Let AB be a chord of a circle with centre O and radius 13 cm such that AB = 10 cm.
From O, draw OL ⊥ AB. Join OA.
Since, the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = LB = ½ AB = 5 cm
Now, in right triangle OLA, we have
OA^{2} = OL^{2} + AL^{2}
⇒ 13^{2} = OL^{2} = 5^{2}
⇒ 13^{2} – 5^{2} = OL^{2}
⇒ OL^{2} = 144
⇒ OL = 12 cm.
Hence, the distance of the chord from the centre is 12 cm.