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 $$\mbox{Area of
Trapezium} = {{ab\over{2}}+ {ab\over{2}}+ {c^2\over{2}}}$$
$$\mbox{Area of Trapezium} = {{ab + ab + c^2\over{2}}}$$
$$\mbox{Area of Trapezium} = {{2ab + c^2\over{2}} }$$ $$\mbox{Area
of Trapezium} = {ab +{ c^2\over{2}} }$$