(i) The length of AS is \(r \tan \theta\). The area of triangle OAB is \(r^2 \tan \theta\) or \(\frac{1}{2} r^2 \tan \theta\). The area of sector OPS is \(\frac{1}{2} r^2 \times 2\theta = r^2 \theta\). Therefore, the shaded area is \(r^2 (\tan \theta - \theta)\).

(ii) Given \(\theta = \frac{1}{3}\pi\) and \(r = 6\), we have \(\cos \frac{\pi}{3} = \frac{6}{OA} \Rightarrow OA = 12\). Thus, \(AP = 6\). The length of AS is \(6 \tan \frac{\pi}{3} \Rightarrow AB = 12\sqrt{3}\). The arc PST is \(12 \times \frac{\pi}{3}\). Therefore, the perimeter is \(12 + 12\sqrt{3} + 4\pi\).

(i) To find the area of the shaded region BPDQ, we calculate the area of sector BPD and subtract the area of triangle BPD. The area of sector BPD is given by \(\frac{1}{2} \times 5^2 \times 1.2\). The area of triangle BPD is \(\frac{1}{2} \times 5^2 \times \sin 1.2\). Therefore, the area of the shaded region is:

\(2 \left[ \frac{1}{2} \times 5^2 \times 1.2 - \frac{1}{2} \times 5^2 \times \sin 1.2 \right] = 6.70\)

(ii) To find the length of PQ, we use the cosine rule in triangle APQ. The length of PQ is given by:

\(5 \cos 0.6\)

Then, subtract from 5:

\(5 - 5 \cos 0.6\)

Multiply by 2:

\(10(1 - \cos 0.6) = 1.75\)

(i) Use the Pythagorean theorem in triangle \(PQR\):

\(RS^2 = PR^2 - PQ^2 = 10^2 - 6^2\)

\(RS^2 = 100 - 36 = 64\)

\(RS = \sqrt{64} = 8 \text{ cm}\)

(ii) Use trigonometry to find \(\angle RPQ\):

\(\sin \theta = \frac{8}{10}\)

\(\theta = \arcsin\left(\frac{8}{10}\right) \approx 0.9273 \text{ radians}\)

(iii) The area of the shaded region is the area of the trapezium minus the areas of the two sectors:

Area of trapezium = \(40 \text{ cm}^2\)

Large sector area = \(\frac{1}{2} \times 8^2 \times 0.9273\)

Small sector angle = \(\pi - 0.9273\)

Small sector area = \(\frac{1}{2} \times 2^2 \times 2.214\)

Shaded area = \(40 - (\text{Large sector} + \text{Small sector}) = 5.90 \text{ cm}^2\)

(i) To find angle DOE, use the sine rule in triangle DOE:

\(\sin \frac{1}{2} \theta = \frac{6}{10}\)

Solving for \(\theta\), we find:

\(\theta = 1.287 \text{ radians}\)

(ii) To find the perimeter of the metal plate, calculate the arc length and add the straight edges:

Arc length of BOD = \(2 \times 10 \times (\pi - 1.287)\)

Perimeter = \(12 + 12 + 2 \times 10 \times (\pi - 1.287) = 61.1 \text{ cm}\)

(iii) To find the area of the metal plate, calculate the area of the sectors and triangles:

Area of sector DOE = \(\frac{1}{2} \times 10^2 \times 1.287\)

Area of triangle DOE = \(\frac{1}{2} \times 10^2 \times \sin 1.287\)

Total area = \(\pi \times 10^2 - (2 \times \text{sectors} - 2 \times \text{triangles})\)

Area = 281 or 282 cm².

(i) To find the length of the arc BD, we first calculate the length of AB using the Pythagorean theorem:

\(AB = \sqrt{6^2 + 6^2} = \sqrt{72}\)

The angle BAD is 45° or \(\frac{\pi}{4}\) radians. The arc length s is given by \(s = r\theta\), where \(r = \sqrt{72}\) and \(\theta = \frac{\pi}{4}\):

\(s = \sqrt{72} \times \frac{\pi}{4} \approx 6.67 \text{ cm}\)

(ii) To find the area of the shaded region, we calculate the area of the sector and subtract the area of triangle OAB. The sector area is given by:

\(\text{Sector area} = \frac{1}{2} r^2 \theta = \frac{1}{2} \times 72 \times \frac{\pi}{4}\)

The area of triangle OAB is:

\(\text{Area of triangle} = \frac{1}{2} \times 6 \times 6 = 18 \text{ cm}^2\)

Thus, the shaded area is:

\(\text{Shaded area} = \frac{1}{2} \times 72 \times \frac{\pi}{4} - 18 \approx 10.3 \text{ cm}^2 \text{ or } 9\pi - 18 \text{ cm}^2\)