In the diagram, rectangle $P Q R S$ has $P S=2$ and $P Q=4$. Points $T, U, V, W$ are positioned so that $R T=R U=P W=P V=a$. If $V U$ and $W T$ pass through the centre of the rectangle, for what value of $a$ is the shaded region $\frac{1}{8}$ the area of $P Q R S$ ?
(A) $\frac{2}{3}$
(D) $\frac{1}{3}$
(B) $\frac{1}{2}$
(E) $\frac{1}{4}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-017.jpg?height=250&width=401&top_left_y=1187&top_left_x=1296)
We begin by joining the centre of the rectangle, $O$, to vertex $P$.

We also draw $O M$ perpendicular to side $P Q$ and $O N$ perpendicular to side $P S$.

Since $O$ is the centre of the rectangle, then $M$ is the midpoint of side $P Q$ and so $P M=\frac{1}{2} \times 4=2$.

Similarly, $N$ is the midpoint of $P S$ and so $P N=\frac{1}{2} \times 2=1$.

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-079.jpg?height=263&width=436&top_left_y=2075&top_left_x=1533)

$\triangle P V O$ has base $P V=a$ and height $O M=1$, and so has area $\frac{1}{2} \times a \times 1=\frac{1}{2} a$.

$\triangle P W O$ has base $P W=a$ and height $O N=2$, and so has area $\frac{1}{2} \times a \times 2=a$.

Thus, quadrilateral $P W O V$ has area equal to the sum of the areas of these two triangles, or $\frac{1}{2} a+a=\frac{3}{2} a$.

Similarly, we can show that quadrilateral $R T O U$ also has area $\frac{3}{2} a$ and so the total area of the shaded region is $2 \times \frac{3}{2} a=3 a$.

The area of rectangle $P Q R S$ is $4 \times 2=8$ and since the area of the shaded region is $\frac{1}{8}$ the area of $P Q R S$, then $3 a=\frac{1}{8} \times 8$ or $3 a=1$ and so $a=\frac{1}{3}$.

ANSWER: (D)