22. The bases of the two touching squares shown lie on the same straight line. The lengths of the diagonals of the larger square and the smaller square are $10 \mathrm{~cm}$ and $8 \mathrm{~cm}$ respectively. $P$ is the centre of the smaller square. What is the area, in $\mathrm{cm}^{2}$, of the shaded triangle $P Q R$ ?
A 18
B 20
C 22
D 24
E 26
Solution
B

Angles $R S T$ and $T S Q$ are each $45^{\circ}$ so triangle $R S Q$ is a right-angled triangle with area, in $\mathrm{cm}^{2}$, equal to $\frac{1}{2} \times R S \times Q S=\frac{1}{2} \times 10 \times 8=40$. $P$ is the midpoint of $Q S$ so the area of triangle $P Q R$ is half of the area of triangle $R S Q$, that is $20 \mathrm{~cm}^{2}$.