#### 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.

#### Solution

From the figure we see that BQ = BR = 27 cm (since length of the tangent segments from an external point are equal)

As BC = 38 cm

⇒ CR = CB − BR = 38 − 27

= 11 cm

Again,

CR = CS = 11cm (length of tangent segments from an external point are equal)

Now, as DC = 25 cm

∴ DS = DC − SC

= 25 − 11

= 14 cm

Now, in quadrilateral DSOP,

`∠`PDS = 90° (given)

`∠` OSD = 90°, `∠`OPD = 90° (since tangent is perpendicular to the

radius through the point of contact)

⇒ DSOP is a parallelogram

⇒ OP ∥ SD and ⇒ PD ∥ OS

Now, as OP = OS (radii of the same circle)

⇒ OPDS is a square. ∴ DS = OP = 14cm

∴ radius of the circle = 14 cm