In the given figure, the circle touches the sides AB, BC, CD and DA of a quadrilateral ABCD at the points P, Q, R and S respectively. If $A B = 11 c m, B C = x c m, C R = 4 c m$ and $A S = 6 c m,$ the value of $x$ is ___ $c m .$

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Solution

Given, $A P = A S = 6 c m$

$\Rightarrow B P = B A - A P = 11 - 6 = 5 c m$

$\Rightarrow B Q = B P = 5 c m$ [Tangents from an external point to the circle are equal in length]

$\Rightarrow C Q = C R = 4 c m$ [Tangents from an external point to the circle are equal in length]

Since, $x = C Q + B Q$

Therefore, $x = 4 + 5 = 9 c m$

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