(i) The arc length AB is given by \(2r\theta\). The tangent segments AT and BT are equal, and each is \(r \tan \theta\) because \(\tan \theta = \frac{AT}{r}\). Therefore, the perimeter \(P\) of the shaded region is:

\(P = 2r\theta + 2r \tan \theta\)

(ii) For r = 5 and \theta = 1.2, the area of triangle AOT is \(\frac{1}{2} \times 5 \times 5 \times \tan 1.2\). The area of sector AOB is \(\frac{1}{2} \times 5^2 \times 2.4\). The shaded area is twice the area of triangle AOT minus the area of sector AOB:

\(\text{Area} = 2 \times \left( \frac{1}{2} \times 5 \times 5 \times \tan 1.2 \right) - \frac{1}{2} \times 25 \times 2.4\)

Calculating gives:

\(\text{Area} = 34.3 \text{ cm}^2\)