25. The diagram shows a circle with radius 2 and a square with sides of length 2 . The centre of the circle lies on the perpendicular bisector of a side of the square, at a distance $x$ from the side, as shown. The shaded region - inside the square but outside the circle - has area 2 .
What is the value of $x$ ?
$$
\begin{array}{l}
\text { A } \frac{\pi}{3}+\frac{\sqrt{3}}{2}-1 \\
\text { B } \frac{\pi}{3}+\frac{\sqrt{3}}{4}-1 \\
\text { C } \frac{\pi}{3}+\frac{1}{2} \\
\text { D } \frac{\pi}{3}+1 \\
\text { E } \frac{\pi}{3} \\
\end{array}
$$
25. A The diagram shows part of the diagram in the question. Point $O$ is the centre of the circle and points $A, B, C, D, E, F$, are as shown. Consider triangle $B O F$ : $B F$ is equal in length to the side of the square so $B F=2$. Also $O B=O F=2$ as they are both radii of the circle. So triangle $B O F$ is equilateral.
Therefore $\angle B O F=60^{\circ}$, so the area of sector $O B F=\frac{60}{360} \times \pi \times 2^{2}=\frac{2 \pi}{3}$.
By Pythagoras' Theorem: $O A=\sqrt{2^{2}-1^{2}}=\sqrt{3}$.

So the area of triangle $O B F=\frac{1}{2} \times 2 \times \sqrt{3}=\sqrt{3}$. Therefore the area of the segment bounded by arc $B F$ and line segment $B F=\frac{2 \pi}{3}-\sqrt{3}$. The area of rectangle $B F E C=B F \times A D=2 \times(\sqrt{3}-x)$.
The shaded region has area 2 , so the area of the above segment + area of rectangle $B F E C=$ area of the given square minus $2=4-2=2$.
Hence $\frac{2 \pi}{3}-\sqrt{3}+2 \sqrt{3}-2 x=2$. So $x=\frac{\pi}{3}+\frac{\sqrt{3}}{2}-1$.