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=29cm, AD=23cm, ∠B=90° and DS=5cm, then the radius of the circle (in cm) is:

The correct option is A
Given : A quadrilateral ABCD in which ∠B=90° and AD=23 cm, AB=29 cm and DS=5 cm.
Then, DR=DS=5 cm
AD=23 cm
or AR+RD=23 cm
or AR+5=23 cm
or AR=23-5=18 cm
AQ=AR [Tangents from an external point]
AR=18 cm
Therefore, AQ=18 cm
or AQ+QB=29 cm
or 18+QB=29 cm
or QB=11 cm
∴ OP and OQ are radii of the circle.
From tangents P and Q, ∠OPB=∠OQB=90°
Now OPBQ is a square.
∴ OP=BQ
Radius(r)=11 cm.