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Area of Trapezium in Circle

CAT 2025 Slot 1 · Quantitative Ability · Hard · Geometry

This is a hard Quantitative Ability question from the CAT 2025 Slot 1 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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In a circle with center CC and radius 62 cm6\sqrt{2}\text{ cm}, PQPQ and SRSR are two parallel chords separated by one of the diameters. If ∠PQC=45∘\angle PQC = 45^\circ, and the ratio of the perpendicular distance of PQPQ and SRSR from CC is 3:23:2, then the area, in sq. cm, of the quadrilateral PQRSPQRS is

  1. A.

    20(3 + \sqrt{14})

  2. B.

    4(3 + \sqrt{14})

  3. C.

    4(3\sqrt{2} + \sqrt{7})

  4. D.

    20(3\sqrt{2} + \sqrt{7})

Answer

A

Explanation

In △PQC\triangle PQC, CP=CQ=R=62 cmCP = CQ = R = 6\sqrt{2}\text{ cm}. Since ∠PQC=45∘\angle PQC = 45^\circ, △PQC\triangle PQC is an isosceles right triangle with ∠PCQ=90∘\angle PCQ = 90^\circ. Perpendicular distance from CC to PQPQ, d1=Rcos⁡45∘=62×12=6 cmd_1 = R \cos 45^\circ = 6\sqrt{2} \times \frac{1}{\sqrt{2}} = 6\text{ cm}. Chord length PQ=2R2−d12=272−36=12 cmPQ = 2 \sqrt{R^2 - d_1^2} = 2 \sqrt{72 - 36} = 12\text{ cm}.

Given ratio d1:d2=3:2  ⟹  6:d2=3:2  ⟹  d2=4 cmd_1 : d_2 = 3 : 2 \implies 6 : d_2 = 3 : 2 \implies d_2 = 4\text{ cm}. Chord length SR=2R2−d22=272−16=256=414 cmSR = 2 \sqrt{R^2 - d_2^2} = 2 \sqrt{72 - 16} = 2 \sqrt{56} = 4\sqrt{14}\text{ cm}.

Height of trapezium PQRS=d1+d2=6+4=10 cmPQRS = d_1 + d_2 = 6 + 4 = 10\text{ cm}. Area=12(PQ+SR)×h=12(12+414)×10=5(12+414)=20(3+14) sq. cm.\text{Area} = \frac{1}{2} (PQ + SR) \times h = \frac{1}{2} (12 + 4\sqrt{14}) \times 10 = 5(12 + 4\sqrt{14}) = 20(3 + \sqrt{14})\text{ sq. cm}.

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