(i) For \(\theta = 1\), angle BOC = \(\pi - \theta\). The area of sector BOC is given by \(\frac{1}{2} r^2 (\pi - \theta)\). Substituting \(r = 8\) and \(\theta = 1\), the area is \(\frac{1}{2} \times 8^2 \times (\pi - 1) = 32(\pi - 1)\) or approximately 68.5.

(ii) The perimeter of sector AOB is \(8 + 8 + 8\theta\) and the perimeter of sector BOC is \(8 + 8 + 8(\pi - \theta)\). Setting \(8 + 8 + 8\theta = \frac{1}{2}(8 + 8 + 8(\pi - \theta))\), solve for \(\theta\):

\(16 + 8\theta = 8 + 8 + 4\pi - 4\theta\)

\(12\theta = 4\pi - 16\)

\(\theta = \frac{1}{3}(\pi - 2)\) or approximately 0.381.

(iii) For \(\theta = \frac{1}{3}\pi\), AB = 8 cm. Using the cosine rule or trigonometry, BC = \(2 \times 8 \sin \frac{\pi}{3} = 8\sqrt{3}\). The perimeter of triangle ABC is \(8 + 8 + 8\sqrt{3} = 24 + 8\sqrt{3}\) cm.

(i) To find the area of the shaded region, we need to calculate the area of triangle OQR and subtract the area of sector OPQ.

The length QR is given by QR = r \tan \theta.

The area of triangle OQR is \(\frac{1}{2} \times r \times r \tan \theta = \frac{1}{2}r^2 \tan \theta\).

The area of sector OPQ is \(\frac{1}{2}r^2 \theta\).

Thus, the area of the shaded region is \(\frac{1}{2}r^2 \tan \theta - \frac{1}{2}r^2 \theta = \frac{1}{2}r^2(\tan \theta - \theta)\).

(ii) Given θ = 0.8 and r = 15, we calculate the perimeter of the shaded region.

The arc length PQ is \(r \theta = 15 \times 0.8 = 12 \text{ cm}\).

The length OR is \(\frac{r}{\cos \theta} = \frac{15}{\cos 0.8} \approx 21.53 \text{ cm}\).

The perimeter of the shaded region is \(r \tan \theta + \text{arc } PQ + (r - r \cos \theta)\).

Substituting the values, we get \(15 \tan 0.8 + 12 + (15 - 15 \cos 0.8) \approx 34.0 \text{ cm}\).

(i) To find angle AOB, use the sine rule in the right triangle formed by the radius and half of AB. The sine of half the angle is given by:

\(\sin(\frac{1}{2} \text{angle}) = \frac{16}{20}\)

Solving for the angle, we find:

\(\text{Required angle} = 1.855 \text{ radians}\)

(ii) The area of sector AXBO is calculated using the formula:

\(\frac{1}{2} r^2 \theta\)

Substituting the values, we get:

\(\frac{1}{2} \times 20^2 \times 1.855 = 371 \text{ cm}^2\)

(iii) The area of the new cross-section is found by subtracting the area of the rectangle and the sector from the area of the circle, and then adding the area of the triangle:

\(\pi r^2 - l \times b - \frac{1}{2} r^2 \theta + \frac{1}{2} bh\)

Substituting the values, we get:

\(\pi \times 20^2 - 32 \times 18 - \frac{1}{2} \times 20^2 \times 1.855 + \frac{1}{2} \times 32 \times 18 = 502 \text{ cm}^2\) (accept 501)

(a) To find the perimeter of the shaded region, we first need to find the length of AC. Using the cosine rule:

\(AC^2 = 6^2 + 6^2 - 2 \times 6 \times 6 \times \cos(1.8)\)

\(AC = 9.40 \text{ cm}\)

Next, calculate the angle CAB:

\(\text{Angle } CAB = \frac{1}{2}(\pi - 1.8)\)

\(\text{Angle } CAB = 0.6708 \text{ radians}\)

Now, find the length of arc CD:

\(\text{Arc } CD = 9.40 \times 0.6708 = 6.306 \text{ cm}\)

The perimeter of the shaded region is:

\(6 + 3.40 + 6.306 = 15.7 \text{ cm}\)

(b) To find the area of the shaded region, calculate the area of sector ADC and subtract the area of triangle ABC:

\(\text{Area of sector } ADC = \frac{1}{2} \times 9.40^2 \times 0.6708\)

\(\text{Area of triangle } ABC = \frac{1}{2} \times 6^2 \times \sin(1.8)\)

\(\text{Area of shaded region} = 29.64 - 17.53 = 12.1 \text{ cm}^2\)

(a) To find angle BOC, we use the formula for arc length: \(s = r\theta\). Given the arc length AB is 2 cm and radius is 10 cm, we have:

\(\theta = \frac{s}{r} = \frac{2}{10} = 0.2 \text{ radians}\)

Angle BOC is the sum of angle AOC and angle AOB:

\(\angle BOC = \angle AOC + \angle AOB = \frac{1}{6} \pi + 0.2 \approx 0.724 \text{ radians}\)

(b) To find the area of the shaded region BPC, we first find the lengths of BP and OP using trigonometry in triangle OBP:

\(BP = 10 \sin \left( \frac{5\pi + 6}{30} \right) \approx 6.6208\)

\(OP = 10 \cos \left( \frac{5\pi + 6}{30} \right) \approx 7.494\)

The area of triangle OBP is:

\(\text{Area of } \triangle OBP = \frac{1}{2} \times 10 \times 10 \times \sin \left( \frac{5\pi + 6}{30} \right) \approx 24.809 \text{ cm}^2\)

The area of sector BOC is:

\(\text{Area of sector BOC} = \frac{1}{2} \times 10^2 \times \left( \frac{5\pi + 6}{30} \right) \approx 36.1799 \text{ cm}^2\)

The area of the shaded region BPC is:

\(\text{Area of region BPC} = \text{Area of sector BOC} - \text{Area of } \triangle OBP \approx 11.4 \text{ cm}^2\)