(i) To show \(\sin \alpha \cos \alpha = \frac{1}{2} \alpha\), consider the area of triangle \(\triangle OAC\). The area is \(\frac{1}{2} r^2 \sin \alpha \cos \alpha\). Since \(AC\) divides the sector into two regions of equal area, the area of \(\triangle OAC\) is half the area of sector \(OAB\), which is \(\frac{1}{2} \times \frac{1}{2} r^2 \alpha\). Equating these gives \(\frac{1}{2} r^2 \sin \alpha \cos \alpha = \frac{1}{2} \times \frac{1}{2} r^2 \alpha\), simplifying to \(\sin \alpha \cos \alpha = \frac{1}{2} \alpha\).

(ii) The perimeter of \(\triangle OAC\) is \(r + r \sin \alpha + r \cos \alpha = 2.4r\). The perimeter of \(ACB\) is \(r\alpha + r \sin \alpha + r - r \cos \alpha = 2.18r\) or \(2.17r\). The ratio is \(\frac{2.4}{2.18} : 1 = 1.1 : 1\).

(iii) Given \(\alpha = 0.9477\) radians, convert to degrees: \(0.9477 \times \frac{180}{\pi} \approx 54.3°\).