CAT 2025 Slot3 Quant: ABCD is a trapezium in which

1. Height of the Trapezium:
A circle inscribed in a trapezium touches the parallel sides. Thus, the height ($AD$) is equal to the diameter of the circle.
$h = AD = 2 \times 3 = 6$ cm.

2. Side Relationships:
Let $DC = x$, then $AB = 3x$.
For any tangential quadrilateral: $AB + DC = AD + BC$
$3x + x = 6 + BC \implies BC = 4x - 6$.

3. Solving for x:
Drop a perpendicular from $C$ to $AB$ at point $M$.
$CM = 6$ and $MB = AB - AM = 3x - x = 2x$.
In right $\triangle CMB$:
$BC^2 = CM^2 + MB^2$
$(4x - 6)^2 = 6^2 + (2x)^2$
$16x^2 - 48x + 36 = 36 + 4x^2$
$12x^2 = 48x \implies x = 4$ (since $x \neq 0$).

4. Final Area:
$AB = 12$, $DC = 4$, $h = 6$.
Area $= \frac{1}{2} \times (12 + 4) \times 6 = \mathbf{48}$ sq. cm.