(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 \(\alpha\):

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

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

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

(i) The area of circle C is \(\frac{1}{2} \times 3^2 \pi = \frac{9}{2} \pi\).

The area of the sector ORS is \(\frac{1}{2} \times 9^2 \theta - \frac{1}{2} \times 3^2 \theta\).

Equating the areas: \(\frac{9}{2} \pi = \frac{1}{2} \times 9^2 \theta - \frac{1}{2} \times 3^2 \theta\).

Simplifying gives \(\theta = \frac{1}{4}\pi\).

(ii) The perimeter of the shaded region includes the arc RS and the straight lines PS and PR.

The arc length RS is \(9 \times \frac{1}{4}\pi\).

The straight lines PS and PR are both 6 cm.

Thus, the perimeter is \(6 + 6 + 9 \times \frac{1}{4}\pi = 12 + 3\pi\) or approximately \(21.4 \text{ cm}\).

(i) To find angle BOC, use the formula for the angle in radians: \(\theta = 2 \tan^{-1} \left( \frac{1}{2} \right)\). Calculating gives \(\theta = 0.9273\) radians.

(ii) First, find the radius OB using the Pythagorean theorem: \(OB = \sqrt{10^2 + 5^2} = \sqrt{125} = 11.2\) cm.

Next, calculate the arc length BXC using \(s = r\theta\), where \(r = \sqrt{125}\) and \(\theta = 0.9273\). Thus, \(s = 11.2 \times 0.9273 = 10.4\) cm.

The perimeter of the shaded region is the arc length plus the side BC of the square: \(10.4 + 10 = 20.4\) cm.

(iii) The area of the sector BXC is given by \(\frac{1}{2} r^2 \theta\). Substituting the values, \(\frac{1}{2} \times 11.2^2 \times 0.9273 = 7.96\) cm².

(i) The area of sector OAB is given by \(\frac{1}{2} \times 8^2 \times \alpha = 32\alpha\).

The area of semicircle OAC is \(\frac{1}{2} \times \pi \times 4^2 = 8\pi\).

According to the problem, \(8\pi = 2 \times 32\alpha\).

Solving for \(\alpha\), we get \(\alpha = \frac{\pi}{8}\).

(ii) The perimeter of the complete figure consists of the straight lines OA and OB, and the arcs AB and AC.

The length of OA and OB is 8 cm each.

The length of arc AB is \(8 \times \frac{\pi}{8} = \pi\).

The length of arc AC is \(\frac{1}{2} \times 8\pi = 4\pi\).

Thus, the total perimeter is \(8 + 8 + \pi + 4\pi = 8 + 5\pi\).

The area of triangle \(ABC\) is calculated using the formula for the area of a right triangle:

\(\text{Area of } \triangle ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2\sqrt{3} \times 2 = 2\sqrt{3}\)

The angle \(A\) can be found using the tangent function:

\(\tan A = \frac{2\sqrt{3}}{2} = \sqrt{3}\)

Thus, \(A = \frac{\pi}{3}\) radians (or 60 degrees).

The area of the sector \(ACD\) is given by:

\(\text{Area of sector } ACD = \frac{1}{2} \times r^2 \times \theta = \frac{1}{2} \times 2^2 \times \frac{\pi}{3} = \frac{2\pi}{3}\)

The area of the shaded region \(BDC\) is the area of triangle \(ABC\) minus the area of sector \(ACD\):

\(\text{Shaded area } BDC = 2\sqrt{3} - \frac{2\pi}{3}\)