20. A square is inscribed inside a quadrant of a circle. The circle has radius 10 .
What is the area of the square?
A $25 \sqrt{2}$
B 36
C $12 \pi$
D 40
E $30 \sqrt{2}$
SOLUTION
D
We let $O$ be the centre of the circle, and $P, Q, R$ and $S$ be the vertices of the square, as shown. We let $O L$ be the perpendicular from $O$ to $Q R$, and $K$ be the point where this perpendicular meets $P S$.

We let $s$ be the side length of the square. It may be checked (see Problem 20.1) that the triangles $O L Q$ and $O L R$ are congruent. It follows that $Q L=\frac{1}{2} Q R=\frac{1}{2} s=P K$.
It may be checked that $O K P$ is a right-angled isosceles triangle (see Problem 20.2). Therefore $O K=P K=\frac{1}{2} s$.
It follows that in the right-angled triangle $O L Q$ we have $Q L=\frac{1}{2} s, O L=O K+K L=\frac{1}{2} s+s=\frac{3}{2} s$ and $O Q=10$.

Therefore, by Pythagoras' Theorem
$$
\left(\frac{1}{2} s\right)^{2}+\left(\frac{3}{2} s\right)^{2}=10^{2}
$$

Hence
$$
\frac{1}{4} s^{2}+\frac{9}{4} s^{2}=100
$$

That is,
$$
\frac{5}{2} s^{2}=100
$$

Hence
$$
s^{2}=\frac{2}{5} \times 100=40 .
$$

It follows that the area of the square is 40 .
FOR INVESTIGATION
20.1 Show that the triangles $O L Q$ and $O L R$ are congruent.
20.2 Show that $O K P$ is a right-angled isosceles triangle.
20.3 The diagram on the right shows one larger square and two smaller squares inscribed in the quadrant of a circle.
The circle has radius 10 .
Find the side length of the smaller squares.