Solution :-
Let AB be a chord of circle with centre O and radius 13cm. Draw OM perpendicular AB and join OA.
In the right triangle OMA, we have
OA^{2 }= OM^{2} + AM^{2}
⇒ 13^{2 }= 12^{2} + AM^{2}
⇒ AM^{2} = 169 - 144 = 25 ⇒ AM = 5cm.
As the perpendicular from the centre of a chord bisects the chord.Therefore,
AB = 2AM = 2 x 5 = 10cm.