(i) To find the perimeter of the shaded region PQT, we need to calculate the lengths of PT, QT, and the arc PQ.

The length of PT is given by the tangent formula: \(PT = r \tan \alpha\).

The length of QT is \(QT = OT - OQ = \frac{r}{\cos \alpha} - r\).

The length of the arc PQ is \(r \alpha\).

Therefore, the perimeter of the shaded region PQT is \(r \tan \alpha + \frac{r}{\cos \alpha} - r + r \alpha\).

(ii) For \(\alpha = \frac{1}{3} \pi\) and \(r = 10\):

The area of triangle POT is \(\frac{1}{2} \times 10 \times 10 \times \tan \frac{\pi}{3} = 50 \sqrt{3}\).

The area of sector POQ is \(\frac{1}{2} \times 10^2 \times \frac{1}{3} \pi = \frac{50 \pi}{3}\).

The area of the shaded region is \(50 \sqrt{3} - \frac{50 \pi}{3} \approx 34\) (to 2 significant figures).

(i) To find the perimeter of the shaded region, we need to calculate the lengths of CD, BD, and the arc CB.

The length of CD is given by \(CD = r \cos \theta\).

The length of BD is given by \(BD = r - r \sin \theta\).

The length of the arc CB is given by \(r \left( \frac{1}{2} \pi - \theta \right)\).

Thus, the perimeter \(P\) is:

\(P = r \cos \theta + r - r \sin \theta + r \left( \frac{1}{2} \pi - \theta \right)\)

(ii) To find the area of the shaded region when r = 5 cm and θ = 0.6, we calculate the area of the sector and subtract the areas of triangle AOC and sector COD.

The area of sector AOB is \(\frac{1}{2} \times 5^2 \times \left( \frac{1}{2} \pi - 0.6 \right) = 12.135\).

The area of triangle AOC is \(\frac{1}{2} \times 5 \times \cos 0.6 \times 5 \times \sin 0.6 = 5.825\).

The area of the shaded region is:

\(12.135 - 5.825 = 6.31\)

(a)(i) Since \(AX\) is a tangent, \(\angle BAO = \angle OBA = \frac{\pi}{2} - \alpha\). Therefore, \(\angle AOB = \pi - (\frac{\pi}{2} - \alpha) - (\frac{\pi}{2} - \alpha) = 2\alpha\).

(a)(ii) The area of the sector \(OAB\) is \(\frac{1}{2} r^2 (2\alpha)\). The area of triangle \(OAB\) is \(\frac{1}{2} r^2 \sin 2\alpha\). Thus, the area of the shaded segment is \(\frac{1}{2} r^2 (2\alpha) - \frac{1}{2} r^2 \sin 2\alpha\).

(b) Use \(\alpha = \frac{\pi}{6}\) and \(r = 4\). The area of one segment is \(\frac{1}{2} (4)^2 \left( \frac{\pi}{3} \right) - \frac{1}{2} (4)^2 \sin \frac{\pi}{3} = \frac{8\pi}{3} - 4\sqrt{3}\).

The area of triangle \(ABC\) is \(\frac{1}{2} (4)^2 \sin \frac{\pi}{3} = 4\sqrt{3}\).

The area of the shaded region is \(3 \left( \frac{8\pi}{3} - 4\sqrt{3} \right) - 4\sqrt{3} = 16\sqrt{3} - 8\pi\).

(i) The area of sector OCD is given by \(\frac{1}{2} (2r)^2 \theta = 2r^2 \theta\).

The area of sectors OAB and OEF is \(2 \times \frac{1}{2} r^2 (\pi - \theta) = r^2 (\pi - \theta)\).

Thus, the total area of the metal plate is \(2r^2 \theta + r^2 (\pi - \theta) = r^2 (\pi + \theta)\).

(ii) The arc length CD is \(2r \theta\).

The arc lengths AB and EF are \(2 \times r (\pi - \theta) = 2r (\pi - \theta)\).

The straight edges AO, OE, BO, and OF sum to \(4r\).

Thus, the total perimeter is \(2r \theta + 2r (\pi - \theta) + 4r = 2\pi r + 4r\), which is independent of \(\theta\).

(i) To find the length of OB, use the cosine rule in the right-angled triangle OAB:

\(OB = \frac{OA}{\cos(\angle AOB)} = \frac{6}{\cos(0.6)} = 7.270 \text{ cm}\)

(ii) To find the perimeter of the metal plate, calculate the lengths of AB and the arc BC:

\(AB = OA \times \tan(\angle AOB) = 6 \times \tan(0.6) = 4.1 \text{ cm}\)

The arc length BC is given by:

\(s = r \theta = 7.27 \times (\frac{\pi}{2} - 0.6) = 7.06 \text{ cm}\)

Thus, the perimeter is:

\(6 + 7.27 + 7.06 + 4.1 = 24.4 \text{ cm}\)

(iii) To find the area of the metal plate, calculate the areas of triangle OAB and sector OBC:

The area of triangle OAB is:

\(\frac{1}{2} \times 6 \times 7.27 \times \sin(0.6) = 12.31 \text{ cm}^2\)

The area of sector OBC is:

\(\frac{1}{2} \times 7.27^2 \times (\frac{\pi}{2} - 0.6) = 25.65 \text{ cm}^2\)

Thus, the total area is:

\(12.31 + 25.65 = 38.0 \text{ cm}^2\)