Area Ratio in Regular Hexagon

Let side length of regular hexagon be $s = 2$. Total area of regular hexagon $A = 6 \times \frac{\sqrt{3}}{4} s^2 = 6 \sqrt{3}$. $P$ is midpoint of $AB \implies PB = 1$. $Q$ is midpoint of $CD \implies CQ = 1$. Trapezium $PBCQ$ has parallel sides $PB = 1$ and $QC = 1$? Wait, $BC = 2$. $P, B, C, Q$ form a quadrilateral / isosceles trapezium. The area of $PBCQ$ can be calculated by dissecting into triangles. The resulting ratio of Area($PBCQ$) to Total Area = $5/24$.