Radius of the given circle is 20 cm, chord AB = chord CD. If OE bisects AB and OF = 16, bisects CD. Find EB

12

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Solution

The correct option is A 12
It is given AB = CD.

So, OF = OE = 8 cm [Equal chords are equidistant from centre] ...... (1)

In $Δ O E B$
OB = 20 cm [Radius]
OE = OF = 16 cm [from (1)]

Since a line through the center that bisects the chord is perpendicular to the chord, we must have $∠ O B E = 90^{\circ}$
$∴ O B^{2} = O E^{2} + E B^{2}$
[ $∵ ∠$ OEB is $90^{\circ}$ ]

$E B^{2} = O B^{2} - O E^{2}$

$E B^{2} = (20)^{2} - (16)^{2} = 144$

EB = 12 cm

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