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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

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

(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

(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

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

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

(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. $1230.30 B. $1261.37 C. $1531.54 D. $1500.00 A B C D 5Min 1-5

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

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

Key Concept 10-1a

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

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

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

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

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

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

Key Concept 10-1b

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

Rationalizing the Denominator B. Product Property of Square Roots Prime factorization Lesson 1 Ex4

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

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

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

Use Conjugates to Rationalize a Denominator is 5 + . (a – b)(a + b) = a2 – b2 Answer: Simplify. Lesson 1 Ex5

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

Concept Summary 10-1c

End of Lesson 1

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

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

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

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

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

(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

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

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

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

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

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

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

Expressions with Unlike Radicands Answer: Lesson 2 Ex2

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

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

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

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

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

End of Lesson 2

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

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

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

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

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

(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

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

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

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

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

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

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

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

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

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

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

End of Lesson 3

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

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

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

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

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

(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. 14.86 m B. 8.32 m C. 4.7 m D. 1.49 m A B C D 5Min 4-5

(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

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

Key Concept 10-4a

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 = 182 + 242 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

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

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 13.23 units Lesson 4 Ex2

Find the length of the missing side. A. about 12 units B. about 22 units C. about 16.25 units D. about 5 units A B C D Lesson 4 CYP2

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

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

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

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

Key Concept 10-4b

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 = 72 + 122 a = 7, b = 12, and c = 15 225 = 49 + 144 Multiply. ? 225 ≠ 193 Add. Answer: Since c2 ≠ a2 + b2, the triangle is not a right triangle. Lesson 4 Ex4

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 = 272 + 362 a = 27, b = 36, and c = 45 2025 = 729 + 1296 Multiply. 2025 = 2025 Add. Answer: Since c2 = a2 + b2, the triangle is a right triangle. Lesson 4 Ex4

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

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

End of Lesson 4

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

(over Lesson 10-4) In the figure show, find the length of the missing side. If necessary, round to the nearest hundredth. A. 30.23 B. 45.75 C. 60.46 D. 84 A B C D 5Min 5-1

(over Lesson 10-4) In the figure shown, find the length of the missing side. If necessary, round to the nearest hundredth. A. 22.14 B. 19.80 C. 14 D. 9.9 A B C D 5Min 5-2

(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. 14.00 B. 10.30 C. 7.48 D. 6.71 A B C D 5Min 5-3

(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. 12.45 B. 11.18 C. 10.63 D. 9.11 A B C D 5Min 5-4

(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 > 482 + 222 B. yes; 522 = 482 + 222 C. no; 52 48 + 22 D. no; 522 > 482 + 222 A B C D 5Min 5-5

(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

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

Key Concept 10-5a

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

Find the distance between the points at (5, 4) and (0, –2). A. 29 units B. 61 units C. 7.81 units D. 10 units A B C D Lesson 5 CYP1

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

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

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. 9.49 in. A B C D Lesson 5 CYP2

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

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 = 6 a = –2 Solve. The value of a is –2 or 6. Answer: –2 or 6 Lesson 5 Ex3

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

End of Lesson 5

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

(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

(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

(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

(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

(over Lesson 10-5) Find the length of the shorter diagonal of parallelogram ABCD shown in the figure. A. 5.00 units B. 6.71 units C. 7.28 units D. 8.54 units A B C D 5Min 6-5

(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

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

Animation: Similar Triangles Key Concept 10-6a

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

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

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

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

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

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

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

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

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

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

Let x = the height of the tree. 1 m = 100 cm = 0.536 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

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. 0.98 m C. 3.2 m D. 1.45 m A B C D Lesson 6 CYP3

End of Lesson 6

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