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

$(a)\;3\qquad(b)\;2\qquad(c)\;\large\frac{3}{2}$$\qquad(d)\;1 1 Answer Comment A) Step 1: Let CD=x then AB=2CD=2x.Let r be the radius of the circle inscribed in the quadrilateral ABCD. Given: Area of quadrilateral ABCD=18 and AB\:is\:|| to CD. \Rightarrow \large\frac{1}{2}$$(x+2x).2r=18$
$\Rightarrow\:3xr=18$
$\Rightarrow\:xr=6$-----(1)
$OP=OM=PD=OQ=AM=r$
$\Rightarrow\:PC=x-r\:\: and \:\: MB=2x-r$
Let $\angle PCO=angle\:OCQ=\theta$ then from right-angled $\Delta OPC$
$\tan\theta=\large\frac{OP}{CP}=\frac{r}{x-r}$-----(2)
$CD \parallel AB$
$\therefore \angle PCB=\angle QOM=2\theta$
Step 2:
$\angle CBA=180^{\large\circ}-2\theta$
$\angle OBM=90^{\large\circ}-\theta$
$\Rightarrow\:$ From $\Delta \:OMB,\tan(90^{\large\circ}-\theta)=\large\frac{OM}{MB}=\frac{r}{2x-r}$
From right angled $\Delta OBM$
$\tan\theta=\large\frac{2x-r}{r}$-----(3)
From (2) & (3)
$\large\frac{r}{x-r}=\frac{2x-r}{r}$
$\Rightarrow 2x^2-3xr=0$
$x(2x-3r)=0$
$x=\large\frac{3r}{2}$-------(4)
From (1) & (4) we get,
$xr=6$
$\large\frac{3rr}{2}=$$6$
$r^2 = 4$
$r=2$
Hence (b) is the correct answer.