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)

${\displaystyle \mathrm{Area}\mathrm{of}\mathrm{Square}=a^2+2\mathrm{ab}+b^2}$

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

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

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

Area of Trapezium as a sum of areas of triangles

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

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

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

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

Proof

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

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

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

${\displaystyle a^2+b^2=c^2(\mathrm{Phythagoras}\mathrm{Theorem})}$