CAT 2025 Slot 3 QA Q3 Geometry

Question 3 of the Quantitative Ability (QA) section, CAT 2025 Slot 3, with CAT's official answer and a worked solution.

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

A
\( 36 \sqrt { 2 } \)
B
\( 54 \)
C
\( 48 \)
D
\( 30 \sqrt { 3 } \)

Answer: (C) 48 Geometry MCQ

Solution

Since \(AD \perp AB\), the height equals the circle's diameter: \(AD = 6\).
Let \(DC = x\), \(AB = 3x\). Slant side \(BC = \sqrt{(2x)^2 + 6^2}\).
A tangential quadrilateral has equal sums of opposite sides: \(AB + DC = AD + BC\), so \(4x - 6 = \sqrt{4x^2 + 36}\).
Squaring: \(16x^2 - 48x + 36 = 4x^2 + 36 \Rightarrow 12x^2 = 48x \Rightarrow x = 4\).
Area \(= \frac{12 + 4}{2} \times 6 = 48\).

Shortcut. Pitot's theorem (sums of opposite sides equal) gives the parallel sides directly.