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

The correct option is 48.

Let us solve the problem step-by-step using the properties of a tangential right-angled trapezium.

Step 1: Determine the height of the trapezium
In trapezium $ABCD$, side $AB$ is parallel to $DC$, and side $AD$ is perpendicular to $AB$ (and thus perpendicular to $DC$ as well).
Since an inscribed circle of radius $r=3\text{cm}$ touches all four sides of the trapezium, the height $h$ of the trapezium (the distance between parallel sides $AB$ and $DC$) is equal to the diameter of the circle.

$AD=h=2r=2\times 3=6\text{cm}$

Step 2: Relate the sides using the tangential quadrilateral property
Let $DC=x$. According to the given condition, $AB=3DC=3x$.
For any tangential quadrilateral (a polygon with an inscribed circle), the sum of the lengths of opposite sides is equal:

$AB+DC=AD+BC$

Substitute the known values into the equation:

$3x+x=6+BC$

$4x=6+BC\Rightarrow BC=4x-6$

Step 3: Calculate the value of $x$ using Pythagoras' theorem
Draw a perpendicular line segment $CE$ from vertex $C$ to side $AB$.
This forms a right-angled triangle $CEB$, where:

$CE=AD=6\text{cm}$

$EB=AB-AE=AB-DC=3x-x=2x$

Applying Pythagoras' theorem in right triangle $CEB$:

$B{C}^2=C{E}^2+E{B}^2$

${(4x-6)}^2=6^2+{\left(2x\right)}^2$

$16x^2-48x+36=36+4x^2$

Subtracting $4x^2+36$ from both sides:

$12x^2-48x=0$

$12x(x-4)=0$

Since side length $x>0$, we have $x=4\text{cm}$.

Therefore:

$DC=4\text{cm}$

$AB=3\times 4=12\text{cm}$

Step 4: Compute the area of the trapezium
The area of a trapezium is given by the formula:

$\text{Area}=\frac{1}{2}\times (AB+DC)\times h$

Substituting the values of $AB$, $DC$, and $h$:

$\text{Area}=\frac{1}{2}\times (12+4)\times 6$

$\text{Area}=\frac{1}{2}\times 16\times 6=48{\text{cm}}^2$

Thus, the area of the trapezium is 48 sq. cm.