In the following figure, a circle is inscribed in the quadrilateral AB

1.	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» 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

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»

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