NUMBAT OER - Open Educational Resources

6. A geometrical proof of Pythagoras' theorem

These seven diagrams illustrate a simple geometric proof of Pythagoras' theorem. This states that for a right-angled triangle, the square of the length of the hypotenuse (the longest side of the triangle, opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

(This illustration is based on an animation produced by the University of Waterloo, Onatario. Visit: http://www.math.uwaterloo.ca/navigation/ideas/grains/pythagoras.shtml)

Figure 1. This right-angled triangle has three sides a, b and c. Side a is the hypotenuse, and Pythagoras' theorem states that:

a2 = b2 + c2

Figure 2. Draw a square with side a along the hypotenuse. The square has an area a2.
Figure 3. Construct a larger square by arranging copies of the triangle around the square on the hypotenuse as shown.
Figure 4. The square contains four copies of the triangle. The area not occupied by the four triangles (unshaded) is still a2.
Figure 5. Wherever the triangles are placed in the new square, the unshaded area will always be equal to a2 (so long as the triangles do not overlap).
Figure 6. This arrangement of the four triangles creates two rectangles. The unshaded area, now broken into two smaller squares, must still be equal to a2.

Figure 7. The sides of the two unshaded squares are equal to the lengths of the non-hypotenuse sides of the original triangle, b and c. Therefore, the areas of the two squares that occupy the non-shaded area are b2 and c2.

The four triangles still account for the same area within the larger square, so these two unshaded squares must occupy the same area as the original unshaded area before the triangles were re-arranged. So:

a2 = b2 + c2