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

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 = 11 cm   ...(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 = 14 cm

∴ Radius of the circle = 14 cm