A circle is inscribed in a quadrilateral ABCD in which B = 90o. If AD = 23 cm, AB = 29 cm and DS = 5 cm. Find the radius of the circle.

Asked by Topperlearning User | 27th Jul, 2017, 01:09: PM

Expert Answer:

Since tangents drawn from an external point to a circle are equal.

DR = DS = 5cm

Now, AR = AD - DR = 23 - 5 = 18cm

But, AR=AQ

AQ = 18cm

Also, BQ = AB - AQ = 29- 18 = 11cm

But, BP = BQ

BP=11cm

Also, Q=P =90o.

In quadrilateral OQBP,

QOP+P+Q+B=360°

QOP = 360° - (P+Q+B)

= 360° - (90°+90°+90°) = 90°

Hence, OQBP is a square.

BQ=OQ=OP=BP=11cm

Hence, radius of the circle is 11cm.

Answered by  | 27th Jul, 2017, 03:09: PM