Question

In quadrilateral ABCD angle D =90*, BC =38cm and DC = 25cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27cm. Find the radius of the circle.

Answer 1

Say the side $BC$ touches the circle at point  $E$, the side $CD$ touches the circle at point $F$ and the side $DA$ touches the circle at point $G$. Also let the center of circle be $O$ :

$27=QB=BE\implies EC=38-27=11$

$\implies CF=EC=11\\\implies FD = 25-11=14$

Observe that $OFDG$ is a square. Thus the radius of circle equals $FD$ 

Answer 2

see diagram.

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

method 2:

   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.