Question

In fig., a circle with centre $O$ is inscribed 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, $\angle B={90}^o$ and $DS=5$ cm, then the radius of the circle (in cm) is:

• A. $11$
• B. $18$
• C. $6$
• D. $15$

Answer 1

The correct option is A $11$

Given : A quadrilateral $ABCD$ in which $\angle B={90}^0$ and $AD=23cm,AB=29cm$ and $DS=5cm$.

Then,

$DR=DS=5cm$

$AD=23cm$ or $AR+RD=23cm$

or $AR+5=23cm$ or $AR=23–5=18cm$

$AQ=AR$ [Tangents from an external point]

$AR=18cm$

Therefore, $AQ=18cm$

or $AQ+QB=29cm$ or $18+QB=29cm$

or $QB=11cm$

$\therefore OP$ and $OQ$ are radii of the circle.

From tangents $P$ and $Q$, $\angle OPB=\angle OQB={90}^0$

Now $OPBQ$ is a square.

$\therefore OP=BQ$

Radius $\left(r\right)=11cm$.