Question

In the given figure, a circle with center O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, ∠B = 90° and DS = 5 cm then find the radius of the circle.

Answer 1

We know that tangent segments to a circle from the same external point are congruent

Now, we have

DS = DR, AR = AQ

 Now  AD = 23 cm 

⇒ AR + RD = 23

⇒ AR = 23 – RD

⇒ AR = 23 – 5   ...[∴ DS = DR = 5]

⇒ AR = 18 CM

Again, AB = 29 cm

⇒ AQ + QB = 29 

⇒ QB = 29 – AQ

⇒ QB = 29 – 18   ...[∵ AR = AQ = 18]

⇒ QB = 11 cm

Since all the angles are in a quadrilateral BQOP are right angles and OP = BQ

Hence, BQOP is a square. We know that all the sides of square are equal.

Therefore, BQ = PO = 11 cm