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Question

In a circle of radius 10 cm given below, chord AB and CD are equal. If OE bisects AB and OF bisects CD and OF = 6 cm, then length EB is .

A
8 cm
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B
6 cm
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C
4 cm
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Solution

The correct option is A 8 cm

It is given AB = CD.

So, OF = OE = 6 cm [Equal chords are equidistant from centre] ...... (1)

In ΔOEB
OB = 10 cm [Radius]
OE = OF = 6 cm [from (1)]

Since a line through the center that bisects the chord is perpendicular to the chord, we must have OBE=90
OB2=OE2+EB2
[OEB is 90]

EB2=OB2OE2

EB2=(10)2(6)2=64

EB = 8 cm


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