NinetynineCAT 2025 Slot 1 QA Q12 Geometry
Question 12 of the Quantitative Ability (QA) section, CAT 2025 Slot 1, with CAT's official answer and a worked solution.
In a circle with center \(C\) and radius \(6 \sqrt{2} \mathrm{~cm}, P Q\) and \(S R\) are two parallel chords separated by one of the diameters. If \(\angle P Q C=45^{\circ}\), and the ratio of the perpendicular distance of \(P Q\) and \(S R\) from \(C\) is \(3: 2\), then the area, in sq. cm , of the quadrilateral \(P Q R S\) is
- A\(4(3+\sqrt{14})\)
- B\(20(3 \sqrt{2}+\sqrt{7})\)
- C\(20(3+\sqrt{14})\)
- D\(4(3 \sqrt{2}+\sqrt{7})\)
Answer: (C) 20(3+√(14)) Geometry MCQ
Solution
- \(CP = CQ\) and \(\angle PQC = 45^\circ\) make \(\angle PCQ = 90^\circ\), so \(PQ = 6\sqrt2 \cdot \sqrt2 = 12\) and its distance from \(C\) is 6.
- From the ratio \(3:2\), the distance of \(SR\) is 4, so half of \(SR\) is \(\sqrt{72 - 16} = 2\sqrt{14}\) and \(SR = 4\sqrt{14}\).
- The chords are on opposite sides of a diameter, so the trapezium's height is \(6 + 4 = 10\).
- Area \(= \frac{12 + 4\sqrt{14}}{2} \times 10 = 60 + 20\sqrt{14} = 20(3 + \sqrt{14})\).