ABCD quadrilateral. Given D = 90 deg. CD = 25 cm. BC = 38 cm and BQ is 27 cm.
Let the circle PQRS be the inscribed circle touching the sides and with center O. So we have radius of circle as R.
As BQ and BR are the tangents to the circle from B, they are equal. So BR = 27 cm. Hence, RC = 38 - 27 = 11 cm. Hence, CP = 11 cm. Hence PD = 25 - 11 = 14 cm.
method 1 :
Thus DS = 14 cm. Since DPS is a right angle triangle and DP = DS, the angles DPS = angle DSP = 45 deg.
So SP = 14 √2 cm.= diagonal.
Sow the triangle OPS, is an isosceles triangle. OT is the altitude of this triangle on to SP.
The angles OPS = angle OSP = 45 deg, because angle OPD = angle OSD = 90 deg.
The triangle OTP is also an isoscles triangle. OT = TP as the angle OTP = 90, angle OPT = = 45 deg., because the angle POT = 90 - 45 = 45 deg.
Thus side PT = 1/2 PS = 7 √2 cm = OT
OPT is a right angle triangle with OP = R = radius.
R = √2 * PT = 7 * √2 * √2 = 14 cm
Here PD= DS = 14 cm. angles OPD = angle PDS = angle DSO = 90 deg. Hence angle POS = 90 deg. Thus the quadrilateral OPDS is a rectangle. But PD = DS , hence it is a square.
Hence, OP = radius = OS = 14 cm.