## 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 *a*^{2} and *b*^{2} and the area of the square on the diagonal is *c*^{2}. We
need to find a formula relating the *a*’s, *b*’s and *c*’s. 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 ½*ab*, and four copies
of it have total area 4(½ *ab*) = 2*ab*. Regardless of the sizes of *a* and *b*, the outer area always contributes 2*ab* 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