Question

In Fig. 8.61, a circle is inscribed in a quadrilateral ABCD in which $\angle B={90}^{\circ }$. If AD =23 cm, AB =29 cm and DS =5 cm, find the radius r of the circle.

Answer 1

$\angle B={90}^{\circ }$

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

Let radius of the incircle be r cm.

RD = DS = 5cm (Tangents from an external point)
$\because$ AD = 23cm
So,
AR + RD = AD
$\Rightarrow$ AR + 5 = 23cm
$\Rightarrow$ AR = 18cm (i)

And,

AQ = AR (Tangents from an external point)
$\because$ AR = 18 cm from(i)
So,
AQ + QB = AB
$\Rightarrow$ 18 + QB = 29cm
$\Rightarrow$ QB = 11cm

Now, OP and OQ are radius of the circle. So,

From tangents P and Q,

$\angle OPB=\angle OQB={90}^{\circ }$
$\Rightarrow$ OPBQ is a square (all angles are right angle)
$\Rightarrow$ OP = QB
$\Rightarrow$ Radius of the circle (r) = QB = 11cm

Hence, radius of the circle is 11 cm.