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 $AB = 29$ cm, $AD = 23$ cm, $\angle B = 90^o$ and $DS = 5$ cm, then the radius of the circle (in cm) is:

Because the circle touches the sides of the quadrilateral, the tangent lengths from the vertices are equal, so DR = DS = 5 cm and AR = AQ. Since AD = 23 cm, AR = 23 - 5 = 18 cm, making AQ = 18 cm. Given AB = 29 cm, BQ = 29 - 18 = 11 cm, and because BP = BQ, BP = 11 cm. At the right angle B, the radius drawn to the tangents equals BQ, so the radius of the circle is 11 cm.

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