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Lesson 8.3 Pythagorean Theorem pp. 323-327

Lesson 8.3 Pythagorean Theorem pp. 323-327. Objectives: 1. To prove the Pythagorean theorem. 2. To apply the Pythagorean theorem to right triangles. 3. To develop a formula for the area of an equilateral triangle. Theorem 8.7

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Lesson 8.3 Pythagorean Theorem pp. 323-327

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  1. Lesson 8.3 Pythagorean Theorem pp. 323-327

  2. Objectives: 1. To prove the Pythagorean theorem. 2. To apply the Pythagorean theorem to right triangles. 3. To develop a formula for the area of an equilateral triangle.

  3. Theorem 8.7 Pythagorean Theorem. In a right triangle, the sum of the squares of the length of the legs is equal to the square of the length of the hypotenuse: a2 + b2 = c2.

  4. The Pythagorean Theorem can also be thought of as leg2 + leg2 = hypotenuse2

  5. The following illustrates the Pythagorean theorem.

  6. b a a c b c c b c a a b

  7. b a a c b c c b c a a b Area of large square: A = (a+b)2 Area of small square: A = c2 Area of triangle: A = ½ab

  8. b a a c b c c b c a a b Area of large square = Area of small square + 4 x Area of triangle.

  9. b a a c b c c b c a a b (a+b)2 = c2 + 4(½ab) a2 + 2ab + b2 = c2 + 2ab a2 + b2 = c2

  10. The converse of the Pythagorean theorem is also true. If the sum of the squares of the lengths of two sides of a triangle equals the square of the length of the third side, then the triangle is a right triangle.

  11. Theorem 8.8 The area of an equilateral triangle is times the square of the length of one side: A = s2 3 3 4 4

  12. C s s h ½s A D B s 3 2 s h2 + (½s)2 = s2 h2 + ¼s2 = s2 4h2 + s2 = 4s2 4h2 = 3s2 h2 = ¾s2 h =

  13. C (s) A = s s h s2 3 4 1 2 A = A D B s 3 2 s A = ½bh

  14. L A = s2 A = 72 7 cm 3 4 3 4 49 3 4 A = sq. cm N M EXAMPLE Find the area of equilateral LMN.

  15. c 80 in. 36 in. Practice: A screen door measures 36 in. by 80 in. Find the length of a diagonal brace for the screen door. 362 + 802= c2 7,696 = c2 c ≈ 87.7 in.

  16. Homework pp. 325-327

  17. 85 5 2 13 x2 + 25 ►A. Exercises Complete the table. Consider ABC to be a right triangle with c as the hypotenuse. a(units) b(units) c(units) 1. 3 5 3. 9 2 5. 2 3 7. 6 4 9. x 5 4

  18. 85 or 9.2 5 or 2.2 52 or 7.2 x2 + 25 ►A. Exercises Complete the table. Consider ABC to be a right triangle with c as the hypotenuse. a(units) b(units) c(units) 1. 3 5 3. 9 2 5. 2 3 7. 6 4 9. x 5 4

  19. 9 6 8 ►A. Exercises Tell which of the following triangles are right triangles. 11. 62 + 82 92

  20. 20 21 29 ►A. Exercises Tell which of the following triangles are right triangles. 13. 202 + 212= 292

  21. 8 5 ►B. Exercises Find each area. 15. A right triangle has a leg measuring 5 inches and a hypotenuse measuring 8 inches.

  22. 15 12 5 ►B. Exercises 17. Find the area.

  23. 5 5 5 ►B. Exercises 19. Find the area.

  24. 6 5 12 ►B. Exercises 21. The bases of an isosceles trapezoid are 6 inches and 12 inches. Find the area if the congruent sides are 5 inches.

  25. ►B. Exercises 22. The perimeter of a rhombus is 52 and one diagonal is 24. Find the area. 24 13

  26. ►B. Exercises 22. The perimeter of a rhombus is 52 and one diagonal is 24. Find the area. 12 5 13

  27. ►B. Exercises 22. The perimeter of a rhombus is 52 and one diagonal is 24. Find the area. A = ½d1d2 = ½(10)(24) = 120 24 10 13

  28. 7 3 ■ Cumulative Review Find the area of each rectangle. 29.

  29. 7 4 ■ Cumulative Review Find the area of each rectangle. 30.

  30. ■ Cumulative Review 31. Give bounds for c if b  5. c b 7

  31. ■ Cumulative Review 32. The consecutive sides of a rectangle have a ratio 4:5. If the area is 5120 m2, what are the dimensions of the rectangle?

  32. ■ Cumulative Review 33. The figure is made up of 8 congruent squares and has a total area of 968 cm2. Find the perimeter.

  33. Analytic Geometry Heron’s Formula

  34. a + b + c . s = 2 Definition The semiperimeter of a triangle is one-half the perimeter of a triangle:

  35. Heron’s Formula If ABC has sides of lengths a, b, and c and semiperimeter s, then the area of the triangle is A = s(s - a)(s - b)(s - c) .

  36. 9 + 10 + 15 s = 2 = 17 Find the area of a triangle with sides 9, 10, and 15.

  37. A = s(s – a)(s – b)(s – c) = 17(17 – 9)(17 – 10)(17 – 15) = 17(8)(7)(2) = 1904 Find the area of a triangle with sides 9, 10, and 15. ≈43.6 sq. units

  38. ►Exercises Find the area of each triangle that has the given side measures. Round you answers to the nearest tenth. 1. 3 units, 8 units, 9 units

  39. ►Exercises Find the area of each triangle that has the given side measures. Round you answers to the nearest tenth. 2. 6 units, 18 units, 21 units

  40. ►Exercises Find the area of each triangle that has the given side measures. Round you answers to the nearest tenth. 3. 27 units, 13 units, 18 units

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