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Question

A chord AB of a circle whose centre is O, is bisected at E by a diameter CD. OC= OD = 15 cm and OE = 9 cm. Find the length of AD.


A

60°

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B

70°

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C

80°

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D

90°

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Solution

The correct option is B

70°


Given OE = 9 cm, OA = 15 cm

Consider OAE, OEA=90o
[The segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord]

Applying Pythagoras' theorem,

AE2 = OA2 OE2

AE2 = 152 92

AE2 = 225 81 =144

AE=12 cm

OD = 15 cm and OE = 9 cm (radius)

DE = 15 - 9 = 6 cm

Consider AED, AED=90o

Applying Pythagoras theorem

AD2 = AE2 + ED2

AD2 = 122 + 62

AD2 = 144 + 36 =180

AD=65 cm


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