The whole shape is made up of a rectangle, two semicircles on $AB$ and on $DC$ together making one circle, and the two semicircles on $AD$ and $BC$ making another circle.

Excellent solutions were sent in by a pupil from Dr Challoner's Grammar School, Amersham and Nisha Doshi ,Y9, The Mount School, York. Here is one of their solutions: Take: $AB=2x, AD=2y$.

\begin{eqnarray} \mbox{Total area of shape}
&=& \pi x^2 + \pi y^2 + (2x \times\ 2y)\\ &=& \pi
x^2 + \pi y^2 + 4xy. \end{eqnarray}

By Pythagoras Theorem
\begin{eqnarray} AC^2 &=& AD^2 + DC^2\\
&=& (2x)^2 +\ (2y)^2\\ &=& 4(x^2 + y^2)\\
\mbox{Area of big circle} &=& \pi(\mbox{AC}/2)^2\\
&=& \pi(x^2 + y^2)\\ \mbox{Area of crescents} &=&
\mbox{Area of shape - Area of big circle}\\ &=& \pi x^2 +
\pi y^2 + 4xy - \pi (x^2 + y^2)\\ &=& 4xy\\ \mbox{Area of
rectangle} &=& 2x \times\ 2y\\ &=& 4xy
\end{eqnarray}

so the sum of the areas of the four crescents is equal in area to
the rectangle $ABCD$.