Question

Let ABCD be a quadrilateral with area 18sq.cm, with side $AB$ parallel to the side $CD$ and $AB=2CD$. 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

The correct option is B $2$

Let $CD=x$ then $AB=2CD=2x.$

Let $r$ be the radius of the circle inscribed in the quadrilateral $ABCD.$

$ar\left(\square ABCD\right)=18$ and $AB\parallel CD.$

$\Rightarrow$ $\frac{1}{2}(x+2x).2r=18$

$\Rightarrow$ $3xr=18$

$\Rightarrow$ $xr=6$ ----- ( 1 )

$OP=OM=PD=OQ=AM=r$

$\Rightarrow$ $PC=x-r$ and $MB=2x-r$

Let $\angle PCO=\angle OCQ=\theta$

In right-angled $△OPC,$

$\tan \theta =\frac{OP}{CP}=\frac{r}{x-r}$ ----- ( 2 )

$CD\parallel AB$

$\therefore$ $\angle PCB=\angle QOM=2\theta$

$\Rightarrow$ $\angle CBA={180}^o-2\theta$

$\Rightarrow$ $\angle OBM={90}^o-\theta$

From $△OMB,$

$\tan ({90}^o-\theta )=\frac{OM}{MB}=\frac{r}{2x-r}$

Right-angled $△OBM,$

$\Rightarrow$ $\tan \theta =\frac{2x-r}{r}$ ------ ( 3 )

From ( 2 ) and ( 3 ),

$\Rightarrow$ $\frac{r}{x-r}=\frac{2x-r}{r}$

$\Rightarrow$ $2x^2-3xr=0$

$\Rightarrow$ $x(2x-3r)=0$

$\Rightarrow$ $x=\frac{3r}{2}$ ---- ( 4 )

From ( 1 ) and ( 4 ), we get

$\Rightarrow$ $xr=6$

$\Rightarrow$ $\frac{3rr}{2}=6$

$\Rightarrow$ $r^2=4$

$\Rightarrow$ $r=2$