Question

$\text{In the given figure, ABCD is a}\text{quadrilateral circumscribing a circle with}\text{centre 'O'. If}\angle B={90}^{\circ },AD=23cm,AB=29cm\text{and}DS=5cm,\text{then find}\text{the radius of the circle.}$

Answer 1

We know that tangent at any point of a circle is perpendicular to the radius through the point of contact.

$\therefore OP\perp BC\text{and}OQ\perp AB\Rightarrow \angle P={90}^{\circ }\text{and}\angle Q={90}^{\circ }$

$\text{Also,}OP=OQ=r\text{Now,}BP=BQ$
[Tangents drawn from an external point to a circle are equal in lengths]

$\text{Given:}\angle B={90}^{\circ }.$
$\text{By angle sum property of a}\text{quadrilateral,}\angle OQB+\angle QBP+\angle BPO+\angle POQ={360}^{\circ }$

${90}^{\circ }+{90}^{\circ }+{90}^{\circ }+\angle POQ={360}^{\circ }\angle POQ={360}^{\circ }-{270}^{\circ }={90}^{\circ }$

$\text{All the angles measure}{90}^{\circ }.\therefore \text{OPBQ is a square}$

$\text{Hence,}BP=BQ=r$
$\text{(Since, all sides of a square are equal}\text{in length)}$

$DR=DS=5cm$
[Tangents drawn from an external point to a circle are equal in lengths]

$\text{Now,}AR=AD-DR=23-5AR=18cmAQ=AR=18cm$
[Tangents drawn from an external point to a circle are equal in lengths]

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