(a) To find the arc length \(AB\), we first need to find the length of \(OA\). Using the cosine rule in triangle \(OAP\):

\(OA^2 = OP^2 + AP^2 - 2 \cdot OP \cdot AP \cdot \cos(\theta)\)

\(OA^2 = x^2 + x^2 - 2x^2 \cos(\theta)\)

\(OA^2 = 2x^2(1 - \cos(\theta))\)

\(OA = x \sqrt{2(1 - \cos(\theta))}\)

Using the identity \(1 - \cos(\theta) = 2\sin^2(\frac{\theta}{2})\), we have:

\(OA = x \sqrt{4\sin^2(\frac{\theta}{2})} = 2x \sin(\frac{\theta}{2})\)

Since \(\sin(\frac{\theta}{2}) = \cos(\frac{\pi}{2} - \frac{\theta}{2})\), we have:

\(OA = 2x \cos(\frac{\pi}{2} - \frac{\theta}{2})\)

Thus, \(OA = 2x \cos \theta\).

The arc length \(AB\) is given by \(\theta \times OA\):

\(AB = \theta \times 2x \cos \theta = 2x\theta \cos \theta\)

(b) The area of the sector \(OAB\) is:

\(\text{Sector area} = \frac{1}{2} \times (2x \cos \theta)^2 \times \theta\)

\(= 2x^2 \cos^2 \theta \times \theta\)

The area of triangle \(OAP\) is:

\(\text{Triangle area} = \frac{1}{2} \times x \times x \times \sin(\pi - 2\theta)\)

\(= \frac{1}{2} x^2 \sin(2\theta)\)

The area of the shaded region \(APB\) is:

\(\text{Area of } APB = \text{Sector area} - \text{Triangle area}\)

\(= 2x^2 \cos^2 \theta \times \theta - \frac{1}{2} x^2 \sin(2\theta)\)

\(= x^2 \left( 2\cos^2 \theta - \frac{1}{2} \sin 2\theta \right)\)

(i) To find the radius \(OE\) of the arc \(CED\), we use the cosine rule in the triangle \(OEC\):

\(\cos(0.9) = \frac{OE}{6}\)

Solving for \(OE\), we get:

\(OE = 6 \cos(0.9) = 3.73 \text{ cm}\)

(ii) To find the area of the shaded region, we calculate the area of the large sector and subtract the area of the small sector:

The angle of the large sector is \(2\pi - 1.8\).

Area of the large sector:

\(\frac{1}{2} \times 6^2 \times (2\pi - 1.8) = 80.7 \text{ cm}^2\)

Area of the small sector:

\(\frac{1}{2} \times 3.73^2 \times 1.8 = 12.52 \text{ cm}^2\)

Total area of the shaded region:

\(80.7 + 12.52 = 93.2 \text{ cm}^2\)

(i) To find the radius of the semicircle, use the cosine rule in triangle BOD:

\(BD = \sqrt{10^2 + 10^2 - 2 \times 10 \times 10 \times \cos(1.2)} = \sqrt{200 - 200 \cos(1.2)} = 11.3 \text{ cm}\)

Since BD is the diameter of the semicircle, the radius \(r\) is:

\(r = \frac{BD}{2} = \frac{11.3}{2} = 5.646 \text{ cm}\)

(ii) The perimeter of the metal plate consists of the major arc of the circle and the semicircle:

The major arc length is given by:

\(10 \times (2\pi - 1.2) = 10 \times 5.083 = 50.832 \text{ cm}\)

The semicircle perimeter is:

\(\pi \times 5.646 = 17.737 \text{ cm}\)

Thus, the total perimeter is:

\(50.832 + 17.737 = 68.6 \text{ cm}\)

(iii) The area of the metal plate consists of the area of the major sector, the semicircle, and triangle OBD:

The area of the major sector is:

\(\frac{1}{2} \times 10^2 \times (2\pi - 1.2) = 254.159 \text{ cm}^2\)

The area of triangle OBD is:

\(\frac{1}{2} \times 10 \times 10 \times \sin(1.2) = 46.602 \text{ cm}^2\)

The area of the semicircle is:

\(\frac{1}{2} \times \pi \times (5.646)^2 = 50.1 \text{ cm}^2\)

Thus, the total area is:

\(254.159 + 46.602 + 50.1 = 351 \text{ cm}^2\)

(i) The perimeter of the shaded region ABDC is given by the sum of the lengths of arcs AB and CD, and the radii OA and OB. The length of arc AB is \(2r\alpha\) and the length of arc CD is \(r\alpha\). Therefore, the perimeter is:

\(2r\alpha + r\alpha + 2r = 4.4r\)

Simplifying, we get:

\(3r\alpha + 2r = 4.4r\)

\(3r\alpha = 2.4r\)

\(\alpha = 0.8\)

(ii) The area of the shaded region is the difference between the area of sector OAB and sector OCD. The area of sector OAB is \(\frac{1}{2}(2r)^2\alpha\) and the area of sector OCD is \(\frac{1}{2}r^2\alpha\). Therefore, the area of the shaded region is:

\(\frac{1}{2}(2r)^2 \times 0.8 - \frac{1}{2}r^2 \times 0.8 = 30\)

\(2r^2 \times 0.8 - 0.5r^2 \times 0.8 = 30\)

\((3/2)r^2 \times 0.8 = 30\)

\(r^2 \times 1.2 = 30\)

\(r^2 = 25\)

\(r = 5\)

First, calculate the angles in triangle ABC:

\(\angle BAC = \sin^{-1}(3/5) \approx 36.87^\circ\)

\(\angle ABC = \sin^{-1}(4/5) \approx 53.13^\circ\)

\(\angle ACB = \pi/2\) (right angle)

Calculate the area of triangle ABC:

\(\Delta ABC = \frac{1}{2} \times 4 \times 3 = 6 \, \text{cm}^2\)

Calculate the areas of the sectors:

\(\text{Sector AEF} = \frac{1}{2} \times 3^2 \times 0.6435\)

\(\text{Sector BEG} = \frac{1}{2} \times 2^2 \times 0.9273\)

\(\text{Sector CFG} = \frac{1}{2} \times 1^2 \times 1.5708\)

Sum of sectors = \(\frac{1}{2} [3^2 \times 0.6435 + 2^2 \times 0.9273 + 1^2 \times 1.5708]\)

Sum of sectors = 5.536 \, \text{cm}^2

Area of shaded region EFG:

\(\text{Area of EFG} = \Delta ABC - \text{Sum of sectors} = 6 - 5.536 = 0.464 \, \text{cm}^2\)