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's analyze the properties of the trapezium $ABCD$ and the inscribed circle step-by-step:

1. Identifying the Trapezium's Dimensions:
We are given that $AB$ is parallel to $DC$, and $AD$ is perpendicular to $AB$. Since $AB$ and $DC$ are parallel, $AD$ is also perpendicular to $DC$. Thus, $\angle DAB=\angle ADC=90°$.
A circle of radius $r=3\text{cm}$ is inscribed inside the trapezium, meaning it touches all four sides: $AB$, $BC$, $CD$, and $DA$.
Since $AD$ is perpendicular to both parallel sides, the distance between $AB$ and $DC$ must be equal to the diameter of the inscribed circle. Therefore, the height of the trapezium is:
$$AD=2r=2\times 3=6\text{cm}$$

2. Using the Tangent Segments Theorem:
Let the points of contact of the circle with the sides $AD$, $AB$, $BC$, and $CD$ be $P$, $Q$, $R$, and $S$ respectively.
Since the angles at vertices $A$ and $D$ are $90°$, the quadrilaterals $AQOP$ and $DSOP$ (where $O$ is the center of the circle) are squares with side lengths equal to the radius $r=3$.
Consequently, we have:
$$AQ=AP=3\text{cm}$$
$$DS=DP=3\text{cm}$$

3. Relating the Side Lengths:
Let $DC=x$. We are given that:
$$AB=3DC=3x$$
Since $DC=DS+SC$, we can write:
$$SC=x-3$$
Similarly, since $AB=AQ+QB$, we have:
$$QB=3x-3$$
By the theorem of tangents from an external point to a circle, the tangents from $C$ and $B$ are equal:
$$CR=SC=x-3$$
$$BR=QB=3x-3$$
Therefore, the length of side $BC$ is:
$$BC=BR+CR=(3x-3)+(x-3)=4x-6$$

4. Setting up the Equation using Pythagoras' Theorem:
Let us draw a perpendicular line segment $CE$ from vertex $C$ to the side $AB$.
Since $ADCE$ forms a rectangle:
$$CE=AD=6\text{cm}$$
$$AE=DC=x$$
Now we calculate $EB$:
$$EB=AB-AE=3x-x=2x$$
Applying Pythagoras' theorem in the right-angled triangle $CEB$:
$$B{C}^2=C{E}^2+E{B}^2$$
Substitute the values we found:
$${(4x-6)}^2=6^2+{\left(2x\right)}^2$$
Expanding both sides:
$$16x^2-48x+36=36+4x^2$$
Subtracting $4x^2+36$ from both sides gives:
$$12x^2-48x=0$$
Factoring out $12x$:
$$12x(x-4)=0$$
Since $x$ represents a length, $x>0$. Therefore, we have:
$$x=4\text{cm}$$

5. Calculating the Area of the Trapezium:
Now we calculate the lengths of the parallel sides:
$$DC=x=4\text{cm}$$
$$AB=3x=12\text{cm}$$
The formula for the area of a trapezium is:
$$\text{Area}=\frac{1}{2}\times \left(\text{sum of parallel sides}\right)\times \text{height}$$
Substituting the known values:
$$\text{Area}=\frac{1}{2}\times (12+4)\times 6$$
$$\text{Area}=\frac{1}{2}$$
$$\text{Area}=\frac{1}{2}\times 16\times 6=8\times 6=48{\text{cm}}^2$$