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9.1 Similar Right Triangles Goals p Solve problems involving similar right triangles formed by the altitude drawn to the hypotenuse of a right triangle. p Use a geometric mean to solve problems. THEOREM 9.1 C If the altitude is drawn to the hypotenuse of a right triangle, then the two triangles formed are similar to the original triangle and to each other. A D B TCBD S TABC, TACD S TABC, and TCBD S TACD Example 1 Finding the Height of a Ramp Ramp Height A ramp has a cross section that is a right triangle. The diagram shows the approximate dimensions of this cross section. Find the height h of the ramp. K 4.7 ft J 11 ft h M 11.7 ft L Solution By Theorem 9.1, T JKL S T KML . Use similar triangles to write a proportion. KM JK h 4.7 KL JL Corresponding side lengths are in proportion. 11 11.7 Substitute. 11.7 h 11 ( 4.7 ) h ≈ 4.4 Cross product property Solve for h. Answer The height of the ramp is about 4.4 feet. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 187 Checkpoint Write similarity statements for the three triangles in the diagram. Then find the given length. Round decimals to the nearest tenth, if necessary. 1. Find AD. 2. Find NQ. C N 26 24 A D 18 Q 10 B Sample answer: TABC S TCBD S TACD; 32 M P Sample answer: TMNP S TQNM S TQMP; 3.8 GEOMETRIC MEAN THEOREMS THEOREM 9.2 C In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. A The length of the altitude is the geometric mean of the lengths of the two segments. D BD CD B CD AD THEOREM 9.3 In a right triangle, the altitude from the right angle to the hypotenuse divides the hypotenuse into two segments. CB AB CB DB The length of each leg of the right triangle is the geometric mean of the lengths of the hypotenuse and the segment of the hypotenuse that is adjacent to the leg. AB AC Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. AC AD Chapter 9 • Geometry Notetaking Guide 188 Using a Geometric Mean Example 2 Find the value of each variable. a. b. 2 x 6 8 y 4 Solution a. Apply Theorem 9.2. 8 b. Apply Theorem 9.3. 62 y x 4 x 8 y 32 x 2 y 2 y 2 x 32 16 y 2 42 x 4 y Checkpoint Find the value of the variable. 3. 4. 6 t 9 17 17 t 315 17 5. z 13 5 310 Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 189 9.2 The Pythagorean Theorem Goals p Prove the Pythagorean Theorem. p Use the Pythagorean Theorem to solve problems. VOCABULARY Pythagorean triple A Pythagorean triple is a set of three positive integers a, b, and c that satisfy the equation c2 a2 b2. THEOREM 9.4: PYTHAGOREAN THEOREM In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. c a b c2 a2 b2 Example 1 Finding the Length of a Hypotenuse Find the length of the hypotenuse of the right triangle. Tell whether the side lengths form a Pythagorean triple. Solution (hypotenuse)2 (leg)2 (leg)2 8 15 Pythagorean Theorem x2 8 2 15 2 Substitute. x2 64 225 Multiply. x2 289 Add. x 17 x Find the positive square root. Answer The length of the hypotenuse is 17 . Because the side lengths 8, 15, and 17 are integers , they form a Pythagorean triple. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 190 Finding the Length of a Leg Example 2 Find the length of the leg of the right triangle. 20 15 Solution (hypotenuse)2 (leg)2 (leg)2 x Pythagorean Theorem 20 2 x2 15 2 Substitute. 400 x2 225 Multiply. 175 x2 Subtract 225 from each side. 175 x Find the positive square root. 25 p 7 x Use product property. 57 x Simplify the radical. Answer The length of the leg is 57 . Checkpoint Find the value of x. Simplify answers that are radicals. Then tell whether the side lengths form a Pythagorean triple. 1. 2. 