$(a)\;3\qquad(b)\;2\qquad(c)\;\large\frac{3}{2}$$\qquad(d)\;1$

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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.

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