24. The top diagram on the right shows a shape that tiles the plane, as shown in the lower diagram. The tile has nine sides, six of which have length 1. It may be divided into three congruent quadrilaterals as shown. What is the area of the tile?
A $\frac{1+2 \sqrt{3}}{2}$
B $\frac{4 \sqrt{3}}{3}$
C $\sqrt{6}$
D $\frac{3+4 \sqrt{3}}{4}$
E $\frac{3 \sqrt{3}}{2}$
Solution: B
The tile is made up of three congruent quadrilaterals one of which, $P Q R S$, is shown in the diagram on the right. We let $T$ be the foot of the perpendicular from $Q$ to $S R$ and we let $U$ be the point shown. Then $S R=U R=1$.
Also $P Q=Q U$. Let their common length be $x$. Hence $Q R=1+x$ and $T R=1-x$. Because the three congruent quadrilaterals fit together at $Q$, we have $\angle P Q R=120^{\circ}$. As $P Q$ and $S R$ are both perpendicular to $P S$, they are parallel and hence $\angle Q R T=60^{\circ}$. Hence, from the right-angled triangle $Q R T$ we have $\frac{T R}{Q R}=\sin 60^{\circ}$, that is, $\frac{1-x}{1+x}=\frac{1}{2}$, and therefore $2-2 x=1+x$. Hence $3 x=1$ and so $x=\frac{1}{3}$. Therefore $T R=\frac{2}{3}$ and $Q R=\frac{4}{3}$. Now $Q T / Q R=\cos 60^{\circ}$ and hence $Q T=Q R \cos 60^{\circ}$ $=\frac{4}{3} \cos 60^{\circ}=\frac{4}{3}\left(\frac{\sqrt{3}}{2}\right)=\frac{2 \sqrt{3}}{3}$. [We could also use Pythagoras' Theorem applied to the triangle $Q R T$ to deduce this length.]
We can now work out the area of the quadrilateral $P Q R S$ in two different ways. Most straightforwardly, area $(P Q R S)=\operatorname{area}(P Q T S)+\operatorname{area}(Q R T)=\frac{2 \sqrt{3}}{3} \times \frac{1}{3}+\frac{1}{2}\left(\frac{2 \sqrt{3}}{3} \times \frac{2}{3}\right)=\frac{4 \sqrt{3}}{9}$. Alternatively, as $P Q$ is parallel to $S R, P Q R S$ is a trapezium. We now use the fact that the area of a trapezium is its height multiplied by the average length of its parallel sides. Hence, $P Q R S$ has area. $\frac{2 \sqrt{3}}{3} \times \frac{1}{2}\left(\frac{1}{3}+1\right)$ $=\frac{2 \sqrt{3}}{3} \times \frac{2}{3}=\frac{4 \sqrt{3}}{9}$. The tile is made up of three copies of $P Q R S$. So its area is $3 \times \frac{4 \sqrt{3}}{9}=\frac{4 \sqrt{3}}{3}$.