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 find the length of EB.

Answer 1

It is given AB = CD.

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

In $\Delta OEB$
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 $\angle OBE={90}^{\circ }$
$\therefore O{B}^2=O{E}^2+E{B}^2$
[$\because \angle$OEB is ${90}^{\circ }$]

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

$E{B}^2=(10{)}^2-(6{)}^2=64$

EB = 8 cm