Let O be the common centre of two concentric circles, and let AB be a chord of the larger circle touching the smaller circle at P. Join OP.
∵ OP is the radius of the smaller circle and AB is tangent to this circle at P,
∴ \( OP \ ⊥ \ AB \).
We know that, the perpendicular drawn from the centre of a circle to any chord of the circle bisects the chord.
So, \( OP \ ⊥ \ AB \) and \( AP \ = \ BP \).
In \( ∆ \ APO \), we have
\( OA^2 \ = \ AP^2 \ + \ OP^2 \ => \ 5^2 \ = \ AP^2 \ + \ 3^2 \)
\( AP^2 \ = \ 5^2 \ - \ 3^2 \ = \ 25 \ - \ 9 \ = \ 16 \)
\( AP \ = \ 4 \)
Now, \( AB \ = \ 2AP \) [∵ \( AP \ = \ BP \) ]
\( AB = \ 2 \ × \ 4 \ = \ 8 \)
∴ The length of the chord of the larger circle which touches the smaller circle is 8 cm.