(a) The perimeter of the cross-section RASB is the sum of the arc lengths ARB and RBS. The arc length ARB is given by \(2.5 \times \frac{4\pi}{3}\) and the arc length RBS is given by \(2.24 \times \frac{5\pi}{6}\). Therefore, the perimeter is:

\(2.5 \times \frac{4\pi}{3} + 2.24 \times \frac{5\pi}{6} = 10.47 + 5.86 = 16.34\)

(b) The area of triangle AOB is \(\frac{1}{2} \times 2.5^2 \times \sin \frac{2\pi}{3} = 2.706\) and the area of triangle APB is \(\frac{1}{2} \times 2.24^2 \times \sin \frac{5\pi}{6} = 1.254\). The difference in area is:

\(2.706 - 1.254 = 1.45\)

(c) The area of the cross-section RASB is the sum of the areas of sectors OARB and PASB plus the area of quadrilateral OAPB. The area of sector OARB is \(\frac{1}{2} \times 2.5^2 \times \frac{4\pi}{3} = 13.09\) and the area of sector PASB is \(\frac{1}{2} \times 2.24^2 \times \frac{5\pi}{6} = 6.57\). Adding the difference in area from part (b), the total area is:

\(13.09 + 6.57 + 1.45 = 21.1\)

(i) The perimeter of R₁ is given by the sum of the two radii and the arc length: \(r + r + r\theta = 2r + r\theta\).

The length of the major arc AB is \(2\pi r - r\theta\).

Equating the two expressions: \(2r + r\theta = 2\pi r - r\theta\).

Solving for \(\theta\):

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

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

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

\(2r(1 + \theta) = 2\pi r\)

\(1 + \theta = \pi\)

\(\theta = \pi - 1\)

(ii) The area of region R₁ is given by \(\frac{1}{2}r^2\theta\).

Given \(\theta = \pi - 1\) and area of R₁ is 30 cm²:

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

\(r^2 = \frac{60}{\pi - 1}\)

\(r = \sqrt{\frac{60}{\pi - 1}} \approx 5.29\)

The area of region R₂ is the area of the circle minus the area of R₁:

\(R_2 = \frac{1}{2}r^2(\pi + 1)\)

\(R_2 = \frac{1}{2} \times \frac{60}{\pi - 1} \times (\pi + 1)\)

\(R_2 \approx 58.0 \text{ cm}^2\)

(i) To find the angle \(\angle POQ\) in radians, use the formula for arc length: \(s = r\theta\). Given \(s = 9\) cm and \(r = 5\) cm, we have:

\(9 = 5\theta\)

\(\theta = \frac{9}{5} = 1.8 \text{ rad}\)

(ii) To find the length of \(PT\), consider triangle \(POT\). The angle \(\angle POT\) is half of \(\angle POQ\), so \(\angle POT = \frac{1.8}{2} = 0.9 \text{ rad}\). Using the tangent function:

\(PT = 5 \tan(0.9) \approx 6.30 \text{ cm}\)

(iii) To find the area of the shaded region, first calculate the area of sector \(POQ\):

\(\text{Area of sector} = \frac{1}{2} \times 5^2 \times 1.8 = 22.5 \text{ cm}^2\)

Next, calculate the area of triangle \(POT\):

\(\text{Area of } POT = \frac{1}{2} \times 5 \times 6.30 = 15.75 \text{ cm}^2\)

The shaded area is twice the area of triangle \(POT\) minus the area of the sector:

\(\text{Shaded area} = 2 \times 15.75 - 22.5 = 9.00 \text{ cm}^2\)

(i) To find the perimeter of the shaded region, we first use the Pythagorean theorem in triangle OPT to find OT:

\(OT = \sqrt{OP^2 + PT^2} = \sqrt{5^2 + 12^2} = 13 \text{ cm}\)

Next, find QT:

\(QT = OT - OQ = 13 - 5 = 8 \text{ cm}\)

Calculate the angle POQ using the tangent function:

\(\angle POQ = \tan^{-1}\left(\frac{12}{5}\right) = 1.176 \text{ radians}\)

Find the arc length S:

\(S = r\theta = 5 \times 1.176 = 5.88 \text{ cm}\)

Thus, the perimeter of the shaded region is:

\(5.88 + 12 + 8 = 25.9 \text{ cm}\)

(ii) To find the area of the shaded region, calculate the area of sector POQ:

\(\text{Area of sector} = \frac{1}{2}r^2\theta = \frac{1}{2} \times 5^2 \times 1.176 = 14.7 \text{ cm}^2\)

Calculate the area of triangle OPT:

\(\text{Area of triangle} = \frac{1}{2} \times 12 \times 5 = 30 \text{ cm}^2\)

The area of the shaded region is:

\(30 - 14.7 = 15.3 \text{ cm}^2\)

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

In \(\triangle OAX\), \(AX = r\sin\theta\) and \(OX = r\cos\theta\).

The area of \(\triangle OAX\) is \(\frac{1}{2} \times OX \times AX = \frac{1}{2}r^2\sin\theta\cos\theta\).

Thus, the area of the shaded region AXB is:

\(A = \frac{1}{2}r^2\theta - \frac{1}{2}r^2\sin\theta\cos\theta = \frac{1}{2}r^2(\theta - \sin\theta\cos\theta)\)

(ii) Given \(r = 12\) and \(\theta = \frac{1}{6}\pi\):

\(AX = 12\sin\frac{1}{6}\pi = 6\)

\(OX = 12\cos\frac{1}{6}\pi = 6\sqrt{3}\)

\(BX = OB - OX = 12 - 6\sqrt{3}\)

The length of arc AB is \(r\theta = 12 \times \frac{1}{6}\pi = 2\pi\).

Therefore, the perimeter of the shaded region AXB is:

\(P = AX + BX + \text{arc } AB = 6 + (12 - 6\sqrt{3}) + 2\pi = 18 - 6\sqrt{3} + 2\pi\)