12.3 The Pythagorean Theorem

1. 12.3 The Pythagorean Theorem CORD Mrs. Spitz Spring 2007

2. Objectives/Assignment • Use Pythagorean Theorem • Assignment: pp. 484-485 #4-39 all • Assignment due today: 12.2 p. 479-480 #5-51

3. History Lesson • Around the 6th century BC, the Greek mathematician Pythagorus founded a school for the study of philosophy, mathematics and science. Many people believe that an early proof of the Pythagorean Theorem came from this school. • Today, the Pythagorean Theorem is one of the most famous theorems in geometry. Over 100 different proofs now exist.

4. Proving the Pythagorean Theorem • In this lesson, you will study one of the most famous theorems in mathematics—the Pythagorean Theorem. The relationship it describes has been known for thousands of years.

5. In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the legs. Theorem 9.4: Pythagorean Theorem c2 = a2 + b2

6. Using the Pythagorean Theorem • A Pythagorean triple is a set of three positive integers a, b, and c that satisfy the equation c2 = a2 + b2For example, the integers 3, 4 and 5 form a Pythagorean Triple because 52 = 32 + 42.

7. Find the length of the hypotenuse of the right triangle. Tell whether the sides lengths form a Pythagorean Triple. Ex. 1: Finding the length of the hypotenuse.

8. (hypotenuse)2 = (leg)2 + (leg)2 x2 = 52 + 122 x2 = 25 + 144 x2 = 169 x = 13 Because the side lengths 5, 12 and 13 are integers, they form a Pythagorean Triple. Many right triangles have side lengths that do not form a Pythagorean Triple as shown next slide. Pythagorean Theorem Substitute values. Multiply Add Find the positive square root. Note: There are no negative square roots until you get to Algebra II and introduced to “imaginary numbers.” Solution:

9. Find the length of the leg of the right triangle. Ex. 2: Finding the Length of a Leg

10. (hypotenuse)2 = (leg)2 + (leg)2 142 = 72 + x2 196 = 49 + x2 147 = x2 √147 = x √49 ∙ √3 = x 7√3 = x Pythagorean Theorem Substitute values. Multiply Subtract 49 from each side Find the positive square root. Use Product property Simplify the radical. Solution: • In example 2, the side length was written as a radical in the simplest form. In real-life problems, it is often more convenient to use a calculator to write a decimal approximation of the side length. For instance, in Example 2, x = 7 ∙√3 ≈ 12.1

11. Note: • Determine if the following lengths can represent the sides of a right triangle. Right ∆ c2 = a2 + b2 Acute ∆ c2 < a2 + b2 c2 > a2 + b2 Obtuse ∆

12. #32. 12, 11, 15 • The measures of the sides of a triangle are given. Determine whether each triangle is a right triangle. c2 = a2 + b2 152 = 112 + 122 225 = 121 +144? 225 ≠ 265 Not a right Triangle

13. c2 = a2 + b2 c2 = √72 + √92 c2 =7 + 9 c2 = 16 c = 4 Note: #14 a = √7 b =√9 c=?

14. The area of a rectangle is 40 square meters. Find the length of a diagonal of the rectangle if its length is 2 meters less than twice its width. Area of a rectangle ? 40 = bh b = 2h + 2

15. Find the area of the triangle to the nearest tenth of a meter. You are given that the base of the triangle is 10 meters, but you do not know the height. Ex. 3: Finding the area of a triangle Because the triangle is isosceles, it can be divided into two congruent triangles with the given dimensions. Use the Pythagorean Theorem to find the value of h.

16. Steps: (hypotenuse)2 = (leg)2 + (leg)2 72 = 52 + h2 49 = 25 + h2 24 = h2 √24 = h Reason: Pythagorean Theorem Substitute values. Multiply Subtract 25 both sides Find the positive square root. Solution: Now find the area of the original triangle.

17. Area of a Triangle Area = ½ bh = ½ (10)(√24) ≈ 24.5 m2 The area of the triangle is about 24.5 m2