Q. 424.7( 3 Votes )

In the given figu A. 11 cm

B. 15 cm

C. 20 cm

D. 21 cm

Answer :

In quadrilateral ORDS


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]


SDR = 90° [AD CD]


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 [1]


Now,


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


In given figure, We have


CQ = CR …[2]


[ 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 [2]


CQ = CR = 11 cm


Now,


CD = CR + DR


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

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