Question

In Fig.2, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches the sides 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 the radius of the circle (in cm.) is:

Options

• (A) 11

• (B) 18

• (C) 6

• (D) 15

Answer 1

We know that, the lengths of the tangents drawn from an external point to a circle are equal.

DS = DR = 5 cm

∴ AR = AD − DR = 23 cm − 5 cm = 18 cm

AQ = AR = 18 cm

∴ QB = AB − AQ = 29 cm − 18 cm = 11 cm

QB = BP = 11 cm

In right Δ PQB, PQ² = QB² + BP² = (11 cm)² + (11 cm)² = 2 × (11 cm)²

PQ = `11sqrt2r` cm … (1)

In right ΔOPQ,

PQ² = OQ² + OP² = r² + r² = 2 r²

PQ = `11sqrt2r`… (2)

From (1) and (2), we get

r = 11 cm

Thus, the radius of the circle is 11 cm.