Let the two concentric circles with centre O.
AB be the chord of the larger circle which touches the smaller circle at point P.
∴ AB is tangent to the smaller circle to the point P.
⇒OP⊥AB
By Pythagoras theorem in Δ OPA,
OA2=AP2+OP2
⇒52=AP2+32
⇒AP2=25−9
⇒AP=4
In Δ OPB,
Since OP ⊥ AB,
AP = PB (∵ Perpendicular from the centre of the circle bisects the chord)
AB = 2AP = 2 × 4 = 8 cm
∴ The length of the chord of the larger circle is 8 cm.