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:

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

Here is Aftab's solution:

Area of the square = (a+b) ² (square of sides a+b)

Area of Square = a^{2} + 2 ab + b^{2}

Area of the Trapezium = Area of square divided by 2 (rotational symmetry)

Area of Trapezium = \frac{a^{2} + 2 ab + b^{2}}{2}

Area of Trapezium = \frac{a^{2}}{2} + ab + \frac{b^{2}}{2}

Area of Trapezium as a sum of areas of triangles

Area of Trapezium = \frac{ab}{2} + \frac{ab}{2} + \frac{c^{2}}{2}

Area of Trapezium = \frac{ab + ab + c^{2}}{2}

Area of Trapezium = \frac{2 ab + c^{2}}{2}

Area of Trapezium = ab + \frac{c^{2}}{2}

Proof

Area of Trapezium derived from square = Area of Trapezium as a sum of areas of three triangles.

\frac{a^{2}}{2} + ab + \frac{b^{2}}{2} = b + \frac{c^{2}}{2} (subtracting ab from  both sides)

\frac{a^{2}}{2} + ab + \frac{b^{2}}{2} - ab = \frac{c^{2}}{2} (multiplying 2 both sides)

a^{2} + b^{2} = c^{2} (Pythagoras Theorem)