Question

Let ABCD be a quadrilateral with area 18, with side AB parallel to the side CD and AB = 2 CD. Let AD be perpendicular to AB and CD. If a circle is drawn inside the quadrilateral ABCD touching all the side, then its radius is

• A. 3
• B. 2
• C. $\frac{3}{2}$
• D. 1

Answer 1

The correct option is B 2

Here $\frac{1}{2}\times 3x_1\times 2r=18$
$\Rightarrow x_1\times r=6$ ........ (i)
Equation of BC is: $y=-\frac{2r}{x_1}(x-2x_1)$
BC is tangent to the circle $(x-r{)}^2+(y-r{)}^2=r^2$
$\therefore$ Perpendicular distance of BC from centre $=$ radius
$\frac{|r+\frac{2r}{x_1}(r-2x_1\left)\right||}{\sqrt{1+\frac{4r^2}{x_1^2}}}=r$
$\Rightarrow \frac{2r^2}{x_1}-3r=r\sqrt{1+4\frac{r^2}{x_1^2}}$
$\Rightarrow (2r-3x_1{)}^2=x_1^2+4r^2$
$\Rightarrow r.x_1=\frac{2}{3}{x}_1^2\Rightarrow 3r=2x_1$ (ii)
From Equations (i) and (ii), $r=2units$