23. $P Q R S$ is a square. The points $T$ and $U$ are the midpoints of $Q R$ and $R S$ respectively. The line $Q S$ cuts $P T$ and $P U$ at $W$ and $V$ respectively.
What fraction of the total area of the square $P Q R S$ is the area of the pentagon RTWVU?
A $\frac{1}{3}$
B $\frac{2}{5}$
C $\frac{3}{7}$
D $\frac{5}{12}$
$\mathrm{E} \frac{4}{15}$
Solution
A The lines $P Q$ and $S R$ are parallel. Hence $\angle U S V=\angle P Q V$, since they are alternate angles. Similarly $\angle S U V=\angle P Q V$. It follows that the triangles $U S V$ and $Q P V$ are similar. Now $S U: P Q=1: 2$ and so the heights of these triangles are in the same ratio. So the height of triangle $U S V$ is $\frac{1}{3}$ of the side-length of the square. The base of this triangle is $\frac{1}{2}$ of the side of the square. Hence the area of this triangle is $\frac{1}{2}\left(\frac{1}{2} \times \frac{1}{3}\right)=\frac{1}{12}$ of the area of the square.
Similarly the area of triangle $Q W T$ is $\frac{1}{12}$ of the area of the square. The area of triangle $P Q S$ is $\frac{1}{2}$ of the area of the square. The area of the pentagon $R T W V U$ is the area of the square minus the total areas of the triangles $U S V, Q W T$ and $P Q S$, so its area, as a fraction of the area of the square $P Q R S$, is $1-\frac{1}{12}-\frac{1}{12}-\frac{1}{2}=\frac{1}{3}$.