Question

15. In the given figure, a circle with centre 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, ZB = 90° and
DS = 5 cm then find the radius of the
circle.
[CBSE 2008,'13]


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Answer 1

Answer:

AB=29cm,AD=23cm,∠B=90 and DS=5cm

Tangents to a circle from an external point are equal in length.

AQ=AR

DS=DR

CP=CS

PB=BQ

Also, AB=AQ+BQ

Since DS=5cm

DR=5cm

So, AR=23−5=18cm

AQ=18cm

and BQ=29−18=11cm

Now, In quadrilateral OPBQ, ∠B=90 ∘ .

Also, OPB=OQB=90 ∘

(tangent is perpendicular to radius at point of contact)

So, ∠POQ=90 ∘

; that is OPBQ is a rectangle.

Further since PA=PB; OPBQ is a square.

Hence, Radius=OP=BQ=11cm. (Sides of a square)