CAT 2025 Question Paper | Quant Slot 1

CAT Questions

CAT 2025 Quant was dominated by Algebra followed by Arithmetic. In Arithmetic, the questions were dominated by topics like Speed-time-distance, Mixture and Alligations. This year, there was a surprise. The questions from Geometry were relatively on the lower side as compared to the previous years. There were 8 TITA Qs this year. Overall this section was at a medium level of difficulty.

Question 12 : 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

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Video Explanation

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Explanatory Answer

Isosceles Property:
\(CP=CQ=6\sqrt{2}\text{ cm}\)
\(r=6\sqrt{2}\)
\(\angle QPC=\angle PQC=45^\circ\)
\(\angle PCQ=90^\circ\)

Perpendicular Distance (\(d_1\)):
Dropping height \(d_1\) splits \(\triangle PQC\) into two identical \(45^\circ-45^\circ-90^\circ\) right isosceles triangles where \(d_1=\frac{PQ}{2}\).

\(d_1=r\sin45^\circ\)
\(=6\sqrt{2}\times\frac{1}{\sqrt{2}}=6\text{ cm}\)

Given \(d_1:d_2=3:2\):
\(\frac{6}{d_2}=\frac{3}{2}\Rightarrow d_2=4\text{ cm}\)

Using the Pythagorean theorem for chord \(SR\):
\(\left(\frac{SR}{2}\right)^2+d_2^2=r^2\)
\(\left(\frac{SR}{2}\right)^2+4^2=(6\sqrt{2})^2\)
\(SR=4\sqrt{14}\text{ cm}\)

Since the two parallel chords lie on opposite sides of a diameter:
Height \(h=d_1+d_2=6+4=10\text{ cm}\)

\(PQ=12\)

Area \(=\frac{1}{2}\times(PQ+SR)\times h\)
\(=\frac{1}{2}\times(12+4\sqrt{14})\times10\)
\(=20(3+\sqrt{14})\text{ sq. cm}\)

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The answer is '\(20(3+\sqrt{14})\)'

Choice 3 is the correct answer.

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