Question

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

• A. \(54\)
• B. \(30\sqrt{3}\)
• C. \(48\)
• D. \(36\sqrt{2}\)

Hint:

For figures with an incircle:
The distance between two parallel tangents equals twice the radius.
In tangential quadrilaterals (those with an incircle), using coordinates with the incenter at convenient positions (like \((r, r)\)) can simplify distance calculations.

Answer 1

The Correct Option is C

Approach: Place the figure on axes, drop the centre at $(r,r) = (3,3)$, and force its distance to the slant side $BC$ to equal $3$.

Step 1: Coordinates. Let $DC = c$, so $AB = 3c$. Put $A(0,0)$, $B(3c,0)$, $D(0,6)$, $C(c,6)$ (height $= 2r = 6$). The circle touches $AB$, $DC$, $AD$, so its centre is $O(3,3)$.

Step 2: Equation of BC. Through $B(3c,0)$ and $C(c,6)$. Direction gives line $6x + 2c\,y - 18c = 0$ (check $B$: $18c - 18c = 0$ ✓; $C$: $6c + 12c - 18c = 0$ ✓).

Step 3: Distance $= r$. \[ \frac{|6(3) + 2c(3) - 18c|}{\sqrt{6^2 + (2c)^2}} = 3 \;\Rightarrow\; \frac{|18 - 12c|}{\sqrt{36 + 4c^2}} = 3. \]

Step 4: Solve. Square: $(18 - 12c)^2 = 9(36 + 4c^2)$, i.e. $324 - 432c + 144c^2 = 324 + 36c^2$, giving $108c^2 = 432c \Rightarrow c = 4$.

Step 5: Area. $AB = 12$, $DC = 4$, height $= 6$: \[ \text{Area} = \tfrac12(12 + 4)(6) = 48\ \text{cm}^2. \] Answer: option (c).