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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.
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.
Abhi1803:
Your answer is incorrect
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you can use both d methods.......
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hope this helps you...!!
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