8 24 x x 25 12 ; No 413 Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. 7; Yes Chapter 9 • Geometry Notetaking Guide 191 Example 3 Finding the Area of a Triangle Find the area of the triangle to the nearest tenth of a square inch. 12 in. Solution You are given that the base of the triangle is 18 inches, but you do not know the height h. 12 in. h 18 in. Because the triangle is isosceles, it can be divided into two congruent right angles with the given dimensions. Use the Pythagorean Theorem to find the value of h. 12 2 9 2 h2 Pythagorean Theorem 144 81 h2 Multiply. 63 h2 12 in. h 9 in. Subtract 81 from each side. 63 h Find the positive square root. 37 h Simplify the radical. Now find the area of the original triangle. 1 A bh 2 Area of a triangle 1 ) ( 18 )( 37 2 Substitute. ≈ 71.4 Use a calculator. 3 7 in. ____ 18 in. Answer The area of the triangle is about 71.4 square inches. Checkpoint Find the area of the triangle. Round your answer to the nearest tenth. 3. 4. 14 cm 9 cm 9 cm 39.6 cm2 Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. 10 ft 21 ft 92.3 ft2 Chapter 9 • Geometry Notetaking Guide 192 9.3 The Converse of the Pythagorean Theorem Goals p Use the Converse of the Pythagorean Theorem to solve problems. p Use side lengths to classify triangles by their angle measures. THEOREM 9.5: CONVERSE OF THE PYTHAGOREAN THEOREM If the square of the length of the longest side of a triangle is equal to the sum of the squares of the lengths of the other two sides, then the triangle is a right triangle. B c a C b A If c2 a2 b2, then TABC is a right triangle. Example 1 Verifying Right Triangles Tell whether the triangle at the right is a right triangle. 8 6 Solution Let c represent the length of the longest side of the triangle. Check to see whether the side lengths satisfy the equation c2 a2 b2. ( 815 )2 ( 86 )2 24 2 8 15 24 8 2 p ( 15 )2 8 2 p ( 6 )2 24 2 64 p 15 64 p 6 576 960 384 576 960 960 Answer The triangle is Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. a right triangle. Chapter 9 • Geometry Notetaking Guide 193 Checkpoint Tell whether the triangle is a right triangle. 1. 2. 33 2 38 65 The triangle is a right triangle. 10 16 56 The triangle is not a right triangle. THEOREM 9.6 If the square of the length of the longest side of a triangle is less than the sum of the squares of the lengths of the other two sides, then the triangle is acute . If c2 < a2 b2, then TABC is acute . A c b C a B THEOREM 9.7 If the square of the length of the longest side of a triangle is greater than the sum of the squares of the lengths of the other two sides, then the triangle is obtuse . If c2 > a2 b2, then TABC is obtuse . Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. A c b C a B Chapter 9 • Geometry Notetaking Guide 194 Classifying Triangles Example 2 Decide whether the set of numbers can represent the side lengths of a triangle. If they can, classify the triangle as right, acute, or obtuse. a. 28, 40, 48 b. 5.7, 12.2, 13.9 Solution Compare the square of the length of the longest side with the sum of the squares of the lengths of the two shorter sides. a. c2 ? a2 b2 Compare c2 with a2 b2. 48 2 ? 282 40 2 Substitute. 2304 ? 784 1600 Multiply. 2304 < 2384 c2 is less than a2 b2. Answer Because c2 < a2 b2, the triangle is acute . b. c2 ? a2 b2 13.9 2 ? Compare c2 with a2 b2. 5.7 2 12.22 Substitute. 193.21 ? 32.49 148.84 Multiply. 193.21 > 181.33 c2 is greater than a2 b2. Answer Because c2 > a2 b2, the triangle is obtuse . Checkpoint Can the numbers represent the side lengths of a triangle? If so, classify the triangle as right, acute, or obtuse. 3. 16, 30, 34 yes; right 4. 8, 13, 22 no Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. 