An Easy Approach to the Theorem of Pythagoras


We have seen that a triangle links two ways of dividing the area of a square into two.  Now we look at what happens in some more complicated example.  First, a 3x3 grid.


With a three-by-three grid, we can view the two squares on the left of the top row as forming a 2x1 rectangle, and add the diagonal to form a triangle with sides of lengths 2 and 1.

 The squares on the sides of the rectangle have areas 1 and 4, so the total area is 5 squares. 

Now we have to find the area of the square on the diagonal, or the hypotenuse of the triangle.  First, construct the square and color it in. 

What is the area of the red square?  It includes 1 whole small square and a lot of bits of the outer squares.  But the outer bits are made up of four copies of the original triangle, and that was just half of the 2x1 rectangle.  The rectangle has area 2, so half of it has area 1 and four copies then have area 4.  The total area is the one central square, plus 4 from the outside to make 5.  The red and the blue areas are the same. 


A 4x4 grid lets us try out a 3x1 rectangle and its diagonal. 

The squares on the long and short sides of the rectangle have 9 and 1 little squares, making 10 total. 

Next, build the square on the diagonal of the rectangle.

This square is made up of a central portion of 4 grid squares, and four copies of the original triangle, so four times half of 3 (the area of the 3x1 rectangle), making 10 grid squares altogether.  Once again, the red and blue areas are the same.

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A 5x5 grid lets you do some more examples, including a 3x2 rectangle.  A pad of squared paper lets you do many examples.  You will find that they always do work out right.  If you are not convinced, here is a simple argument to show that it works in general using algebra.


Last modified: 16 May 2005