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 = 29cm, AD = 23cm, ∠B = 90° and DS=5cm 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 = 11CM

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