(i) The length of the arc AC is given by the formula \(r \theta = 6\), where \(r = OA\) and \(\theta = \frac{3}{8} \pi\). Solving for \(r\), we have:

\(OA \times \frac{3}{8} \pi = 6\)

\(OA = \frac{6 \times 8}{3 \pi} = \frac{16}{\pi} \approx 5.093 \text{ cm}\)

(ii) The perimeter of the shaded region is the sum of the lengths of AB, BC, and the arc AC. Since AB and BC are tangents, \(AB = BC\). Using the tangent formula:

\(AB = OA \times \tan\left(\frac{3}{16} \pi\right)\)

\(AB = 5.093 \times \tan\left(\frac{3}{16} \pi\right) \approx 3.403 \text{ cm}\)

The perimeter is:

\(2 \times 3.403 + 6 = 12.8 \text{ cm}\)

(iii) The area of the shaded region is the difference between the area of triangle OABC and the area of sector OAC. The area of triangle OABC is:

\(\text{Area of } OABC = \frac{1}{2} \times OA \times AB = \frac{1}{2} \times 5.093 \times 3.403 \approx 8.665 \text{ cm}^2\)

The area of sector OAC is:

\(\text{Area of sector } OAC = \frac{1}{2} \times (5.093)^2 \times \frac{3}{8} \pi \approx 6.615 \text{ cm}^2\)

The shaded area is:

\(8.665 - 6.615 = 2.05 \text{ cm}^2\)

(i) To find the length of AD, we use the angle EAD = ACD = \frac{3\pi}{10} radians. Using the sine rule in triangle ACD, we have:

\(AD = 8 \sin\left(\frac{3\pi}{10}\right)\)

Calculating this gives \(AD \approx 6.47\) cm.

(ii) To find the area of the shaded region, we calculate the area of sector ADE and subtract the area of triangle ADC.

The area of sector ADE is:

\(\frac{1}{2} \times (\text{their } AD)^2 \times \frac{\pi}{5}\)

The area of triangle ADC is:

\(\frac{1}{2} \times 8 \times (\text{their } AD) \times \sin\left(\frac{\pi}{5}\right)\)

Calculating these areas and subtracting gives the shaded area \(\approx 35.0 \text{ or } 34.9 \text{ cm}^2\).

The perimeter of sector \(AOC\) is given by the arc length plus the radii, which is \(2r + r\theta\).

The angle \(COB\) is \(\pi - \theta\) because the total angle in a semicircle is \(\pi\) radians.

The perimeter of sector \(BOC\) is \(2r + r(\pi - \theta)\).

According to the problem, the perimeter of \(BOC\) is twice the perimeter of \(AOC\):

\(2r + r(\pi - \theta) = 2(2r + r\theta)\)

Simplifying, we have:

\(2r + r\pi - r\theta = 4r + 2r\theta\)

\(r\pi - r\theta = 2r\theta + 2r\)

\(r\pi = 3r\theta + 2r\)

\(\pi = 3\theta + 2\)

\(\theta = \frac{\pi - 2}{3}\)

Calculating \(\theta\) gives \(\theta \approx 0.38\) to 2 significant figures.

To find the perimeter of the sector, we need to express it in terms of the radius \(r\) and the area \(A\).

The area of a sector is given by \(A = \frac{1}{2} r^2 \theta\), where \(\theta\) is the angle in radians.

Rearranging for \(\theta\), we have:

\(\theta = \frac{2A}{r^2}\)

The perimeter \(P\) of the sector is the sum of the two radii and the arc length:

\(P = r + r + r\theta\)

Substituting \(\theta\) gives:

\(P = 2r + r \left( \frac{2A}{r^2} \right)\)

Simplifying, we get:

\(P = 2r + \frac{2A}{r}\)

1. Calculate angle CBA using the sine rule: \(\angle CBA = \sin^{-1}\left(\frac{7}{8}\right) \approx 1.0654 \text{ radians}\).

2. Calculate the area of sector BCYD: \(\frac{1}{2} \times 8^2 \times 2 \times 1.0654 \approx 68.19 \text{ cm}^2\).

3. Calculate the area of triangle BCD: \(7 \times \sqrt{8^2 - 7^2} \approx 27.11 \text{ cm}^2\).

4. Calculate the area of the semicircle CXD: \(\frac{1}{2} \pi \times 7^2 \approx 76.97 \text{ cm}^2\).

5. Total area enclosed by the arcs: \(68.19 - 27.11 + 76.97 \approx 118.0 \text{ to } 118.1 \text{ cm}^2\).