ABCD is a trapezium in which AB is parallel to \( \mathrm { DC } , \mathrm { AD } \) is perpendicular to AB , and \( \mathrm { AB } = 3 \mathrm { DC } \). 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
Answer: C
Key idea: A quadrilateral with an incircle has equal sums of opposite sides, and here the perpendicular side AD equals the diameter.
1. AD is perpendicular to both parallel sides, so AD = height = \( 2r = 6 \) cm.
2. Let DC = \( x \), so AB = \( 3x \). The slant side BC spans a horizontal gap of \( 3x - x = 2x \) and a vertical gap of 6, so \( BC = \sqrt{4x^2 + 36} \).
3. Tangential condition: \( AB + DC = AD + BC \Rightarrow 4x = 6 + \sqrt{4x^2 + 36} \).
4. Squaring \( 4x - 6 = \sqrt{4x^2+36} \): \( 16x^2 - 48x + 36 = 4x^2 + 36 \Rightarrow 12x^2 = 48x \Rightarrow x = 4 \).
5. So DC = 4, AB = 12, and the area \( = \frac{1}{2}(12 + 4) \times 6 = 48 \) sq. cm.
Check: BC = \( \sqrt{64 + 36} = 10 \), and 12 + 4 = 6 + 10.
Hence, the answer is (C) 48.