CAT 2025 Slot 2 · Quant · Question 6

Let each side be a. A regular hexagon has circumradius a, and opposite sides are parallel.

BC is a full side, so BC=a.

Placing the centre at the origin, P and Q are mirror images in the vertical axis and each lies \frac{3a}{4} from it, so PQ=\frac{3a}{2}.

BC sits at height \frac{a\sqrt{3}}{2} from the centre and PQ at \frac{a\sqrt{3}}{4}, so the perpendicular distance between them is \frac{a\sqrt{3}}{4}.

The trapezium area is \frac{1}{2}\left( a+\frac{3a}{2}\right)\times \frac{a\sqrt{3}}{4}=\frac{5\sqrt{3}{a}^{2}}{16}.

The hexagon area is \frac{3\sqrt{3}}{2}{a}^{2}=\frac{24\sqrt{3}{a}^{2}}{16}.

Dividing, the ratio is 5:24.

Shortcut. Cut the hexagon into 24 identical small triangles by joining the centre to every vertex and every side-midpoint. Counting the ones inside PBCQ gives 5 directly, so the ratio is 5:24 with no algebra at all.