##### In the given figure, a circle is inscribed in a quadrilateral ABCD touching its sides AB, BC, CD and AD at P, Q, R and S respectively. lithe radius of the circle is 10 cm, BC = 38 cm, PB = 27 cm and AD⏊ CD then the length of CD is ORD = 90°

[ Tangent at any point on the circle is perpendicular to the radius through point of contact]

OSD = 90°

[ Tangent at any point on the circle is perpendicular to the radius through point of contact]

By angle sum property of quadrilateral PQOB

ORD + OSD + SDR + SOR = 360°

90° + 90° + 90° + SOR = 360°

SOR = 90°

As all angles of this quadrilaterals are 90° The quadrilateral is a rectangle

Also, OS = OR = r

i.e. adjacent sides are equal, and we know that a rectangle with adjacent sides equal is a square

POQB is a square

And OS = OR = DR = DS = r = 10 cm 

Now,

As we know that tangents drawn from an external point to a circle are equal

In given figure, We have

CQ = CR …

[ tangents from point C]

PB = BQ = 27 cm

[Tangents from point B and PB = 27 cm is given]

BC = 38 cm [Given]

BQ + CQ = 38

27 + CQ = 38

CQ = 11 cm

From 

CQ = CR = 11 cm

Now,

CD = CR + DR

CD = 11 + 10 = 21 cm [from 1, DR = 10 cm]

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