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 sides, then its radius is

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

Answer 1

Answer: (B)

2

Answer 2

$18=\frac{1}{2}(3\alpha)(2r)\Rightarrow\alpha r=6$
Line, $y=-\frac{2r}{\alpha}(x-2\alpha)$ is tangent to circle $(x-r)^2+(y-r)^2=r^2$ $2\alpha =3r , \alpha r=6$ and $r=2$
Alternate Solution ${\frac{1}{2}}(x+2x)\times2r=18\: x r=6$....(i)
In $\triangle$ AOB, $\tan\theta=\frac {x-r}{r}$ and in $\triangle$ DOC,
$\tan(90^\circ-\theta)=\frac {2x-r}{r}$
$\therefore$ $\frac {x-r}{r}=\frac {r}{2x-r}$
$\Rightarrow$ x(2x-3r)=0
$\Rightarrow$ $x=\frac{3r}{2}$
From Eqs. (i) and (ii), we get $r = 2$