CAT 2023 Slot 1 · Quant · Question 19

The right angle is at B, so the area of \triangle ABC is \frac{1}{2}\times 5\times 12=30 sq cm.

The area of \triangle ABC is 1.5 times the area of \triangle ABP, so the latter is \frac{30}{1.5}=20 sq cm.

The three areas — 20, the area of \triangle ABQ, and 30 — are in arithmetic progression, so the middle term is the average of the outer two.

So the area of \triangle ABQ is \frac{20+30}{2}=25 sq cm.

Every one of these triangles has the same altitude AB from the line BC, so area is proportional to the base along BC.

For \triangle ABP, \frac{1}{2}\times 5\times BP=20, giving BP=8 cm.

For \triangle ABQ, \frac{1}{2}\times 5\times BQ=25, giving BQ=10 cm.

Both P and Q lie on BC, with BP=8 and BQ=10, both within the 12 cm base.

So PQ=10-8=2 cm.

Shortcut. With the altitude fixed, areas and bases are in the same ratio, so the areas 20,25,30 correspond to bases 8,10,12. The step in the area AP is 5, which translates to a step of 2 cm along the base — that step is PQ.