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

Since $AD \perp AB$, place the trapezium on a coordinate plane:

$A = (0, 0)$, $B = (3a, 0)$, $D = (0, h)$, $C = (a, h)$

where $DC = a$, $AB = 3a$ (since $AB = 3 \cdot DC$), and $h = AD$.

---

For any quadrilateral with an inscribed circle, the sum of opposite sides are equal:

$AB + CD = BC + DA$

First, find $BC$ using the distance formula:

$BC = \sqrt{(3a - a)^2 + h^2} = \sqrt{4a^2 + h^2}$

Applying the property:

$3a + a = \sqrt{4a^2 + h^2} + h$

$4a - h = \sqrt{4a^2 + h^2}$

Squaring both sides:

$16a^2 - 8ah + h^2 = 4a^2 + h^2$

$12a^2 = 8ah$

$h = \dfrac{3a}{2}$

---

With $h = \dfrac{3a}{2}$:

$BC = \sqrt{4a^2 + \dfrac{9a^2}{4}} = \dfrac{5a}{2}$

So the sides are: $AB = 3a$, $BC = \dfrac{5a}{2}$, $CD = a$, $DA = \dfrac{3a}{2}$

Perimeter $= 3a + \dfrac{5a}{2} + a + \dfrac{3a}{2} = 8a$

Semi-perimeter $= 4a$

---

For a tangential polygon, $\text{Area} = \text{semi-perimeter} \times r$.

Using the trapezium area formula:

$\text{Area} = \dfrac{1}{2}(AB + CD) \times h = \dfrac{1}{2}(3a + a) \times \dfrac{3a}{2} = 3a^2$

Using the incircle formula with $r = 3$:

$\text{Area} = 4a \times 3 = 12a$

Setting them equal:

$3a^2 = 12a$

$a = 4$ cm

---

$\text{Area} = 3a^2 = 3 \times 16 = 48$ sq. cm

---

The area of the trapezium is $48$ sq. cm.