25. The diagram shows six points $P, Q, R, S, T$ and $U$ equally spaced around a circle of radius $2 \mathrm{~cm}$. The inner circle has radius $1 \mathrm{~cm}$. The shaded region has three lines of symmetry.
Which of the following gives the area, in $\mathrm{cm}^{2}$, of the shaded region?
A $2 \pi+3$
B $3 \pi+2$
C $\frac{4 \pi+3}{2}$
D $3(\pi+2)$
E $4 \pi+3$
25. A The diagram shows exactly one third of the shaded area in the original diagram. It is made up of the quadrilateral $U O P X$, together with a sector of the outer circle, $P O Q$, where $O$ is the centre of the original circle.
Since $P$ and $Q$ are two of the six points equally spaced around the circle, $\angle P O Q=\frac{1}{6} \times 360^{\circ}=60^{\circ}$. The outer circle has radius $2 \mathrm{~cm}$, so the area, in $\mathrm{cm}^{2}$, of sector $P O Q$ is $\frac{1}{6} \times \pi \times 2^{2}=\frac{2 \pi}{3}$.

Since $U$ and $P$ are also equally spaced, $\angle U O P=60^{\circ}$. So $\angle U O X=30^{\circ}$.
Hence the area of triangle $U O X$, in $\mathrm{cm}^{2}$, is $\frac{1}{2} \times 1 \times 2 \times \sin 30^{\circ}=\frac{1}{2}$.
So the area of quadrilateral $U O P X$ is $1 \mathrm{~cm}^{2}$.
Therefore the total shaded area, in $\mathrm{cm}^{2}$, in the original diagram is $3 \times\left(\frac{2 \pi}{3}+1\right)=2 \pi+3$.