David of Madras College, St Andrew's had a novel way of looking at this. He first rearranged the semicircles into three circles touching at $A$. He then pointed out that, when working out the areas, all the answers will be multiples of $\pi$ and ultimately, when comparing the areas, it is necessary to factorise out $\pi$ anyway. David's geometrical representation of the results "when divided by $\pi$'', compares the equivalent areas of squares.
Thomas of Simon Langton School, Canterbury also sent in a good solution.