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Area of Inscribed Circle Trapezium

CAT 2025 Slot 3 · Quantitative Ability · Hard · Geometry

This is a hard Quantitative Ability question from the CAT 2025 Slot 3 paper. It tests Geometry. The full answer key and a step-by-step explanation are below — try it yourself first, then reveal the solution.

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ABCD is a trapezium in which AB is parallel to DC, AD is perpendicular to AB, and AB = 3DC. 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

  1. A.

    48

  2. B.

    30330\sqrt{3}

  3. C.

    36236\sqrt{2}

  4. D.

    54

Answer

A

Explanation

Since AD⊥ABAD \perp AB and the circle of radius r=3r = 3 touches all sides, height of trapezium AD=2r=6AD = 2r = 6 cm. Let DC=xDC = x, then AB=3xAB = 3x. Extend non-parallel sides to intersect at EE. △DCE∼△ABE\triangle DCE \sim \triangle ABE. Since AD=6AD = 6, DE=3DE = 3 and AE=9AE = 9. CE=x2+9CE = \sqrt{x^2 + 9} and BE=3x2+9BE = 3\sqrt{x^2 + 9}. BC=2x2+9BC = 2\sqrt{x^2 + 9}. In right-angled triangle ABEABE, s1+s2−hypotenuse=2r  ⟹  9+3x−3x2+9=6  ⟹  3x+3=3x2+9  ⟹  x+1=x2+9  ⟹  x2+2x+1=x2+9  ⟹  x=4s_1 + s_2 - \text{hypotenuse} = 2r \implies 9 + 3x - 3\sqrt{x^2 + 9} = 6 \implies 3x + 3 = 3\sqrt{x^2 + 9} \implies x + 1 = \sqrt{x^2 + 9} \implies x^2 + 2x + 1 = x^2 + 9 \implies x = 4. Then DC=4DC = 4, AB=12AB = 12. Area of trapezium =12(AB+DC)×AD=12(12+4)×6=48= \frac{1}{2}(AB + DC) \times AD = \frac{1}{2}(12 + 4) \times 6 = 48 sq. cm.

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