21. The diagram shows a square of side $4 \mathrm{~cm}$ with four identical semi-circles drawn with their centres at the mid-points of the sides. The four semi-circles each touch two other semi-circles, as shown.
What is the shaded area, in $\mathrm{cm}^{2}$ ?

A $8-\pi$
B $\pi$
C $\pi-2$
D $\pi-\sqrt{2}$
E $8-2 \pi$
21. E Let $A, B, C, D, P, Q, R, S$ be the the points shown and let the radius of each semicircle be $r$.
Note that $P Q$ passes through the touching point of the two semicircles with centres $P$ and $Q$. Now consider triangle $P A Q$. As $P$ and $Q$ are the midpoints of $A D$ and $A B$ respectively, $A P=A Q=2$. Therefore $P A Q$ is an isosceles right-angled triangle and $\angle A P Q=45^{\circ}$. Similarly, $\angle D P S=45^{\circ}$. Hence $\angle Q P S=(180-2 \times 45)^{\circ}=90^{\circ}$. So from the symmetry of the figure we see that $P Q R S$ is a square which contains the shaded area and four quarter-circles of radius $r$.
By Pythagoras' Theorem, $P Q^{2}=P A^{2}+A Q^{2}$. Therefore $(2 r)^{2}=2^{2}+2^{2}$. Hence $4 r^{2}=8$, that is $r^{2}=2$.
So the shaded area is $(2 r)^{2}-4 \times \frac{1}{4} \times \pi \times r^{2}=4 r^{2}-\pi r^{2}=8-2 \pi$.