Given: ABCD is a quadrilateral in which
\(\angle B=90^\circ\)
\(AD=23cm, DS=5cm\,and\,AB=29cm\)
Let the radius of the incircle be r cm
AD = DS = 5cm (tangent from an external point)
Since AD=23 cm
So,
AR + RD + AD
AR + 5 = 23 cm
AR = 18 cm ------(i)
and AQ = AR
since AR = 18 cm
So, AQ + QB = AB
Now OP and OQ are radius of the circle. So from tangent P and Q
\(\angle OPB=\angle OQB=90^\circ\)
\(OPBQ\) is a square
\(OP=QB\)
Radius of the circle = 11cm