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

Place the regular hexagon $ABCDEF$ with side length $1$, centered at the origin. A regular hexagon has all sides equal and all interior angles equal to $120°$. The vertices are:

$A = \left(1,\ 0\right)$

$B = \left(\dfrac{1}{2},\ \dfrac{\sqrt{3}}{2}\right)$

$C = \left(-\dfrac{1}{2},\ \dfrac{\sqrt{3}}{2}\right)$

$D = \left(-1,\ 0\right)$

$E = \left(-\dfrac{1}{2},\ -\dfrac{\sqrt{3}}{2}\right)$

$F = \left(\dfrac{1}{2},\ -\dfrac{\sqrt{3}}{2}\right)$

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$P$ is the midpoint of $AB$:

$P = \dfrac{A + B}{2} = \left(\dfrac{3}{4},\ \dfrac{\sqrt{3}}{4}\right)$

$Q$ is the midpoint of $CD$:

$Q = \dfrac{C + D}{2} = \left(-\dfrac{3}{4},\ \dfrac{\sqrt{3}}{4}\right)$

Notice that $B$ and $C$ share the same $y$-coordinate $\left(\dfrac{\sqrt{3}}{2}\right)$, and $P$ and $Q$ share the same $y$-coordinate $\left(\dfrac{\sqrt{3}}{4}\right)$. Since $BC \parallel PQ$, the quadrilateral $PBCQ$ is a trapezium.

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Using the trapezium area formula:

$\text{Area} = \dfrac{1}{2} \times (b_1 + b_2) \times h$

where $b_1$ and $b_2$ are the parallel sides and $h$ is the perpendicular distance between them.

$BC = \left|\dfrac{1}{2} - \left(-\dfrac{1}{2}\right)\right| = 1$

$PQ = \left|\dfrac{3}{4} - \left(-\dfrac{3}{4}\right)\right| = \dfrac{3}{2}$

$h = \dfrac{\sqrt{3}}{2} - \dfrac{\sqrt{3}}{4} = \dfrac{\sqrt{3}}{4}$

So,

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

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The area of a regular hexagon with side $a$ is $\dfrac{3\sqrt{3}}{2}a^2$. For $a = 1$:

$\text{Area}_{\text{hexagon}} = \dfrac{3\sqrt{3}}{2}$

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$\text{Ratio} = \dfrac{\text{Area}_{PBCQ}}{\text{Area}_{\text{hexagon}}} = \dfrac{\dfrac{5\sqrt{3}}{16}}{\dfrac{3\sqrt{3}}{2}} = \dfrac{5\sqrt{3}}{16} \times \dfrac{2}{3\sqrt{3}} = \dfrac{10}{48} = \boxed{\dfrac{5}{24}}$