Q. 154.5( 15 Votes )

In the given figure, a circle with center O, is inscribed in a quadrilateral ABCD such that it touches the side BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, LB = 90° and DS = 5 cm then find the radius of the circle.


Answer :

In quadrilateral POQB


OPB = 90°


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


OQB = 90°


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


PQB = 90° [Given]


By angle sum property of quadrilateral PQOB


OPB + OQB + PBQ + POQ = 360°


90° + 90° + 90° + POQ = 360°


POQ = 90°


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


Also, OP = OQ = 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 OP = PB = BQ = OQ = r [1]


Now,


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


In given figure, We have


DS = DR = 5 cm


[Tangents from point D and DS = 5 cm is given]


AD = 23 cm [Given]


AR + DR = 23


AR + 5 = 23


AR = 18 cm


Now,


AR = AQ = 18 cm


[Tangents from point A]


AB = 29 cm [Given]


AQ + QB = 29


18 + QB = 29


QB = 11 cm


From [1]


QB = r = 11 cm


Hence Radius of circle is 11 cm.


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