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Let ABCD be a quadrilateral in which AB is parallel to CD and perpendicular to AD, AB = 3CD, the area of quadrilateral is 4 sq. unit. If a circle can be drawn touching all the sides of the quadrilateral, then the radius of the circle is 

(a) 2 units 

(b) \(\sqrt{3}\) units 

(c) \(\frac{\sqrt{3}}{2}\) units  

(d) 2\(\sqrt{3}\) units

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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.

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