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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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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


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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$
$\Rightarrow\:PC=x-r\:\: and \:\: MB=2x-r$
Let $\angle PCO=angle\:OCQ=\theta$ then from right-angled $\Delta OPC$
$CD \parallel AB$
$\therefore \angle PCB=\angle QOM=2\theta$
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 $\Delta OBM$
From (2) & (3)
$\Rightarrow 2x^2-3xr=0$
From (1) & (4) we get,
$r^2 = 4$
Hence (b) is the correct answer.
answered Nov 19, 2013 by sreemathi.v
edited Mar 25, 2014 by rvidyagovindarajan_1

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