CAT 2025 Slot 3 · Quant · Question 3

AD is perpendicular to both parallel sides, so AD is the height of the trapezium.

The circle touches AB and DC, which are parallel, so the distance between them is two radii: AD=2\times 3=6 cm.

Write DC=x, so AB=3x.

A quadrilateral with an inscribed circle touching all four sides satisfies AB+DC=AD+BC, so 3x+x=6+BC and BC=4x-6.

Placing A at the origin, B=\left( 3x,0 \right), D=\left( 0,6 \right) and C=\left( x,6 \right), the slant side gives BC=\sqrt{{\left( 3x-x \right)}^{2}+{6}^{2}}=\sqrt{4{x}^{2}+36}.

Equating, {\left( 4x-6 \right)}^{2}=4{x}^{2}+36, so 16{x}^{2}-48x+36=4{x}^{2}+36 and 12{x}^{2}=48x.

Since x>0, this gives x=4, so DC=4 and AB=12.

The area is \frac{1}{2}\left( 12+4 \right)\times 6=48 sq cm.

Shortcut. Two facts settle almost everything before any algebra: the height of a trapezium with an inscribed circle is 2r, and the two pairs of opposite sides have equal sums. Only the slant side then needs Pythagoras.