Answers: Answer: Example 7-1a

Answers: Answer: Example 7-1a

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

65 Find cos 65° to the nearest ten thousandth. Keystrokes COS ENTER .4226182617 Answer: Rounded to the nearest ten thousandth, Example 7-2a

Find tan 32° to the nearest ten thousandth. Answer: 0.6249 Example 7-2b

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

[COS–1] ENTER 53.13010235 12 Keystrokes 20 2nd Answer: To the nearest degree, the measure of is 53°. Example 7-3a

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

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

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

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

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

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

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

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

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

End of Lesson 7

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