Consider ∆XOB and using Pythagoras theorem

13\(^2\) = 12\(^2\) + h\(^2\)

⇒ 169 = 144 + h\(^2\)

⇒ 169 - 144  =h\(^2\)

⇒ 25 = h\(^2\)

⇒ h = \(\sqrt25\) = 5cm

tan θ = \(\frac {opp}{adj}\)

⇒ tan θ = \(\frac{12}{5}\) = 2.4

⇒ θ = tan\(^{-1}\)(2.4)

⇒ θ = 67.38\(^0\)

∠AOB = 2θ = 2 x 67.38\(^o\) = 134.76\(^o\)

L = \(\frac{θ}{360^o} \times 2\pi r\)

⇒ L = \(\frac {134.76}{360} \times 2 \times \frac {22}{7} \times 13 = \frac {77082.72}{2520}\)

∴ L = 30.6cm (to 3 s.f)