1. Chapter 10 Measurement Section 10.4 The Pythagorean Theorem

2. The Pythagorean Theorem is an idea that relates the sides of a right triangle that is thousands of years old. There are some words to identify certain sides of a right triangle that are used. The hypotenuse of a right triangle is the side opposite the right angle. The legs of a right triangle are the sides that form the right angle. The Pythagorean Theorem The Pythagorean Theorem states that the squares of the legs of any right triangle added together is equal to the square of the hypotenuse. Labeling the lengths of the legs as a and b and the length of the hypotenuse as c we get the relation: a 2 + b 2 = c 2. b a c The justification for this (some people refer to this as a proof) has a long history. Many different people have contributed to this and seeking logical justification of the Pythagorean Theorem was one of the driving forces in the development of mathematics. There are books filled with hundreds of arguments (proofs) as to why the Pythagorean Theorem is true. We will give one here. This is very old some people attribute it to Pythagoras himself. The method used here is to compute the area in two different ways and apply some algebra.

3. c a b a a a a b b b b c c c c 1. Start with any right triangle with sides of length a, b (legs) and c (hypotenuse). 2. Make 4 identical (congruent) copies of this triangle and place them vertex to vertex to make a square. 3. Each angle of the inside figure is a right angle since: m  1 + m  2 + 90  = m  1 + m  2 + m  3 90  = m  3 4. Compute the area of the big square by squaring the length of the side. 5. Compute the area of the big square by adding the area of 4 triangles and the little square. 6. The two expression both compute the same area. Set them equal and cancel the term 2ab from each side. 4. Area of Big Square = (a+b) 2 = (a+b)(a+b) = a 2 + 2ab + b 2 5. Area of Big Square = Area of 4 triangles + Area of small square = 4·(½)·a·b+ c 2 = 2ab + c 2 Set equal: a 2 +2ab+b 2 = 2ab + c 2 a 2 + b 2 = c 2 1 21 3 90 

4. Square Roots The way the Pythagorean Theorem is most often applied is when two of the sides of a right triangle are known solving for the remaining side. This often requires solving and equation that looks like: x 2 = number. The solution to such and equation is given by x = or the square root of the number. The square root of a number is the number that when you square it you get the number you are taking the square root of. The square roots to the right show the square roots of numbers that are perfect squares. If you are taking the square root of a number that is between any of them the result is between the two perfect squares. For example: is between 1 and 2 because the number 2 is between 2 and 4. Most often we leave the symbol to stand for the value.

5. Find the length of the missing side labeled x in each of the right triangles below. The idea is to apply the Pythagorean Theorem to get an equation relating the sides, then use algebra to solve the equation. 6 8 x xx 10 x 13 12 8 2 + 6 2 = x 2 64 + 36 = x 2 100 = x 2 x 2 + 12 2 = 13 2 x 2 + 144 =169 x 2 = 25 x 2 + x 2 = 10 2 2x 2 =100 x 2 = 50 Perimeter of the triangle is: 8 + 6 + 10 = 24 Perimeter of the triangle is: 5 + 12 + 13 = 30 Perimeter of the triangle is:

6. On a geoboard we can find the length of a line segment between any pair of diagonal points by thinking of a right triangle that has the line segment as the hypotenuse. For example to find the length of the segment pictured to the left we do the following: x 3 4 x 2 = 3 2 + 4 2 x 2 = 9 + 16 x 2 = 25 Find the perimeter of the trapezoid pictured on the geoboard to the left. Perimeter = 2 y 2 x 2 3 x 2 = 2 2 + 2 2 x 2 = 4 + 4 x 2 = 8 y 2 = 2 2 + 3 2 y 2 = 4 + 9 y 2 = 13 Find the perimeter of the star pictured to the left. First find the length of one side (x). 1 2 x x 2 = 2 2 + 1 2 x 2 = 4 + 1 x 2 = 5 Since all 8 sides are the same Perimeter =

7. The area of an isosceles triangle can be found if the lengths of the sides are known by applying the Pythagorean Theorem. Use this to find the area of the isosceles triangle to the right. 9 9 6 1. Start by finding the height (h) 2. The height bisects the base. h 2 + 3 2 = 9 2 h 2 + 9 =81 h 2 = 72 3 h 3. Area = (½)·b·h = In the last section we found the area of the triangle to the right by enclosing it in a rectangle and saw it was 6. The diagonal distance between 2 consecutive points on a geoboard is. The area can be found by the following: