In the given figure, a circle is inscribed in a quadrilateral ABC… - Vidyadip

MCQ

In the given figure, a circle is inscribed in a quadrilateral $\text{ABCD}$ touching its sides $\text{AB, BC, CD}$ and $\text{AD}$ at $\text{P, Q, R}$ and $S$ respectively. If the radius of the circle is $10\ cm, \text{BC} = 38\ cm, \text{PB} = 27\ cm$ and $\text{AD} \perp \text{CD}$ then the length of $\text{CD}$ is:

A
$11\ cm$
B
$15\ cm$
C
$20\ cm$
✓
$21\ cm$

✓

Answer

Correct option: D.

$21\ cm$

We know that tangles from an external point to the circle are equal.
$\text{BQ = PB} = 27\ cm$
So, $\text{CQ = BC - BQ} = 38 - 27 = 11\ cm$
$\Rightarrow \text{CR = CQ} = 11\ cm$
In quad. $\text{SORD,}$
$\angle\text{SDR}=90^\circ....(\therefore\text{AD}\perp\text{CD})$
$\Rightarrow\angle\text{OSD}=\angle\text{ORD}=90^\circ$
Also, $\text{OS = OR}$ and $\text{SD = SR}$
So, quad. $\text{SORD}$ is a square.
Thus, $\text{DR = SO} = 10\ cm$
Hence, $\text{CD = DR + CR} = 10 + 11 = 21cm.$

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