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Square Roots and Irrational Numbers

a. 144. 144 = 12. b.   – 81. – 81 = – 9. Square Roots and Irrational Numbers. Lesson 11-1. Additional Examples. Simplify each square root. Use the formula. d = 1.5 h. Replace h with 20. d = 1.5(20). Multiply. d =. <. 25. <. 30. 30. 36.

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Square Roots and Irrational Numbers

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  1. a. 144 144 = 12 b.  – 81 – 81 = – 9 Square Roots and Irrational Numbers Lesson 11-1 Additional Examples Simplify each square root.

  2. Use the formula. d = 1.5h Replace h with 20. d = 1.5(20) Multiply. d = < 25 < 30 30 36 Find perfect squares close to 30. 25 = 5 Find the square root of the closest perfect square. Square Roots and Irrational Numbers Lesson 11-1 Additional Examples You can use the formula d = 1.5h to estimate the distance d, in miles, to a horizon line when your eyes are h feet above the ground. Estimate the distance to the horizon seen by a lifeguard whose eyes are 20 feet above the ground. The lifeguard can see about 5 miles to the horizon.

  3. a.  49 c.  3 e. – 15 Square Roots and Irrational Numbers Lesson 11-1 Additional Examples Identify each number as rational or irrational. Explain. rational, because 49 is a perfect square b. 0.16 rational, because it is a terminating decimal irrational, because 3 is not a perfect square d. 0.3333 . . . rational, because it is a repeating decimal irrational, because 15 is not a perfect square f. 12.69 rational, because it is a terminating decimal g. 0.1234567 . . . irrational, because it neither terminates nor repeats

  4. c2 = a2 + b2 Use the Pythagorean Theorem. c2 = 282 + 212 Replace a with 28, and b with 21. Simplify. c2 = 1,225 Find the positive square root of each side. c = 1,225 = 35 The Pythagorean Theorem Lesson 11-2 Additional Examples Find c, the length of the hypotenuse. The length of the hypotenuse is 35 cm.

  5. a2 + b2 =c2 Use the Pythagorean Theorem. 72 + x2 =142 Replace a with 7, b with x, and c with 14. 49 + x2 =196 Simplify. x2 =147 Subtract 49 from each side. x=147 Find the positive square root of each side. The Pythagorean Theorem Lesson 11-2 Additional Examples Find the value of x in the triangle. Round to the nearest tenth.

  6. Then use one of the two methods below to approximate . 147 Method 1: Use a calculator. A calculator value for is 12.124356. Round to the nearest tenth. Method 2: Use a table of square roots. Use the table on page 800. Find the number closest to 147 in the N2 column. Then find the corresponding value in the N column. It is a little over 12. x 12.1  x 12.1  147 Estimate the nearest tenth. The Pythagorean Theorem Lesson 11-2 Additional Examples (continued) The value of x is about 12.1 in.

  7. c2 = a2 + b2 Use the Pythagorean Theorem. c2 = 102 + 102 Replace a with 10 (half the span), and b with 10. c2 = 100 + 100 Square 10. c2 = 200 Add. c = 200 Find the positive square root. c 14.1 Round to the nearest tenth. The Pythagorean Theorem Lesson 11-2 Additional Examples The carpentry terms span, rise, and rafter length are illustrated in the diagram. A carpenter wants to make a roof that has a span of 20 ft and a rise of 10 ft. What should the rafter length be? The rafter length should be about 14.1 ft.

  8. a2 + b2 = c2 Write the equation for the Pythagorean Theorem. Replace a and b with the shorter lengths and c with the longest length. 102 + 242262 Simplify. 100 + 576 676 676 = 676 The Pythagorean Theorem Lesson 11-2 Additional Examples Is a triangle with sides 10 cm, 24 cm, and 26 cm a right triangle? The triangle is a right triangle.

  9. d = (x2 – x1)2 + (y2 – y1)2 Use the Distance Formula. d = (8 – 3)2 + (3 – (–2 ))2 Replace (x2, y2) with (8, 3) and (x1, y1) with (3, –2). 52 + 52 d = Simplify. 50 d = d 7.1 Find the exact distance. Round to the nearest tenth. Distance and Midpoint Formulas Lesson 11-3 Additional Examples Find the distance between T(3, –2) and V(8, 3). The distance between T and V is about 7.1 units.

  10. WX= (–2 – (–3))2 + (–1 – 2)2 = 1 + 9 = 10 Replace (x2, y2) with (–2, –1) and (x1, y1) with (–3, 2). XY= (4 – (–2))2 + (0 – (–1)2 Replace (x2, y2) with (4, 0) and (x1, y1) with (–2, –1). 37 Simplify. = 36 + 1 = Simplify. Distance and Midpoint Formulas Lesson 11-3 Additional Examples Find the perimeter of WXYZ. The points are W (–3, 2), X (–2, –1), Y (4, 0), Z (1, 5). Use the Distance Formula to find the side lengths.

