Question

In fig. 2, 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 $AB=29cm,AD=23$,
$\angle B={90}^{\circ }$ and $DS=5cm$, 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

$AB=29cm,AD=23cm,\angle B=90$ and $DS=5cm$

Tangents to a circle from an external point are equal in length.

$AQ=AR$

$DS=DR$

$CP=CS$

$PB=BQ$

Also, $AB=AQ+BQ$

Since $DS=5cm$

$DR=5cm$

So, $AR=23-5=18cm$

$AQ=18cm$

and $BQ=29-18=11cm$

Now, In quadrilateral OPBQ, $\angle B={90}^{\circ }$.

Also, $OPB=OQB={90}^{\circ }$ (tangent is perpendicular to radius at point of contact)

So, $\angle POQ={90}^{\circ }$; that is OPBQ is a rectangle.

Further since $PA=PB$; OPBQ is a square.

Hence, $Radius=OP=BQ=11cm$. (Sides of a square)

hence answer will be option A