Answer : (c) \(\frac{\sqrt{3}}{2}\) units
Let the radius of the circle drawn inside the quadrilateral ABCD be r.
∵ AB || CD, ∴ ABCD is a trapezium.
Let CD = x, then AB = 3 CD = 3x
Draw a perpendicular CM from C on AB.
Area of a trapezium = \(\frac{1}{2}\) (sum of parallel sides) × height
⇒ \(\frac{1}{2}\) × (AB + CD) × AD = 4 (Area = 4 sq. units, given)
⇒ \(\frac{1}{2}\) × (3x + x) × 2r = 4
⇒ 4xr = 4 ⇒ x = \(\frac{1}{r}\)
As all the sides i.e., AB, BC, CD and DA touch the incircle,
DC + AB = DP + PC + AQ + QB
= DR + CS + AR + SB . [Using, tangents from the same external point are equal]
= DR + AR + CS + SB = AD + BC
⇒ x + 3x = 2r + \(\sqrt{(2r)^2 +(2x)^2}\)
(∵ ∠M = 90°, CM = PQ = 2r, MB = AB – CD = 2x)
⇒ 4x = 2r + \(2\sqrt{r^2+x^2}\)
⇒ 2x – r = \(\sqrt{r^2+x^2}\)
On squaring both the sides, we have
4x2 – 4rx + r2 = r2 + x2
⇒ 3x2 = 4xr
⇒ \(\frac{3}{r^2} =\frac{4}{r}\times r\)
⇒ \(r^2 = \frac{3}{4}\)
⇒ r = \(\frac{\sqrt{3}}{2}\) units.