Distance between equal chords, AB and CD of a circle with centre, O and radius, 5 cm is 8 cm. Find the length of the the chords.
6 cm
Consider the figure:
We know that equal chords are equidistant from the center.
Since AB=CD, we must have OE=OF.
Now, EF=8 cm [Given]
OE+OF=EF=8 cm
⇒2OE=8
⇒OE=4 cm=OF
In △OCF, applying Pythagoras theorem,
OC2=OF2+CF2
⇒52=42+CF2
⇒CF2=52−42=25−16=9
⇒CF=3 cm
We know that perpendicular from the centre to the chord bisects the chord.
So, CF=FD=3 cm
⇒CD=2×CF=2×3=6 cm
So, length of the chords is 6 cm.