5. 6, 9, 12 yes; obtuse Chapter 9 • Geometry Notetaking Guide 195 9.4 Special Right Triangles Goals p Find the side lengths of special right triangles. p Use special right triangles to solve real-life problems. VOCABULARY Special right triangles Special right triangles are right triangles whose angle measures are 45-45-90 or 30-60-90. THEOREM 9.8: 45-45-90 TRIANGLE THEOREM In a 45-45-90 triangle, the hypotenuse is 2 times as long as each leg. x Hypotenuse 2 p leg 45 2x 45 x THEOREM 9.9: 30-60-90 TRIANGLE THEOREM In a 30-60-90 triangle, the hypotenuse is twice as long as the shorter leg, and the longer leg is 3 times as long as the shorter leg. Hypotenuse 2 p shorter leg Longer leg 3 p shorter leg Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. x 60 2x 30 3x Chapter 9 • Geometry Notetaking Guide 196 Example 1 Finding the Hypotenuse in a 45- 45- 90 Triangle Find the value of x. By the Triangle Sum Theorem, the measure of the third angle is 45 . The triangle is a 45 - 45 -90 right triangle, so the length x times the length of the hypotenuse is 2 of a leg. p leg Hypotenuse 2 Example 2 x 45 7 7 45 - 45 -90 Triangle Theorem x 2 p 7 Substitute. x 72 Simplify. Side Lengths in a 30- 60- 90 Triangle Find the values of s and t. 9 30 Because the triangle is a 30-60-90 triangle, the longer leg is 3 times the length of the shorter leg. p shorter leg Longer leg 3 9 3 ps 9 s 3 3 9 p s 3 3 33 s t s 60 30-60-90 Triangle Theorem Substitute. Divide each side by 3 . Multiply numerator and . denominator by 3 Simplify. The length of the hypotenuse is twice the length of the shorter leg. Hypotenuse 2 p shorter leg 30-60-90 Triangle Theorem t 2 p 33 Substitute. t 63 Simplify. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 197 Checkpoint Find the values of the variables. 1. 2. y x 45 10 3 45 4 3. a 60 m n 60 15 b 30 x4 a 10 m 153 y 42 b 20 n 30 Example 3 Finding the Area of a Window Construction The window is a square. Find the area of the window. 36 2 in. Solution First find the side length s of the window by dividing it into two 45 - 45 -90 triangles. The length of the hypotenuse is inches. Use this length to find s. 362 362 2 ps 36 s 45 - 45 -90 Triangle Theorem Divide each side by 2 . Use s 36 to find the area of the window. A s2 Area of a square 36 2 Substitute. 1296 Multiply. Answer The area of the window is 1296 square inches. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 198 9.5 Trigonometric Ratios Goals p Find the sine, the cosine, and the tangent of an acute angle. p Use trigonometric ratios to solve real-life problems. VOCABULARY Trigonometric ratio A trigonometric ratio is a ratio of the lengths of two sides of a right triangle. Sine A sine is a trigonometric ratio, abbreviated as sin. Cosine A cosine is a trigonometric ratio, abbreviated as cos. Tangent A tangent is a trigonometric ratio, abbreviated as tan. Angle of elevation An angle of elevation is the angle that your line of sight makes with a horizontal line when you stand and look up at a point in the distance. TRIGONOMETRIC RATIOS Let TABC be a right triangle. The sine, the cosine, and the tangent of acute aA are defined as follows. a side opposite aA sin A c hypotenuse side adjacent to aA b cos A c hypotenuse a side opposite aA tan A side adjacent to aA b Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. hypotenuse c B side a opposite aA A b C side adjacent to aA Chapter 9 • Geometry Notetaking Guide 199 Example 1 Finding Trigonometric Ratios Find the sine, the cosine, and the tangent of aP. P 8 17 Solution R 15 The length of the hypotenuse is 17 . The length of the side opposite aP is 15 , and the length of the side adjacent to aP is 8 . Q 15 opposite sin P ≈ 0.8824 17 hypotenuse adjacent 8 cos P ≈ 0.4706 hypotenuse 17 opposite 15 tan P 1.875 adjacent 8 Example 2 Trigonometric Ratios for 60 Find the sine, the cosine, and the tangent of 60. Solution Begin by sketching a 30-60-90 triangle as shown at the right. To make the calculations simple, choose 1 as the length of the shorter leg. From the 30-60-90 Triangle Theorem, it follows that the and the length of the length of the longer leg is 3 hypotenuse is 2 . Label these lengths in the diagram. __ 2 __3 60 1 3 opp. sin 60 ≈ 0.8660 2 hyp. 1 adj. cos 60 0.5 2 hyp. opp. tan 60 3 ≈ 1.7321 adj. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 200 Checkpoint Use the diagram at the right to find the trigonometric ratio. 1. sin X X 4 0.8 5 12 Z 16 20 Y 2. cos X 3 0.6 5 3. tan Y 3 0.75 4 Example 3 Indirect Measurement Flag Pole You are measuring the height of a flag pole. You stand 19 feet from the base of the pole. You measure the angle of elevation from a point on the ground to the top of the pole to be 64. Estimate the height of the pole. h 64 19 ft Solution opposite tan 64 adjacent Write trigonometric ratio. h tan 64 19 Substitute. 19 tan 64 h 19 ( 2.0503 ) h 38.9557 ≈ h Multiply each side by 19 . Evaluate tan 64. Simplify. Answer The height of the flag pole is about 39 feet. Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 201 9.6 Solving Right Triangles Goals p Solve a right triangle. p Use right triangles to solve real-life problems. VOCABULARY Solve a right triangle To solve a right triangle means to determine the measures of all three angles and the lengths of all three sides. Solving a Right Triangle Example 1 Solve the right triangle. Round decimals to the nearest tenth. B Solution Use the Pythagorean Theorem to find the length of the hypotenuse c. (hypotenuse)2 (leg)2 (leg)2 Pythagorean Theorem c2 4 2 8 2 Substitute. c2 80 Simplify. c A 8 4 C c 45 Find the positive square root. c ≈ 8.9 Use a calculator to approximate. Use a calculator to find the measure of aB. ( 4 8 ) 2nd TAN ≈ 26.6 aA and aB are complementary. The sum of their measures is 90 . maA maB 90 maA 26.6 90 maA 63.4 aA and aB are complementary. Substitute for maB. Subtract. Answer The side lengths are 4 , 8 , and 8.9 . The angle measures are 26.6 , 63.4 , and 90 . Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 202 Solving a Right Triangle Example 2 Q Solve the right triangle. Round decimals to the nearest tenth. 30 Solution Use trigonometric ratios to find the values of p and q. opp. sin Q hyp. q sin 53 30 P 53 q p R adj. cos Q hyp. cos 53 30 sin 53 q 30 cos 53 p 30 ( 0.7986 ) ≈ q 30 ( 0.6018 ) ≈ p 24.0 ≈ q 18.1 ≈ p p 30 aP and aQ are complementary. The sum of their measures is 90 . maP maQ 90 aP and aQ are complementary. maP 53 90 Substitute for maQ. maP 37 Subtract. Answer The side lengths of the triangle are 18.1 , 24.0 , and 30 . The angle measures are 37 , 53 , and 90 . Checkpoint Solve the right triangle. Round decimals to the nearest tenth. 1. 2. D C 80 a e A 89 31 B F Side lengths: 39, 80, 89 Angle measures: 26.0, 64.0, 90 Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. f 8 E Side lengths: 4.8, 8, 9.3 Angle measures: 31, 59, 90 Chapter 9 • Geometry Notetaking Guide 203 Example 3 Solving a Right Triangle 7 4 ft Sports When a hockey player is 35 feet from the goal line, he shoots the puck 35 ft directly at the goal. The angle of elevation at which the puck leaves the ice is 7. The height of the goal is 4 feet. Will the player score a goal? Solution Begin by finding the height h of the puck at the goal line. Use a trigonometric ratio. opp. tan 7 adj. tan 7 h Write trigonometric ratio. Substitute. 