ABCD is a quadrilateral such that …

Since tangent to a circle is perpendicular to the radius through the point.
{tex} \therefore \quad \angle O R D = \angle O S D{/tex} = 90°
It is given that {tex} \angle D{/tex} = 90 °. Also, OR = OS. Therefore, ORDS is a square.
Since tangents from an exterior point to a circle are equal in length.
{tex} \therefore{/tex}BP = BQ
CQ = CR
and, DR = DS.
Now,
BP = BQ
{tex}\Rightarrow{/tex}BQ = 27 [{tex}\because{/tex} BP = 27cm(Given)]
{tex}\Rightarrow{/tex} BC - CQ = 27
38 -C Q = 27{tex} [ \because B C = 38 \mathrm { cm } ]{/tex}
CQ = 11cm
CR = 11cm{tex} [ \because C R = C Q ]{/tex}
CD - DR = 11
{tex}\Rightarrow{/tex} 25 -D R = 11 {tex} [ \because C D = 25 \mathrm { cm } ]{/tex}
{tex}\Rightarrow{/tex} DR = 14 cm
But, ORDS is a square. Therefore, OR = DR = 14 cm.
Hence, r = 14 cm.