$( A) \ 11$
$( B) \ 18$
$( C) \ 6$
$( D) \ 15$"
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In fig., a circle with center O is inscirbed in a quadrilateral ABCD such that, it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively, if $AB=29\ cm,AD=23\ cm\ and\ \angle B=90^{o}$ and $DS=5\ cm$, then the radius of the circle $( in\ cm)$ is:

$( A) \ 11$
$( B) \ 18$
$( C) \ 6$
$( D) \ 15$"

Given: A circle with center O inscribing a quadrilateral ABCD touching its sides AB, BC, CD and DA at P, Q, R and S respectively.

To do: To find the radius of the circle.

Solution: 
Here as shown in the figure,
Quadrilateral ABCD, touches the circle at P, Q, R and S.

thus AR and AQ are tangents to the circle at R and Q respectively.

BP and BQ are tangents to the circle at P and Q respectively.

CP and CS are tangents to the circle at P and S.

DR and DS are tangents to the circle at R and S.

$\because$ Tangents drawn to a circle from an external point have equal length.

$\therefore AR=AQ  .......( 1)$

$BP=BQ .............( 2)$

$CP=CS ..............( 3)$

$DR=DS ..............( 4)$

Let us join PQ.

Here given $AB=29\ cm,\ AD=23\ cm,\ \angle B=90^{o} \ and\ DS=5\ cm$.
$\Rightarrow DR=DS=5\ cm$

$AR=AD-DR=23-5=18\ cm$

From $( 1) ,\ AR=AQ=18\ cm$

Here given $AB=29\ cm$

$\therefore BQ=AB-AQ=29-18=11\ cm$

From $( 2)$,

$BP=BQ=11\ cm$

In $\vartriangle PBQ,$

$BP=BQ=11\ cm$

$\angle B=90^{o}$

It is a right triangle,

Using pythagoras theorem,

$PQ^{2} =BP^{2} +BQ^{2}$

$\Rightarrow PQ^{2} =11^{2} +11^{2} =121+121=242=11\sqrt{2} \ cm$

In $\vartriangle OPQ$,

OP and OQ are radius of the circle.

$\therefore OP=OQ=r$

And $PQ=11\sqrt{2} \ cm$

$\therefore$ Using pythagoras theorem,

$PQ^{2} =OP^{2} +OQ^{2}$

$\Rightarrow \left( 11\sqrt{2}\right)^{2} =r^{2} +r^{2} =2r^{2}$

$\Rightarrow r^{2} =121$

$\Rightarrow r=11\ cm$

$\therefore$ The radius of the circle is $11\ cm$.

$\therefore$ Option $( A)$ is correct.
 

Updated on: 10-Oct-2022

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