24. The diagram shows a regular hexagon $R S T U V W$.
The area of the shaded pentagon $R S T P Q$ is one quarter of the area of the hexagon. Jay and Kay walk around the hexagon from $P$ to $Q$, Jay going clockwise and Kay anticlockwise.
What is the ratio of the distance Jay walks to the distance Kay walks?

A $1: 2$
B $2: 3$
С $3: 5$
D $4: 7$
E $5: 8$
24. C Let the side of the hexagon be of length $2 x$. Also, let $P T$ and $T R$ have length $y$ and $2 h$ respectively. Let $O$ be the centre of the hexagon.
The area of the hexagon may be divided into six equilateral triangles of side-length $2 x$. Also, the height of each triangle is $h$. Then the area of the hexagon is $6 \times \frac{1}{2} \times 2 x \times h=6 x h$.
The shaded area is the sum of the area of rectangle $P Q R T$ and the area of triangle RST. Now the area of triangle $R S T$ is equal to the area of triangle $O S T$, so the shaded area is $y \times 2 h+\frac{1}{2} \times 2 x \times h=2 h y+x h$.
Therefore $2 h y+x h=\frac{1}{4} \times(6 x h)$, that is $8 h y+4 x h=6 x h$.
So $8 y=2 x$, as $h \neq 0$. Hence $y=\frac{x}{4}$. Now as Jay walks clockwise around the figure, she walks a distance $2 y+4 x$, that is $2 \times \frac{x}{4}+4 x$. So Jay walks $\frac{9 x}{2}$ between $P$ and $Q$.
Therefore the distance which Kay walks between $P$ and $Q$ is $6 \times 2 x-\frac{9 x}{2}=\frac{15 x}{2}$. Hence the required ratio is $\frac{9 x}{2}: \frac{15 x}{2}=9: 15=3: 5$.