18. A square has two of its vertices on a semicircle and the other two on the diameter of the semicircle as shown. The radius of the circle is 1 . What is the area of the square?
A $\frac{4}{5}$
B $\frac{\pi}{4}$
C 1
$\mathrm{D} \frac{4}{3}$
$\mathrm{E} \frac{2}{\sqrt{3}}$
18. A Let $O$ be the centre of the circle, and $P, Q, P^{\prime}$ and $Q^{\prime}$ the vertices of the square. The triangles $O P Q$ and $O P^{\prime} Q^{\prime}$ are congruent since they are right-angled and have two equal sides ( $P Q=P^{\prime} Q^{\prime}$ since they are edges of a square, and $O P=O P^{\prime}$ because each is a radius). Hence $O Q=O Q^{\prime}$, and $O$ is thus the midpoint of the edge of the square.
Let the side of the square be $2 x$. And $O P=1$ since it is a radius. And by Pythagoras' Theorem on triangle $O P Q$ we have $(2 x)^{2}+x^{2}=1^{2}$, so $4 x^{2}+x^{2}=1$, and $5 x^{2}=1$. This gives $x^{2}=\frac{1}{5}$. The area of the square is $(2 x)^{2}=4 x^{2}=4 \times \frac{1}{5}=\frac{4}{5}$.