35 35 tan 7 h Multiply each side by 35 . 35 ( 0.1228 ) ≈ h Use a calculator. 4.3 ≈ h Multiply. Answer Because the height of the puck at the goal line ( 4.3 feet) is greater than the height of the goal (4 feet), the player will not score a goal. Checkpoint Complete the following exercise. 3. A hockey player is 27 feet from the goal line. He shoots the puck directly at the goal. The height of the goal is 4 feet. What is the maximum angle of elevation at which the player can shoot the puck and still score a goal? 4 ft x 27 ft 8.4 Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 204 9.7 Vectors Goals p Find the magnitude and direction of a vector. p Add vectors. VOCABULARY Magnitude of a vector The magnitude of a vector is the distance from the initial point to the terminal point. Direction of a vector The direction of a vector is determined by the angle that the vector makes with a horizontal line. Equal vectors Two vectors are equal when they have the same magnitude and direction. Parallel vectors Two vectors are parallel when they have the same or opposite direction. Sum of vectors The sum of two vectors is a vector that joins the initial point of the first vector and the terminal point of the second vector. Example 1 Finding the Magnitude of a Vector P(4, 3) and Q(2, 1) are the initial and terminal points of a vector. Draw PQ &) in a coordinate plane. Then find its magnitude. Solution Component form x2 x1, y2 y1 PQ &) 2 (4) , 1 3 y P 3 6 , 4 Use the Distance Formula to find the magnitude. PQ &) 1 3 1 x Q [ 2 ( 4 )] 2 ( 1 3 )2 52 ≈ 7.2 Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. Chapter 9 • Geometry Notetaking Guide 205 Describing the Direction of a Vector Example 2 The vector CD &) describes the velocity of a moving hot air balloon. The scale on each axis is in miles per hour. D N y a. Find the speed of the balloon. 5 b. Find the direction it is traveling relative to east. W C 5 x E S Solution a. The magnitude of the vector CD &) represents the balloon’s speed. Use the Distance Formula. CD &) (30 5)2 (25 5)2 1025 ≈ 32.0 Answer The speed of the balloon is about 32 miles per hour. b. The tangent of the angle formed by the vector and a line drawn parallel 20 to the x-axis is , or 0.8 . Use a 25 calculator to find the angle measure. 0.8 2nd TAN D N y 20 ? 5 W ≈ 38.7 C 25 5 x E S Answer The balloon is traveling in a direction about 38.7 north of east. Checkpoint Complete the following exercise. 1. The vector represents the velocity of a moving hot air balloon. The scale on each axis is in miles per hour. Find the balloon’s speed and direction relative to west. 25.5 miles per hour; 78.7 south of west Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. N J W 5 y 5 x K E S Chapter 9 • Geometry Notetaking Guide 206 Example 3 Identifying Equal and Parallel Vectors In the diagram, these vectors have the ****) , CD &*) , IJ **) . same direction: AB ****) , IJ **) . These vectors are equal: AB y F B D G A 1 These vectors are parallel: C 1 1 ****) , CD &*) , GH &*) , IJ **) . AB E H 1 3 x J I ADDING VECTORS Sum of Two Vectors The sum of u *) a1, b1 and *v) a2, b2 is u *) *v) a1 a2 , b1 b2 . Example 4 Finding the Sum of Two Vectors Let u *) 6, 2 and *v) 8, 7. Find the sum u *) *v). y 5 3 Solution To find the sum, add the horizontal components and add the vertical components of u and v. uv 1 1 v 3 5 x u u *) *v) 6 (8) , 2 7 2 , 5 Checkpoint Find the sum of the vectors. 2. 6, 0, 1, 3 7, 3 Copyright © McDougal Littell/Houghton Mifflin Company All rights reserved. 3. 5, 2, 7, 6 2, 4 Chapter 9 • Geometry Notetaking Guide 207