In Fig. 1, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. If the radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD ⊥ CD, then calculate the length of CD.

Given:

Radius = OS = OR = 10 cm,

BC = 38 cm,

PB = 27cm, ∠ADC = 90°

∠ORD = ∠OSD = 90° [Tangent to a circle is perpendicular to the radius at the point of contact]

∠ADC = 90° [given]

OS = OR [Radius of the same circle]

DS = DR [Tangents to the same circle from an external point are equal]

⸫ OSDR is a square.

BP = BQ = 27cm [Tangents to the same circle from an external point are equal]

CQ = CR [Tangents to the same circle from an external point are equal]

Now,

BQ = 27

⇒ BC – CQ = 27

⇒ 38 – CQ = 27

⸫ CQ = 11 cm

⸫ CR = 11 cm [⸪ CQ = CR]

⇒ CD – DR = 11

⇒ CD = 11 + DR

[⸪ OSDR is a square, DR = OS = OR]

⸫ CD = 11 + 10 = 21 cm