23. The length of a rectangular piece of paper is three times its width. The paper is folded so that one vertex lies on top of the opposite vertex, thus forming a pentagonal shape.
What is the area of the pentagon as a fraction of the area of the original rectangle?
A $\frac{2}{3}$
B $\frac{11}{16}$
C $\frac{12}{17}$
D $\frac{13}{18}$
$\mathrm{E} \frac{14}{19}$
SolUtion
D

Let $P, Q, R$ and $S$ be the vertices of the rectangular piece of paper. We choose units so that $P Q=1$ and $P S=3$.
Because the fold moves $Q$ to $S$, the fold line goes through the midpoint of $Q S$ which is the centre of the rectangle. We let $U$ and $V$ be the points where this fold line meets the edges of the rectangle, as shown in the diagram.
We let $x$ be the length of $V R$. The rectangle has a rotational symmetry that interchanges $U$ and $V$. Therefore, $P U=V R=x$. Hence $U S=3-x$.
We let $T$ be the point where the vertex $Q$ ends up after the fold. The quadrilateral $S T U V$ is the reflection of $Q P U V$ in the line through $U$ and $V$. It follows that $T S=1, U T=x$ and $\angle U T S$ is a right angle.

We need to find the area of the pentagon RSTUV. This pentagon is made up of the trapezium $R S U V$ and the triangle $S T U$.

It follows from the rotational symmetry of the rectangle that the area of the trapezium RSUV is half the area of the rectangle $P Q R S$. Therefore the area of the trapezium is $\frac{1}{2}(3 \times 1)=\frac{3}{2}$.

The area of the right-angled triangle $S T U$ is $\frac{1}{2}(1 \times x)=\frac{x}{2}$.
By Pythagoras' Theorem applied to the triangle $U S T$, we have $1+x^{2}=(3-x)^{2}$. Therefore $1+x^{2}=9-6 x+x^{2}$. It follows that $6 x=8$. Hence $x=\frac{8}{6}=\frac{4}{3}$.
It follows that the area of the pentagon $R S T U V$ is given by $\frac{3}{2}+\frac{4 / 3}{2}=\frac{3}{2}+\frac{2}{3}=\frac{13}{6}$.
The area of rectangle $P Q R S$ is 3 . Therefore the area of the pentagon as a fraction of the area of the original rectangle is $\frac{13 / 6}{3}=\frac{13}{18}$.