Let AB be the chord of the given circle with centre O and a radius of 5 cm.
From O, draw OM perpendicular to AB.
Then OM = 3 cm and OB = 5 cm
Processing math: 1%From the right Δ OMB, we have:
OB ^2 = OM^ 2 + MB ^2
(Pythagoras theorem)
⇒ 5^ 2 = 3^ 2 + MB^ 2
⇒ 25 = 9 + MB ^2
⇒ MB ^2 = (25 − 9) = 16
⇒ MB=16 cm=4 cm
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
AB = 2 × MB = (2 × 4) cm = 8 cm
Hence, the required length of the chord is 8 cm.