23. The diagram shows a semicircle with centre $O$ and radius 2 and a semicircular arc with diameter $P R$. Angle $P O R$ is a right angle.
What is the area of the shaded region?
A $\pi-2$
B 2
C $\pi$
D 3
23. B By Pythagoras' Theorem, $P R^{2}=P O^{2}+R O^{2}=2^{2}+2^{2}=4+4=8$.
So $P R=\sqrt{8}=2 \sqrt{2}$.
Therefore the semicircle with diameter $P R$ has radius $\sqrt{2}$.
Its area is $\frac{1}{2} \times \pi \times(\sqrt{2})^{2}=\pi$.
The area of the quarter-circle bounded by $P O, R O$ and arc $P R$ is $\frac{1}{4} \times \pi \times 2^{2}=\pi$.
The area of triangle $P O R$ is $\frac{1}{2} \times 2 \times 2=2$.
So the area of the segment bounded by $P R$ and arc $P R$ is $\pi-2$.
Hence the area of the shaded region is $\pi-(\pi-2)=2$.