In Fig. 10.100, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches 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

Question

In Fig. 10.100, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches 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

Philoid · Accepted Answer

Given:

AB = 29 cm

AD = 23 cm

_∠B = 90°

DS = 5 cm

Property 1: If two tangents are drawn to a circle from one external point, then their tangent segments (lines joining the external point and the points of tangency on circle) are equal.

Property 2: The tangent at a point on a circle is at right angles to the radius obtained by joining center and the point of tangency.

Property 3: Sum of all angles of a quadrilateral = 360°.

By property 1,

BP = BQ (tangent from B)

DS = DR = 5 cm (tangent from D)

AR = AQ (tangent from A)

Also,

OQ = OP (radius)

By property 2, ∆OQB is right-angled at ∠OQB (i.e., ∠OQB = 90°) and ∆OPB is right-angled at ∠OPB (i.e., ∠OPB = 90°).

Now by property 3,

∠PBC + ∠BQO + ∠QOP + ∠OPB = 360°

⇒ 90° + 90° + ∠QOP + 90°= 360°

⇒ 270° + ∠QOP = 360°

⇒ ∠QOP = 360° - 270°

⇒ ∠QOP = 90°

∵ adjacent sides (i.e., BP = BQ and OQ = OP) are equal and all angles are 90°

∴ quadrilateral OPBQ is a square

Now,

AD = 23 cm

⇒ AR + RD = 23 cm [∵ AD = AR + RD]

⇒ AR + 5 cm = 23 cm

⇒ AR = 23 cm – 5 cm

⇒ AR = 18 cm

⇒ AQ = AR = 18 cm

Now,

AB = 29 cm

⇒ AQ + QB = 29 cm [∵ AD = AR + RD]

⇒ 18 cm + QB = 29 cm

⇒ QB = 29 cm – 18 cm

⇒ QB = 11 cm

∵ OPBQ is a square

∴ OP = BQ = 11 cm

Hence, radius = 11 cm