First, find the angle \(\theta\) of the sector using the formula for the area of a sector:

\(\frac{1}{2} \times 8^2 \times \theta = \frac{16}{3} \pi\)

Solving for \(\theta\):

\(32 \theta = \frac{16}{3} \pi\)

\(\theta = \frac{\pi}{6}\)

Next, calculate the arc length \(BC\):

\(\text{Arc length} = 8 \times \frac{\pi}{6} = \frac{4\pi}{3} \approx 4.1887 \text{ cm}\)

Now, find the length of \(BC\) using the sine rule or cosine rule. Using the cosine rule:

\(BC^2 = 8^2 + 8^2 - 2 \times 8 \times 8 \times \cos\left(\frac{\pi}{6}\right)\)

\(BC^2 = 128 - 64 \sqrt{3}\)

\(BC \approx 4.1411 \text{ cm}\)

Finally, the perimeter of segment \(BCD\) is:

\(\text{Perimeter} = BC + \text{Arc length} = 4.1411 + 4.1887 \approx 8.33 \text{ cm}\)

(i) To find the perimeter of the shaded region, we first calculate the length of the major arc. The angle of the major arc is given by:

\(2\pi - 2.2 = 4.083\) radians.

The length of the major arc is:

\(s = r\theta = 6 \times 4.083 = 24.5\) cm.

The perimeter of the shaded region is the sum of the lengths of the two radii and the major arc:

\(6 + 6 + 24.5 = 36.5\) cm.

(ii) The area of the major sector is given by:

\(\frac{1}{2} r^2 \theta = \frac{1}{2} \times 6^2 \times 4.083 = 73.49\) cm².

The area of triangle AOB is given by:

\(\frac{1}{2} \times 6^2 \times \sin 2.2 = 14.55\) cm².

The ratio of the area of the shaded region to the area of triangle AOB is:

\(\frac{73.49}{14.55} = 5.05\).

Thus, the ratio is 5.05 : 1.

(i) The area of sector OAB is \(\frac{1}{2} \times 11^2 \times \alpha\).

The area of sector OCD is \(\frac{1}{2} \times 5^2 \times \alpha\).

The area of the shaded region ABDC is \(\frac{1}{2} \times 11^2 \times \alpha - \frac{1}{2} \times 5^2 \times \alpha\).

Thus, \(k = \frac{\frac{1}{2} \times 11^2 \times \alpha - \frac{1}{2} \times 5^2 \times \alpha}{\frac{1}{2} \times 5^2 \times \alpha} = \frac{11^2 - 5^2}{5^2} = \frac{96}{25}\) or 3.84.

(ii) The perimeter of the shaded region ABDC is \(11\alpha + 5\alpha + 6 + 6 = 16\alpha + 12\).

The perimeter of the unshaded region OCD is \(5\alpha + 5 + 5 = 5\alpha + 10\).

Setting the perimeter of the shaded region equal to twice the perimeter of the unshaded region gives:

\(16\alpha + 12 = 2(5\alpha + 10)\)

\(16\alpha + 12 = 10\alpha + 20\)

\(6\alpha = 8\)

\(\alpha = \frac{4}{3}\) or 1.33.

(i) Since angle AOC and angle BOC are both 2.3 radians, angle AOB is calculated as:

\(\angle AOB = 2\pi - 2 \times 2.3 = 1.683 \text{ radians (correct to 4 significant figures)}\)

(ii) The area of the shaded region ACBP can be found by calculating the area of the sector AOB and subtracting the areas of triangles AOC and BOC.

The area of triangle AOC is given by:

\(\frac{1}{2} \times 3^2 \times \sin(2.3)\)

The area of triangle BOC is the same as triangle AOC.

The area of sector AOB is:

\(\frac{1}{2} \times 3^2 \times 1.683\)

Thus, the area of the shaded region ACBP is:

\(2 \times \frac{1}{2} \times 3^2 \times \sin(2.3) + \frac{1}{2} \times 3^2 \times 1.683 = 14.3 \text{ (correct to 3 significant figures)}\)

(i) The area of the shaded region is the difference between the areas of sectors OAB and OCD.

The area of sector OAB is \(\frac{1}{2} \times 16^2 \times 0.8\).

The area of sector OCD is \(\frac{1}{2} \times 10^2 \times 0.8\).

Thus, the area of the shaded region is \(\frac{1}{2} \times 16^2 \times 0.8 - \frac{1}{2} \times 10^2 \times 0.8 = 62.4 \text{ cm}^2\).

(ii) The perimeter of the shaded region consists of the arcs AC and BD and the straight lines AB and CD.

The length of arc AC is \(10 \alpha\) and the length of arc BD is \(16 \alpha\).

The lengths of AB and CD are both 6 cm.

Thus, the perimeter is \(10 \alpha + 16 \alpha + 6 + 6 = 28.9\).

Solving \(26 \alpha + 12 = 28.9\) gives \(\alpha = 0.65\).