Question

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q.11

Answer 1

Answer:r=14 cm

Step-by-step explanation:

AB, BC, CD and AD are tangents to the circle with centre O at Q, P, S and R respectively.

Given that, AB = 30 cm, AD = 24, DS = 8 cm and ∠B = 90°.

The lengths of the tangents drawn from an external point to a circle are equal. DS = DR = 8 cm

∴ AR = AD – DR = 24−8 = 16 cm AQ = AR = 16 cm

∴ QB = AB – AQ = 30 − 16 = 14 cm

QB = BP = 14 cm. ∠PBQ = 90° [Given]

∠OPB = 90° [Angle between the tangent and the radius at the point of contact is a right angle.]

∠OQB = 90° [Angle between the tangent and the radius at the point of contact is a right angle.]

∠POQ = 90° [Angle sum property of a quadrilateral.] So, OQBP is a square.

∴QB = BP = r = 14 cm ∴ the radius of the circle is 14 cm.

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Answer 2

Answer:

Step-by-step explanation:

here we can see that OQBP is a quadrilateral.

∠QBP = 90°                             (given)

∠OPB = 90°         (angle b/w tangent and radius)

∠BQO = 90°         (angle b/w tangent and radius)

∠POQ = 90°         (3 angles are 90° so rest of one angle will be 90°)

So,

by all these lines BQOP is a square.

QB=BP=29/2      (tangent from ext. point B)

BQ=QO=OP=BP=29/2

Radius of circle=14.5cm

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