An Easy Approach to the Theorem of Pythagoras

 

The earlier examples have shown that whenever we try to calculate the area of the square on the diagonal of a rectangle, the result is the same as the sum of the squares on the long and short sides. Here is a simple algebraic proof that the result holds in general. Here is an algebra triangle.

The areas of the squares on the long and short sides are a2 and b2 and the area of the square on the diagonal is c2. We need to find a formula relating the as, bs and cs. The way we calculated the area of the square on the diagonal, c, in terms of grid-squares, was by breaking it up into two parts, the interior square, and four outer triangles:

is made up of and .

 

 

The outer triangles are all copies of the original, and the area of the rectangle formed by a and b is ab. The area of the triangle is half the area of the rectangle or ab, and four copies of it have total area 4( ab) = 2ab. Regardless of the sizes of a and b, the outer area always contributes 2ab to the total. We need to know how large the inner square is, and to do that we need to know how long its sides are. The sides of the inner square are made up of the long side of the rectangle minus the short side, that is, a-b. So the area of the inner square is . The total red area is

 


We have recovered the well-known formula, and shown that the red and blue areas will always be the same.

 


Last modified: 16 May 2005