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.

Answer 1

 
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