Square the lengths of the sides of each triangle. What do you notice?
1.
2.
3.
4.
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
Solutions
1. 32 = (3)(3) = 9; 42 = (4)(4) = 16; 52 = (5)(5) = 25; 9 + 16 = 25 2. 52 = (5)(5) = 25; 122 = (12)(12) = 144; 132 = (13)(13) = 169; 25 + 144 = 169 3. 62 = (6)(6) = 36; 82 = (8)(8) = 64; 102 = (10)(10) = 100; 36 + 64 = 100 4. 42 = (4)(4) = 16; (4
2)2 = (4
2)(4
2) = 16
(2)(2) = 16(2) = 32; 16 + 16 = 32.
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
A right triangle has legs of length 16 and 30. Find the length of the hypotenuse. Do the lengths of the sides form a Pythagorean triple? a2 + b2 = c2 162 + 302 = c2 256 + 900 = c2 1156 = c2 34 = c The length of the hypotenuse is 34.
The lengths of the sides, 16, 30, and 34, form a Pythagorean triple because they are whole numbers that satisfy a2 + b2 = c2. Notice that each length is twice the common Pythagorean triple of 8, 15, and 17.
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
Find the value of x. Leave your answer in simplest radical form.
a2 + b2 = c2
Use the Pythagorean Theorem.
x2 + 102 = 122
Substitute x for a, 10 for b, and 12 for c.
x2 + 100 = 144
Simplify.
x2 = 44
Subtract 100 from each side.
x =
Take the square root of each side.
x =2
4(11) 11
Simplify.
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
A baseball diamond is a square with 90-ft sides. Home plate and second base are at opposite vertices of the square. About how far is home plate from second base? Use the information to draw a baseball diamond. a2 + b2 = c2 902 + 902 = c2 8100 + 8100 = c2 16,200 = c2 c= c
16,200 127.27922
The distance to home plate from second base is about 127 ft.
The hypotenuse of an isosceles right triangle has length 20 cm. Find the area. Use the information to draw the triangle. Because it is isosceles, its two legs must be congruent. a2 + b2 = c2
Use the Pythagorean Theorem. Substitute x for a x for b and 20 for c. Simplify. Divide each side by 2.
x2 + x2 = 202 2x2 = 400 x2 = 200 x=
200
x=
100(2) = 10
Take the square root. 2
Each leg is 10 2 cm. Because the legs are perpendicular, one leg is a base and one leg is an altitude.
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
(continued)
Let b = 10 1 bh 2 1 A = 2 (10
2 and h = 10
2 , and use the formula for the area of a triangle.
A=
Area of a triangle 2)(10
1
A = 2 (10)(10)(
2) 2)(
Substitute 10 2)
A = 50(2) = 100 The area of the triangle is 100 cm2.
Simplify.
2 for b and for h.
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
Is this triangle a right triangle?
a2 + b2
c2
42 + 62
72
Substitute 4 for a, 6 for b, and 7 for c.
49
Simplify.
16 + 36
52 =/ 49 Because a2 + b2 =/ c2, the triangle is not a right triangle.
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
The numbers represent the lengths of the sides of a triangle. Classify each triangle as acute, obtuse, or right. a. 15, 20, 25 c2
a2 + b2
b. 10, 15, 20 Compare c2 with a2 + b2. Substitute the greatest length for c.
252
152 + 202
625
225 + 400 Simplify.
c2
a2 + b2
Compare c2 with a2 + b2. Substitute the greatest length for c.
202
102 + 152
400
100 + 225 Simplify.
625 = 625
400 > 325
Because c2 = a2 + b2, the triangle is a right triangle.
Because c2 > a2 + b2, the triangle is obtuse.
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
Pages 360-364 Exercises 1. 10
9. Yes; 152 + 202 = 252.
17. 17.0 m
2. 7
10.
41
18. 9 3 m2
3. 34
11.
33
19. 12
7 cm2
4. 12
12. 3
11
20. 25
3 in.2
5. 65
13. 2
89
21. No; 192 + 202 =/ 282.
6. 8
14. 3
2
22. No; 82 + 242 =/ 252.
7. No; 42 + 52 =/ 62.
15. 5
2
23. Yes; 332 + 562 = 652.
8. Yes; 102 + 242 = 262.
16. 14 ft
2
24. obtuse
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
25. right
34. right
40. 10.5 in.2
26. acute
35. acute
41. 168 ft2
27. right
36. 10
42. 14 m2
28. obtuse
37. 8
5
43. 32 in.2
29. acute
38. 2
2
44. 4.2 in.
30. obtuse
39. Answers may vary. Sample: Have three people hold the rope 3 units, 4 units, and 5 units apart in the shape of a triangle.
31. right 32. acute 33. obtuse
45. Yes; 72 + 242 = 252, so RST is a rt. .
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
46. a. |x2 – x1|; |y2 – y1|
b. PQ2 = (x2 – x1)2 + (y2 – y1)2 c. PQ =
(x2 – x1)2 + (y2 – y1)2
47. Answers may vary. Sample: Using 2 segments of length 1, construct the hyp. of the right formed by these segments. Using the hyp. found as one leg and a segment of length 1 as the other leg, construct the hyp. of the formed by those legs. Continue this process until constructing a hypotenuse of length n.
48. 29
49. 50 50. 84 51. 35 52–59. Answers may vary. Samples are given. 52. 6; 7 53. 4; 5 54. 8; 11
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
55. 11; 12
62. 12 cm
56. 8; 10
63. 12.5 cm
57. 14; 16
64. 17.9 cm
58. 18; 19
65. a. Answers may vary. Sample: n = 6; 12, 35, 37
59. 39; 42 60. a. c2
b. 122 + 352 = 372 66. a. 5 in.
b. 2ab + (b – a)2
b.
c. c2 = a2 + b2
c. d2 =
29
d. 34 in.
61. 2830 km
BD2 + AC2 + BC2
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
67.
14
73. 9
68.
61
74. 37
69.
17
75. 14.1 76. 72 cm2
70. Draw right FDE with legs DE of length a and EF of length b and hyp. of length x. Then a2 + b2 = x2 by the Pythagorean Thm. We are given ABC with sides of length a, b, c and a2 + b2 = c2. By subst., c2 = x2, so c = x. Since all side lengths of ABC and FDE are the same, ABC FDE by SSS. C E by CPCTC, so m C = 90. Therefore, ABC is a right .
77. 15 ft 78. 7; 33
79. 3; 20 80. 10 81. 3
71. 26.2
82. 50
72. 30
83. 7
Pythagorean Theorem and its Converse GEOMETRY LESSON 7-2
1. Find the value of x.
3. The town of Elena is 24 mi north and 8 mi west of Holberg. A train runs on a straight track between the two towns. How many miles does it cover? Round your answer to the nearest whole number. 25 mi
15
2. Find the value of x. Leave your answer in simplest radical form.
4. Find the area of the shaded region. Leave your answer in simplest radical form. 10
2
33
21 in.2
5. The lengths of the sides of a triangle are 5 cm, 8 cm, and 10 cm. Is it acute, right, or obtuse? obtuse