A9. The triangle $P Q R$ is isosceles with $P R=Q R$. Angle $P R Q=90^{\circ}$ and length $P Q=2 \mathrm{~cm}$. Two arcs of radius $1 \mathrm{~cm}$ are drawn inside triangle $P Q R$. One arc has its centre at $P$ and intersects $P R$ and $P Q$. The other arc has its centre at $Q$ and intersects $Q R$ and $P Q$.
What is the area of the shaded region, in $\mathrm{cm}^{2}$ ?
Solution
Using Pythagoras' Theorem, $P R=Q R=\sqrt{2}$. The area of the triangle, in $\mathrm{cm}^{2}$, is $\frac{1}{2} \times \sqrt{2} \times \sqrt{2}=1$. The total area, in $\mathrm{cm}^{2}$, of the unshaded regions, which is equivalent to the area of a quarter of a circle with radius $1 \mathrm{~cm}$, is $\frac{\pi}{4}$. Therefore, the area of the shaded region, in $\mathrm{cm}^{2}$, is $\left(1-\frac{\pi}{4}\right)$.