25. A semicircle is inscribed in a quarter circle as shown. What fraction of the quarter circle is shaded?
A $\frac{1}{3}$
B $\frac{1}{\sqrt{3}}$
C $\frac{2}{3}$
D $\frac{\sqrt{3}}{2}$
E $\frac{1}{\sqrt{2}}$
Solution
C
Let $O$ be the centre of the quarter circle, $A$ and $B$ be the ends of the diameter of the semicircle and $M$ be the centre of the semicircle, and hence the midpoint of $A B$.

Let $P$ and $Q$ be the points on the straight edges of the quarter circle where the quarter circle is tangent to the semicircle.

We let the radius of the quarter circle be $s$ and let the radius of the semicircle be $r$.

Then $O A=s$.
As a tangent to a circle is perpendicular to its radius, $\angle O P M=\angle O Q M=90^{\circ}$. Because also $\angle P O Q=90^{\circ}$, all the angle of the quadrilateral $O Q M P$ are right angles. Also $P M=r=Q M$. Hence $O P M Q$ is a square with side length $r$.

Applying Pythagoras' theorem to the right-angled triangle $O Q M$ gives $O M^{2}=r^{2}+r^{2}=2 r^{2}$.
The triangle $A M O$ has a right angle at $M$. Therefore, by Pythagoras' theorem applied to this triangle, we have
$$
A O^{2}=O M^{2}+M A^{2}
$$
and hence
$$
\begin{aligned}
s^{2} & =2 r^{2}+r^{2} \\
& =3 r^{2} .
\end{aligned}
$$

The area of the quarter circle is $\frac{1}{4} \pi s^{2}$. The area of the shaded semicircle is $\frac{1}{2} \pi r^{2}$.
Therefore, the fraction of the quarter circle which is shaded is given by
$$
\frac{\frac{1}{2} \pi r^{2}}{\frac{1}{4} \pi s^{2}}=\frac{2 r^{2}}{s^{2}}=\frac{2 r^{2}}{3 r^{2}}=\frac{2}{3}
$$