17. A circle of radius $r$ and a right-angled isosceles triangle are drawn such that one of the shorter sides of the triangle is a diameter of the circle.
What is the shaded area?
A $\sqrt{2} r$
B $r^{2}$
C $2 \pi r$
D $\frac{\pi r^{2}}{4}$
$\mathrm{E}(\sqrt{2}-1) \pi r^{2}$
Solution
B

Let $O$ be the centre of the circle and let $P, Q, R$ and $S$ be the points as shown in the diagram.
Because $P Q R$ is a right-angled isosceles triangle, $\angle P R Q=$ $\angle R P Q=45^{\circ}$, and $P Q=Q R=2 r$.
Because the angle in a semicircle is a right angle [this is Thales' theorem], $\angle R S Q=90^{\circ}$. Therefore, because the sum of the angles in a triangle is $180^{\circ}$, we have $\angle R Q S=45^{\circ}$.

We therefore have $\angle S R Q=45^{\circ}=\angle S Q R$. It follows that $S Q=S R$.
Because $S Q=S R$ the segments of the circle cut off by these lines, shown as hatched in the diagram, are congruent. Hence they have the same area.
It follows that the shaded area is the same as the area of the triangle $P Q S$.
$P Q S$ is a right-angled isosceles triangle with hypotenuse $P Q$ of length $2 r$. Therefore the triangle $P Q S$ has area $r^{2}$.
Therefore the shaded area is $r^{2}$.