(i) The perimeter of the metal plate consists of the arc lengths and the straight edges. The arc length of OABCD is \(2r(2\pi - \alpha)\) and the arc length of OAED is \(r\alpha\). The straight edges are \(2r\) and \(r\). Therefore, the perimeter is:

\(2r(2\pi - \alpha) + r\alpha + 2r + r = 2\pi r + r\alpha + 2r\)

(ii) The area of the metal plate is the sum of the areas of the two sectors. The area of sector OABCD is \(\frac{1}{2}(2r)^2\alpha\) and the area of sector OAED is \(\pi r^2 - \frac{1}{2}r^2\alpha\). Therefore, the total area is:

\(\frac{1}{2}(2r)^2\alpha + \pi r^2 - \frac{1}{2}r^2\alpha = \frac{3r^2\alpha}{2} + \pi r^2\)

(iii) Given that the shaded and unshaded pieces are equal in area, we equate the areas:

\(\pi r^2 - \frac{1}{2}r^2\alpha = \frac{1}{2}(2r)^2\alpha\)

Solving for \(\alpha\):

\(\pi r^2 - \frac{1}{2}r^2\alpha = 2r^2\alpha\)

\(\pi r^2 = \frac{5}{2}r^2\alpha\)

\(\alpha = \frac{2}{5}\pi\)