CAT 2025 Question Paper | Quant Slot 2

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 6 : Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is

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Video Explanation

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Explanatory Answer

Let the side length of the regular hexagon be \(a\).
Length \(PB=\frac{a}{2}\)
Length \(QC=\frac{a}{2}\)
\(BC=a\)

Trapezium \(PBCQ\) has parallel sides \(BC=a\) and median line \(PQ\).

The distance from center to midpoint \(PQ\) gives \(PQ=1.5a\).

Height of trapezium \(PBCQ=\frac{\sqrt{3}}{4}a\)

Area of trapezium \(PBCQ\)
\(=\frac{1}{2}\times(a+1.5a)\times\frac{\sqrt{3}}{4}a\)
\(=\frac{1}{2}\times2.5a\times\frac{\sqrt{3}}{4}a\)
\(=\frac{5\sqrt{3}}{16}a^2\)

Area of Hexagon \(ABCDEF\)
\(=6\times\left(\frac{\sqrt{3}}{4}a^2\right)\)
\(=\frac{6\sqrt{3}}{4}a^2=\frac{24\sqrt{3}}{16}a^2\)

Ratio
\(=\frac{\frac{5\sqrt{3}}{16}a^2}{\frac{24\sqrt{3}}{16}a^2}\)
\(=\frac{5}{24}\)

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The question is " Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is "

Hence, the answer is '5 : 24'

Choice 1 is the correct answer.

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