Question

In the following figure, if the chords AB and CD are equal to 6 cm then find the distance of the chords from the centre of the circle of radius 5 cm.

Answer 1

Given two chords AB and CD.
Draw perpendicular from the centre to the chords.

We know the property that perpendicular drawn from the centre to the chord bisects the chord.

Therefore, AE = BE and DF = CF.

Apply Pythagoras' theorem in $\Delta$OEB and $\Delta$OFD

In $\Delta$OEB,
${OE}^2$ + ${BE}^2$ = ${OB}^2$
${OE}^2$ = 16
$\Rightarrow$OE = 4 cm

Similarly, in $\Delta$OEB
${OF}^2$ + ${FD}^2$ = ${OD}^2$
${OF}^2$ = 16
$\Rightarrow$OF = 4 cm

Thus, we see that the distance of equal chords from centre is also equal.