Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is

The correct option is 5:24.

To find the ratio of the area of trapezium $PBCQ$ to the area of the regular hexagon $ABCDEF$, let us break down the geometry step-by-step.

Let the side length of the regular hexagon $ABCDEF$ be $s$.

Step 1: Calculate the total area of the regular hexagon $ABCDEF$
A regular hexagon with side length $s$ consists of 6 congruent equilateral triangles, each having a side length of $s$.

$\text{Area}\left(ABCDEF\right)=6\times \frac{\sqrt{3}}{4}{s}^2=\frac{3\sqrt{3}}{2}{s}^2$

Step 2: Determine the dimensions of trapezium $PBCQ$
Let us set up a Cartesian coordinate system to find the lengths of the parallel sides and the height of trapezium $PBCQ$.
Let the side $BC$ lie along the x-axis such that vertex $B=(-\frac{s}{2},0)$ and vertex $C=(\frac{s}{2},0)$.
The length of side $BC$ is equal to $s$.

Each interior angle of a regular hexagon is 120°.
- Vector $BA$ makes an angle of 60° relative to the negative x-axis, so $A=(-s,\frac{\sqrt{3}}{2}s)$.
- Since $P$ is the midpoint of $AB$:

$P=(\frac{-s-\frac{s}{2}}{2},\frac{\frac{\sqrt{3}}{2}s+0}{2})=(-\frac{3s}{4},\frac{\sqrt{3}}{4}s)$

- Similarly, vector $CD$ makes an angle of 60° relative to the positive x-axis, so $D=(s,\frac{\sqrt{3}}{2}s)$.
- Since $Q$ is the midpoint of $CD$:

$Q=(\frac{s+\frac{s}{2}}{2},\frac{\frac{\sqrt{3}}{2}s+0}{2})=(\frac{3s}{4},\frac{\sqrt{3}}{4}s)$

From the coordinates of $P$ and $Q$:
1. The line segment $PQ$ is parallel to $BC$ and has length:

$PQ=\frac{3s}{4}-(-\frac{3s}{4})=\frac{3s}{2}$

2. The perpendicular height $h$ of trapezium $PBCQ$ is the y-coordinate of $P$ and $Q$:

$h=\frac{\sqrt{3}}{4}s$

Step 3: Calculate the area of trapezium $PBCQ$
The area of a trapezium is given by $\frac{1}{2}\times \left(\text{sum of parallel sides}\right)\times \text{height}$:

$\text{Area}\left(PBCQ\right)=\frac{1}{2}\times (BC+PQ)\times h$

$\text{Area}\left(PBCQ\right)=\frac{1}{2}\times (s+\frac{3s}{2})\times \frac{\sqrt{3}}{4}s=\frac{1}{2}\times \frac{5s}{2}\times \frac{\sqrt{3}}{4}s=\frac{5\sqrt{3}}{16}{s}^2$

Step 4: Find the ratio of the areas

$\text{Ratio}=\frac{\text{Area}\left(PBCQ\right)}{\text{Area}\left(ABCDEF\right)}=\frac{\frac{5\sqrt{3}}{16}{s}^2}{\frac{3\sqrt{3}}{2}{s}^2}=\frac{5}{16}\times \frac{2}{3}=\frac{5}{24}$

Thus, the required ratio is 5:24.