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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.
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.