Let AB be a chord of a circle with centre O and radius 13 cm. Draw OL ⊥ AB.
Join OA. Clearly, OL = 5 cm and OA = 13 cm.
In the right triangle OLA, we have
OA^{2} = OL^{2} + AL^{2}
^{}
⇒ 13^{2} = 5^{2} + AL^{2}
⇒ AL^{2} = 144 cm^{2}
⇒ AL = 12 cm
Since, the perpendicular from the centre to the chord bisects the chord. Therefore,
AB = 2AL = (2 × 12) cm
= 24 cm.
Let AB be a chord of a circle with centre O and radius 13 cm. Draw OL ⊥ AB.
Join OA. Clearly, OL = 5 cm and OA = 13 cm.
In the right triangle OLA, we have
OA^{2} = OL^{2} + AL^{2}
⇒ 13^{2} = 5^{2} + AL^{2}
⇒ AL^{2} = 144 cm^{2}
⇒ AL = 12 cm
Since, the perpendicular from the centre to the chord bisects the chord. Therefore,
AB = 2AL = (2 × 12) cm
= 24 cm.