Pythagorean Theorem:

Proof 3 found on

Step One

First I started with four identical right triangles.  Then I rotated the second triangle 90°, the third triangle 180°, and the fourth triangle 270° (based off of the position of the first triangle) , I kept rotating the triangles 90°.  The individual triangles have an area of (ab/2) because the area of a triangle is base (x) height (\) 2.  [(x) = multiply and (\) = divide]

Step 2

I then put all four triangles together to form a square, they are already rotated, so they fit together perfectly.  When I put the squares together, I made sure that the longest side, or (c), which is the hypotenuse was on the outside of the square.  (C) then formed the sides of my triangle.  When the four triangles came together, there was an empty space in the middle, where all of the triangles meet.  The empty space is in the middle forms a square.  The square is formed from where (a) and (b) come together in the center.  To figure out the length of each side of the new triangle, I took the (a) and subtracted the distance of (b), which gave me (a-b).  

Step 3

The equation for the area of the new figure is the same area as before, but this time, I multiplied it by 4 because of the four different triangles. The equation in now (4)(a x b)\2. The area of the new inside square is (a-b)² and 2ab. We get 2ab by foiling (4)(a x b)\2.  

Final Step

The area of the whole square is c².  I then set c² equal to the area of the smaller square which is (a-b)² and 2ab.  

This should look like:

= (a - b)² + 2ab

Then we have to foil the ² to both the (a) and (b).  When you foil them together you get

a² - ab - ab + b² + 2ab -------> Which simplifies down to be...

= a² - 2ab + b² + 2ab

The -2ab and the + 2ab cancel each other out, so then in the end you are left with...

= a² + b²