Question

In the following figure, a circle is inscribed in the quadrilateral ABCD.

If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.

Answer 1

ABCD quadrilateral. Given $\angle D={90}^{0}CD=25cmBC=38cmandBQis27cm.$
Let the circle PQRS be the inscribed circle touching the sides and with centre O. So we have the radius of the 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, CS = 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 $\angle DPS=\angle DSP={45}^0$

So $SP=14\sqrt{2}cm.=diagonal.$

Sow the triangle OPS, is an isosceles triangle. OT is the altitude of this triangle on to SP.
$\angle OPS=\angle OSP={45}^0,because\angle OPD=\angle OSD={90}^0.$

The triangle OTP is also an isoscles triangle. OT = TP as $\angle OTP={90}^0,\angle OPT=={45}^0.,because\angle POT=90-45={45}^0$

Thus side $PT=\frac{1}{2}PS=7\sqrt{2}cm=OT$

OPT is a right angle triangle with OP = R = radius.

$R=\sqrt{2}\times PT=7\times \sqrt{2}\times \sqrt{2}=14cm$

method 2:

Here PD= DS = 14 cm. $\angle OPD=\angle PDS=\angle DSO={90}^0$

Hence $\angle POS={90}^0$

Thus the quadrilateral OPDS is a rectangle. But PD = DS , hence it is a square.

Hence, OP = radius = OS = 14 cm.