CAT 2025 Slot 1 · Quant · Question 12

Since CP and CQ are both radii, triangle PCQ is isosceles, so \angle CPQ=\angle PQC={45}^{∘} and therefore \angle PCQ={90}^{∘}.

The perpendicular from C to PQ bisects it and makes a {45}^{∘} right triangle, so the distance is 6\sqrt{2}\cos{45}^{∘}=6 cm and the half-chord is also 6, giving PQ=12.

The distances are in the ratio 3:2, so SR is at distance 4 cm from C. Its half-length is \sqrt{{\left( 6\sqrt{2}\right)}^{2}-{4}^{2}}=\sqrt{72-16}=\sqrt{56}=2\sqrt{14}, so SR=4\sqrt{14}.

The chords are on opposite sides of a diameter, so the gap between them is 6+4=10 cm — this is the step to get right, as adding is what separated by a diameter means.

PQRS is a trapezium with parallel sides 12 and 4\sqrt{14}:

Area =\frac{1}{2}\left( 12+4\sqrt{14}\right)\left( 10 \right)=60+20\sqrt{14}=20\left( 3+\sqrt{14}\right) sq cm.