(i) To find the radius of the semicircle, use the cosine rule in triangle BOD:

\(BD = \sqrt{10^2 + 10^2 - 2 \times 10 \times 10 \times \cos(1.2)} = \sqrt{200 - 200 \cos(1.2)} = 11.3 \text{ cm}\)

Since BD is the diameter of the semicircle, the radius \(r\) is:

\(r = \frac{BD}{2} = \frac{11.3}{2} = 5.646 \text{ cm}\)

(ii) The perimeter of the metal plate consists of the major arc of the circle and the semicircle:

The major arc length is given by:

\(10 \times (2\pi - 1.2) = 10 \times 5.083 = 50.832 \text{ cm}\)

The semicircle perimeter is:

\(\pi \times 5.646 = 17.737 \text{ cm}\)

Thus, the total perimeter is:

\(50.832 + 17.737 = 68.6 \text{ cm}\)

(iii) The area of the metal plate consists of the area of the major sector, the semicircle, and triangle OBD:

The area of the major sector is:

\(\frac{1}{2} \times 10^2 \times (2\pi - 1.2) = 254.159 \text{ cm}^2\)

The area of triangle OBD is:

\(\frac{1}{2} \times 10 \times 10 \times \sin(1.2) = 46.602 \text{ cm}^2\)

The area of the semicircle is:

\(\frac{1}{2} \times \pi \times (5.646)^2 = 50.1 \text{ cm}^2\)

Thus, the total area is:

\(254.159 + 46.602 + 50.1 = 351 \text{ cm}^2\)