Two concentric circles are of radii 13 cm and 5 cm. Find the length of the chord of the outer circle which touches the inner circle.

From figure,

AB is the chord of the outer circle which touches the inner circle at P.

OP is the radius of the inner circle and APB is the tangent to the inner circle.

In the right angled triangle OPB, by pythagoras theorem,

$\Rightarrow OB^2 = OP^2 + PB^2 \\[1em] \Rightarrow 13^2 = 5^2 + PB^2 \\[1em] \Rightarrow 169 = 25 + PB^2 \\[1em] \Rightarrow PB^2 = 169 - 25 \\[1em] \Rightarrow PB^2 = 144 \\[1em] \Rightarrow PB = \sqrt{144} \text{ cm} \\[1em] \Rightarrow PB = 12 \text{ cm}.$

As perpendicular line from centre bisects the chord of the circle so,

AP = PB = 12 cm.

AB = AP + PB = 12 + 12 = 24 cm.

Hence, the length of chord = 24 cm.