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 ________ .
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=OB2−OE2
EB2=(10)2−(6)2=64
EB = 8 cm