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.

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