Garfield's Proof

What a clever little piece of mathematics this is. It is a much neater proof of Pythagoras Theorem than the one I was shown at school. There were also some very well laid out solutions with clear explanations, so well done.

Complete solutions were received from:

Michael - (home educated),
Andrei (School number 205, Bucharest),
Charles (Shrewsbury House School),
Aftab - whose solution is given below.

First Proof

Proof Area of Trapezium derived from square = Area of Trapezium as a sum of areas of three triangles. $${a^2\over{2}}+ ab + {b^2\over{2}} = {b + {c^2\over{2}}}$$ (subtracting ab from both sides) $${{a^2\over{2}}+ ab + {b^2\over{2}}- ab} = {c^2\over{2}}$$ (multiplying 2 both sides) $${a^2 + b^2} = {c^2}$$ (Pythagoras Theorem)

Here is Aftab's solution:

Area of the square = (a+b) ² (square of sides a+b) $$\mbox{Area of Square} = {{a^2 + 2ab + b^2}}$$

Area of the Trapezium = Area of square divided by 2 (rotational symmetry) $$\mbox{Area of Trapezium} = {{a^2 + 2ab + b^2}\over{2}}$$ $$\mbox{Area of Trapezium} = {{a^2\over{2}}+ ab + {b^2\over{2}}}$$

Area of Trapezium as a sum of areas of triangles $$\mbox{Area of Trapezium} = {{ab\over{2}}+ {ab\over{2}}+ {c^2\over{2}}}$$ $$\mbox{Area of Trapezium} = {{ab + ab + c^2\over{2}}}$$ $$\mbox{Area of Trapezium} = {{2ab + c^2\over{2}} }$$ $$\mbox{Area of Trapezium} = {ab +{ c^2\over{2}} }$$