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 answer is 5:24.

To find the ratio of the areas, let's denote the side length of the regular hexagon ABCDEF as $a$. The area of a regular hexagon can be calculated by dividing it into $6$ equilateral triangles, each with side length $a$.

Let $T$ be the area of one such equilateral triangle. The area of an equilateral triangle with side length $a$ is given by:

$T=\frac{\sqrt{3}}{4}{a}^2$

Therefore, the total area of the regular hexagon ABCDEF is $6T$.

Now, let's find the area of the trapezium PBCQ. We can determine its area by splitting it into two separate triangles by drawing an imaginary line segment from P to C. This gives us $△PBC$ and $△PCQ$. Let's calculate the area of each triangle.

1. Area of $△PBC$:

In $△PBC$, we know the lengths of the two sides that form the interior angle at vertex B. Since P is the midpoint of AB, the length $PB=\frac{a}{2}$. The side $BC=a$. In a regular hexagon, every interior angle is $120°$, meaning $\angle PBC=120°$.

Using the sine formula for the area of a triangle, we get:

$\text{Area}(△PBC)=\frac{1}{2}\cdot PB\cdot BC\cdot \sin (120°)$

$\text{Area}(△PBC)=\frac{1}{2}\cdot \left(\frac{a}{2}\right)\cdot a\cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{8}{a}^2$

We can express this area in terms of $T$. Notice that $\frac{\sqrt{3}}{8}{a}^2$ is exactly half of $T$. Thus, the Area of $△PBC=\frac{1}{2}T$.

2. Area of $△PCQ$:

For $△PCQ$, let's consider CQ as the base. Since Q is the midpoint of CD, the base $CQ=\frac{a}{2}$. The height of this triangle will be the perpendicular distance from vertex P to the line containing the segment CD.

To find this perpendicular distance, we can look at the distances from vertices A and B to the line CD. The distance from B to CD is the altitude of an isosceles triangle with sides $a$, $a$ and angle $120°$, which computes to $a\sin (60°)=\frac{a\sqrt{3}}{2}$. Meanwhile, the distance from A to CD represents the total distance between the two parallel sides of the hexagon (AF and CD), which is known geometrically to be $a\sqrt{3}$.

Because point P is exactly halfway between A and B (the midpoint), its distance to the line CD is simply the average of the distances from A and B to CD:

$\text{Height from P to CD}=\frac{1}{2}\left(a\sqrt{3}+\frac{a\sqrt{3}}{2}\right)=\frac{3a\sqrt{3}}{4}$

Now, substitute this height and the base $CQ$ into the triangle area formula:

$\text{Area}(△PCQ)=\frac{1}{2}\cdot \text{Base}\cdot \text{Height}=\frac{1}{2}\cdot \left(\frac{a}{2}\right)\cdot \left(\frac{3a\sqrt{3}}{4}\right)=\frac{3\sqrt{3}}{16}{a}^2$

Expressing this in terms of $T$, we have $\frac{3}{4}\left(\frac{\sqrt{3}}{4}{a}^2\right)=\frac{3}{4}T$.

3. Total Ratio Calculation:

Adding the two parts together gives the total area for trapezium PBCQ:

$\text{Area}\left(PBCQ\right)=\text{Area}(△PBC)+\text{Area}(△PCQ)=\frac{1}{2}T+\frac{3}{4}T=\frac{5}{4}T$

Finally, we find the ratio of the area of the trapezium PBCQ to the area of the whole hexagon ABCDEF:

$\text{Ratio}=\frac{\frac{5}{4}T}{6T}=\frac{5}{4\cdot 6}=\frac{5}{24}$

Therefore, the ratio of the area of the trapezium PBCQ to that of the hexagon is 5:24.