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Question

In the given figure, ABCD is a quadrilateral circumscribing a circle with centre 'O'. If B=90,AD=23 cm,AB=29 cm and DS=5 cm,then find the radius of the circle.


A
11 cm
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B
12 cm
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C
14 cm
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D
9 cm
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Solution

The correct option is A 11 cm

We know that tangent at any point of a circle is perpendicular to the radius through the point of contact.

OPBC and OQABP=90 and Q=90

Also, OP=OQ=r Now, BP=BQ
[Tangents drawn from an external point to a circle are equal in lengths]

Given: B=90.
By angle sum property of a quadrilateral,OQB+QBP+BPO+POQ=360

90+90+90+POQ=360POQ=360270=90

All the angles measure 90.OPBQ is a square

Hence, BP=BQ=r
(Since, all sides of a square are equal in length)

DR=DS=5 cm
[Tangents drawn from an external point to a circle are equal in lengths]

Now, AR=ADDR=235AR=18 cmAQ=AR=18 cm
[Tangents drawn from an external point to a circle are equal in lengths]

BQ=ABAQ=2918=11 cmBQ=r=11 cm

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