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
OA2 = OM2 + AM2
⇒ 132 = 122 + AM2
⇒ AM2 = 169 - 144 = 25 ⇒ AM = 5cm.
As the perpendicular from the centre of a chord bisects the chord.Therefore,
AB = 2AM = 2 x 5 = 10cm.