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Question
In quadrilateral ABCD; angles D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm, Find the radius of the circle.
Solution
BQ and BR are the tangents from B to the circle.
Therefore, BR = BQ = 27 cm.
Also RC = (38 − 27) = 11 cm
Since CR and CS are the tangents from C to the circle
Therefore, CS = CR = 11 cm
So, DS = (25 − 11) = 14 cm
Now DS and DP are the tangents to the circle
Therefore, DS = DP
Now, `∠`PDS = 90° ...(Given)
And OP ⊥ AD, OS ⊥ DC
Therefore, radius = DS = 14 cm
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