Q6
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
Answer: A
Key idea: In a regular hexagon, BC is parallel to AD, and the segment joining the midpoints of AB and CD is the midline of trapezium ABCD.
Take side \(= 1\). Then \(AD = 2\) (a long diagonal) and the distance between BC and AD is \(\dfrac{\sqrt3}{2}\).
PQ is the midline of trapezium ABCD, so \(PQ = \dfrac{BC + AD}{2} = \dfrac{1+2}{2} = \dfrac32\), and it lies halfway between BC and AD, so the height of PBCQ is \(\dfrac{\sqrt3}{4}\).
\[\text{Area}(PBCQ) = \frac12\left(1 + \frac32\right)\cdot\frac{\sqrt3}{4} = \frac{5\sqrt3}{16}.\] \[\text{Area}(ABCDEF) = \frac{3\sqrt3}{2} = \frac{24\sqrt3}{16}.\]Ratio \(= 5 : 24\).
Hence, the answer is (A) 5 : 24.