In the figure, a circle is inscribed in a quadrilateral $ A B C D $ in which $ \angle B=90^{\circ} $. If $ A D=23 \mathrm{~cm}, A B=29 \mathrm{~cm} $ and $ D S=5 \mathrm{~cm} $, find the radius $ r $ of the circle."

Given:

In the figure, a circle is inscribed in a quadrilateral \( A B C D \) in which \( \angle B=90^{\circ} \).

\( A D=23 \mathrm{~cm}, A B=29 \mathrm{~cm} \) and \( D S=5 \mathrm{~cm} \).

To do:

We have to find the radius \( r \) of the circle.

Solution:

From the figure,

$OQ=OP=r$

$AB$ and $BC$ are tangents to the circle and $OP$ and $OQ$ are radii of the circle.

$OP\ perp\ BC$ and $OQ\ perp\ AB$

$\angle OPB = \angle OQB = 90^o$

$PBQO$ is a square.

$DS$ and $DR$ are tangents to the circle.

This implies,

$DR = DS = 5\ cm$

$AR = AD - DR$

$= 23 - 5$

$= 18\ cm$

$AR$ and $AQ$ are the tangents to the circle.

$AQ = AR = 18\ cm$

$AB = 29\ cm$

$BQ = AB - AQ$

$= 29 - 18$

$= 11\ cm$

Therefore,

Side of the square $PBQO$ is 11 cm.

This implies,

$OP = OQ = 11\ cm$

The radius of the circle is 11 cm.

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Updated on: 10-Oct-2022

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