CAT 2025 Question Paper | Quant Slot 3

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 3 : ABCD is a trapezium in which AB is parallel to \( \mathrm { DC } , \mathrm { AD } \) is perpendicular to AB , and \( \mathrm { AB } = 3 \mathrm { DC } \). If a circle inscribed in the trapezium touching all the sides has a radius of 3 cm , then the area, in sq. cm , of the trapezium is

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Video Explanation

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Explanatory Answer

Since the circle has radius \(3\) cm and touches both parallel sides, the height of the trapezium is the diameter:
\(AD=2\times3=6\) cm.

Let \(DC=c\). Given \(AB=3DC\):
\(DC=c,\quad AB=3c\).

Take \(B(3c,0)\) and \(C(c,6)\). The slope of \(BC\) is:
\[ \frac{6-0}{c-3c}=-\frac{3}{c} \]
Therefore, the equation of \(BC\) is:
\[ 3x+cy-9c=0 \]

The centre of the circle is \(O(3,3)\). Since the circle touches \(BC\), the perpendicular distance from \(O\) to \(BC\) equals the radius \(3\):
\[ \frac{|3(3)+c(3)-9c|}{\sqrt{3^2+c^2}}=3 \]
\[ \frac{|9-6c|}{\sqrt{9+c^2}}=3 \]

Dividing by \(3\) and squaring:
\[ (3-2c)^2=9+c^2 \]
\[ 3c(c-4)=0 \]
Since \(c>0\), \(c=4\).

Therefore:
\(DC=4\) cm and \(AB=12\) cm.

Area of trapezium:
\[ \frac{1}{2}(AB+DC)\times AD =\frac{1}{2}(12+4)\times6 =48\text{ cm}^2 \]

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The answer is '\( 48 \)'

Choice 3 is the correct answer.

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