(i) To find the radius of the larger circle, consider the right triangle ABC where AB = r and AC = r. By the Pythagorean theorem:

\(BC^2 = AB^2 + AC^2 = r^2 + r^2 = 2r^2\)

\(BC = \sqrt{2r^2} = r\sqrt{2}\)

Thus, the radius of the larger circle is \(r\sqrt{2}\).

(ii) To find the area of the shaded region:

1. Calculate the area of sector BCFD of the larger circle:

\(\text{Area of sector BCFD} = \frac{1}{4} \pi (r\sqrt{2})^2 = \frac{1}{2} \pi r^2\)

2. Calculate the area of triangle BCAD:

\(\text{Area of } \triangle BCAD = \frac{1}{2} (2r)r = r^2\)

3. Calculate the area of segment CFDA:

\(\text{Area of segment CFDA} = \frac{1}{2} \pi r^2 - r^2\)

4. Calculate the area of semicircle CADE:

\(\text{Area of semicircle CADE} = \frac{1}{2} \pi r^2\)

5. Calculate the shaded area:

\(\text{Shaded area} = \frac{1}{2} \pi r^2 - \left( \frac{1}{2} \pi r^2 - r^2 \right)\)

\(= r^2\)

Thus, the area of the shaded region is \(r^2\).

(i) To show \(\sin \alpha \cos \alpha = \frac{1}{2} \alpha\), consider the area of triangle \(\triangle OAC\). The area is \(\frac{1}{2} r^2 \sin \alpha \cos \alpha\). Since \(AC\) divides the sector into two regions of equal area, the area of \(\triangle OAC\) is half the area of sector \(OAB\), which is \(\frac{1}{2} \times \frac{1}{2} r^2 \alpha\). Equating these gives \(\frac{1}{2} r^2 \sin \alpha \cos \alpha = \frac{1}{2} \times \frac{1}{2} r^2 \alpha\), simplifying to \(\sin \alpha \cos \alpha = \frac{1}{2} \alpha\).

(ii) The perimeter of \(\triangle OAC\) is \(r + r \sin \alpha + r \cos \alpha = 2.4r\). The perimeter of \(ACB\) is \(r\alpha + r \sin \alpha + r - r \cos \alpha = 2.18r\) or \(2.17r\). The ratio is \(\frac{2.4}{2.18} : 1 = 1.1 : 1\).

(iii) Given \(\alpha = 0.9477\) radians, convert to degrees: \(0.9477 \times \frac{180}{\pi} \approx 54.3°\).

The radius of the semicircle is \(\frac{1}{2} AB = r \sin \theta\).

The area of the semicircle is \(\frac{1}{2} \pi r^2 \sin^2 \theta = A_1\).

The shaded area is the area of the semicircle minus the area of the segment.

Shaded area = \(A_1 - \frac{1}{2} r^2 \theta + \frac{1}{2} r^2 \sin 2\theta\).

(i) To find the perimeter of the shaded region, we need to calculate the lengths of AB, BC, and the arc AC.

The length of AB or CB is given by \(\frac{3}{\tan \frac{\pi}{6}} = 3 \tan \frac{\pi}{3} = 3\sqrt{3}\).

The length of the arc AC is \(3 \times \frac{2\pi}{3} = 2\pi\).

Therefore, the perimeter of the shaded region is \(3\sqrt{3} + 3\sqrt{3} + 2\pi = 6\sqrt{3} + 2\pi\).

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

The area of triangle OABC is \(2 \times \frac{1}{2} \times 3 \times 3\sqrt{3} = 9\sqrt{3}\).

The area of sector OADC is \(\frac{1}{2} \times 3^2 \times \frac{2\pi}{3} = 3\pi\).

Thus, the area of the shaded region is \(9\sqrt{3} - 3\pi\).

(i) To find angle BAO, use the tangent function: \(\tan \theta = \frac{5}{12}\). Solving for \(\theta\), we get \(\theta = \tan^{-1}\left(\frac{5}{12}\right) \approx 0.3948\) radians.

(ii) The area of triangle AOB is \(\frac{1}{2} \times 12 \times 5 = 30\) cm².

The shaded area is the sum of two sectors minus the triangle area. The area of the sector with center A is \(\frac{1}{2} \times 12^2 \times 0.3948\).

The other angle in the triangle is \(\frac{\pi}{2} - 0.3948\). The area of the sector with center B is \(\frac{1}{2} \times 5^2 \times \left(\frac{\pi}{2} - 0.3948\right)\).

Thus, the shaded area is:

\(\frac{1}{2} \times 12^2 \times 0.3948 + \frac{1}{2} \times 5^2 \times \left(\frac{\pi}{2} - 0.3948\right) - 30\)

\(= 28.43 + 14.70 - 30 = 13.1\) cm².