  11. YZ= (1 – 4)2 + (5 – 0)2 Replace (x2, y2) with (1, 5) and (x1, y1) with (4, 0). = 9 + 25 = 34 Simplify. Replace (x2, y2) with (–3, 2) and (x1, y1) with (1, 5). ZW= (–3 – 1)2 + (2 – 5)2 = 16 + 9 = 25 = 5 Simplify. Distance and Midpoint Formulas Lesson 11-3 Additional Examples (continued)

  12. perimeter = + + + 520.1 10 37 34 Distance and Midpoint Formulas Lesson 11-3 Additional Examples (continued) The perimeter is about 20.1 units.

  13. 1 2 1 2 1 2 1 2 4 + 9 2 y1 + y2 2 x1 + x2 2 , Use the Midpoint Formula. –3 + 2 2 = , Replace (x1, y1) with (4, –3) and (x2, y2) with (9, 2). = , Simplify the numerators. = 6 , – Write the fractions in simplest form. 13 2 –1 2 The coordinates of the midpoint of TV are 6 , – . Distance and Midpoint Formulas Lesson 11-3 Additional Examples Find the midpoint of TV.

  14. hypotenuse = leg • 2 Use the 45°-45°-90° relationship. y = 10 • 2 The length of the leg is 10. 14.1 Use a calculator. Special Right Triangles Lesson 11-5 Additional Examples Find the length of the hypotenuse in the triangle. The length of the hypotenuse is about 14.1 cm.

  15. hypotenuse = leg • 2 Use the 45°-45°-90° relationship. y = 20 • 2 The length of the leg is 20. 28.3 Use a calculator. Special Right Triangles Lesson 11-5 Additional Examples Patrice folds square napkins diagonally to put on a table. The side length of each napkin is 20 in. How long is the diagonal? The diagonal length is about 28.3 in.

  16. hypotenuse = 2 • shorter leg 14 = 2 • bThe length of the hypotenuse is 14. = Divide each side by 2. 7 = bSimplify. longer leg = shorter leg • 3 a = 7 • 3 The length of the shorter leg is 7. a 12.1 Use a calculator. 14 2 2b 2 Special Right Triangles Lesson 11-5 Additional Examples Find the missing lengths in the triangle. The length of the shorter leg is 7 ft. The length of the longer leg is about 12.1 ft.

  17. 3 5 4 5 3 4 opposite hypotenuse sin A = = = adjacent hypotenuse cos A = = = opposite adjacent tan A = = = 12 20 16 20 12 16 Sine, Cosine, and Tangent Ratios Lesson 11-6 Additional Examples Find the sine, cosine, and tangent of A.

  18. sin 18° 0.3090 Scientific calculator: Enter 18 and press the key labeled SIN, COS, or TAN. cos 18° 0.9511 Table: Find 18° in the first column. Look across to find the appropriate ratio. tan 18° 0.3249 Sine, Cosine, and Tangent Ratios Lesson 11-6 Additional Examples Find the trigonometric ratios of 18° using a scientific calculator or the table on page 779. Round to four decimal places.

  19. sin A = Use the sine ratio. opposite hypotenuse sin 40° = Substitute 40° for the angle, 10 for the height, and w for the hypotenuse. 10 sin 40° w = Divide each side by sin 40°. w 15.6 Use a calculator. 10 w Sine, Cosine, and Tangent Ratios Lesson 11-6 Additional Examples The diagram shows a doorstop in the shape of a wedge. What is the length of the hypotenuse of the doorstop? You know the angle and the side opposite the angle. You want to find w, the length of the hypotenuse. w(sin 40°) = 10 Multiply each side by w. The hypotenuse is about 15.6 cm long.

  20. Draw a picture. opposite hypotenuse sin A = Choose an appropriate trigonometric ratio. sin 52° = Substitute. 24 hSimplify. h 30 Angles of Elevation and Depression Lesson 11-7 Additional Examples Janine is flying a kite. She lets out 30 yd of string and anchors it to the ground. She determines that the angle of elevation of the kite is 52°. What is the height h of the kite from the ground? 30(sin 52°) = hMultiply each side by 30. The kite is about 24 yd from the ground.

  21. Draw a picture. opposite adjacent Choose an appropriate trigonometric ratio. tan A = Substitute 61 for the angle measure and 30 for the adjacent side. tan 61° = 54 hUse a calculator or a table. h 30 Angles of Elevation and Depression Lesson 11-7 Additional Examples Greg wants to find the height of a tree. From his position 30 ft from the base of the tree, he sees the top of the tree at an angle of elevation of 61°. Greg’s eyes are 6 ft from the ground. How tall is the tree, to the nearest foot? 30(tan 61°) = hMultiply each side by 30. 54 + 6 = 60 Add 6 to account for the height of Greg’s eyes from the ground. The tree is about 60 ft tall.

  22. tan 3° = Choose an appropriate trigonometric ratio. 1.5 d Angles of Elevation and Depression Lesson 11-7 Additional Examples An airplane is flying 1.5 mi above the ground. If the pilot must begin a 3° descent to an airport runway at that altitude, how far is the airplane from the beginning of the runway (in ground distance)? Draw a picture (not to scale). d • tan 3° = 1.5 Multiply each side by d.

  23. d • tan 3° tan 3° 1.5 tan 3° 1.5 tan 3° = Divide each side by tan 3°. d 28.6 Use a calculator. d = Simplify. Angles of Elevation and Depression Lesson 11-7 Additional Examples (continued) The airplane is about 28.6 mi from the airport.

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