(i) The arc length of AB is given by \(4\alpha\) since the radius is 4 cm and the angle is \(\alpha\) radians.

The arc length of DC is \((4 \cos \alpha) \alpha\) because the radius of the arc DC is \(4 \cos \alpha\).

The length AC (or DB) is \(4 - 4 \cos \alpha\) since AD is perpendicular to OB.

Thus, the perimeter of the shaded region ABDC is \(4\alpha \cos \alpha + 4\alpha + 8 - 8\cos \alpha\).

(ii) For \(\alpha = \frac{1}{6}\pi\), the length OD is \(4 \cos \frac{\pi}{6} = 2\sqrt{3}\).

The area of the sector AOB is \(\frac{1}{2} \times 4^2 \times \frac{\pi}{6}\).

The area of the sector ODC is \(-\frac{1}{2} (2\sqrt{3})^2 \times \frac{\pi}{6}\).

The shaded area is \(\frac{\pi}{3}\), which can be expressed as \(k\pi\) where \(k = \frac{1}{3}\).

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

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

The area of the segment AXB is \(\frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin \theta\).

Setting the areas equal: \(\frac{1}{2} r^2 \sin \theta = \frac{1}{2} r^2 \theta - \frac{1}{2} r^2 \sin \theta\).

Simplifying gives \(2 \sin \theta = \theta\), so \(p = 2\).

(ii) The chord length AB is \(2 \times 8 \sin \frac{2.4}{2} = 8 \sin 1.2 \times 2 \approx 14.9\).

The arc length AXB is \(2.4 \times 8 = 19.2\).

The perimeter of the segment AXB is the sum of these: \(14.9 + 19.2 = 34.1\).

To find the area of the shaded region, we need to calculate the area of triangle ABC and subtract the area of sector ADE.

The area of triangle ABC is given by:

\(\text{Area of } \triangle ABC = \frac{1}{2} \times AB \times BC = \frac{1}{2} \times 4 \times 4 \tan \alpha = 8 \tan \alpha\)

The area of sector ADE is given by:

\(\text{Area of sector } ADE = \frac{1}{2} \times 2^2 \times \alpha = 2 \alpha\)

Therefore, the area of the shaded region is:

\(8 \tan \alpha - 2 \alpha\)

(i) To find \(\theta\), first calculate the slant height of the cone using the Pythagorean theorem: \(l = \sqrt{6^2 + 8^2} = 10\) cm.

The circumference of the base of the cone is \(2\pi \times 6 = 12\pi\) cm.

The arc length of the sector is equal to the circumference of the base: \(10\theta = 12\pi\).

Solving for \(\theta\), we get \(\theta = \frac{12\pi}{10} = 1.2\pi\) or approximately 3.77 radians.

(ii) The area of the sector (paper) is given by \(\frac{1}{2} r^2 \theta\), where \(r = 10\) cm and \(\theta = 1.2\pi\).

Thus, the area is \(\frac{1}{2} \times 10^2 \times 1.2\pi = 60\pi\) cm\(^2\) or approximately 188.5 cm\(^2\).

(a) To find the perimeter of the shaded region, we first need to find the angle \\ \angle ADC \\ using trigonometry. We have:

\\ \tan BDC = \frac{4}{3} \\

Thus, \\ BDC = 0.927 \\ radians.

Then, \\ ADC = \pi - 0.927 = 2.214 \\ radians.

The arc length \\ AC \\ is given by:

\\ AC = 5 \times 2.214 = 11.07 \\ cm.

The length \\ AC \\ in the triangle is:

\\ AC = \sqrt{8^2 + 4^2} = 8.94 \\ cm.

Therefore, the perimeter of the shaded region is:

\\ 11.07 + 8.94 = 20.0 \\ cm.

(b) To find the area of the shaded region, we calculate the area of sector \\ ACD \\ and subtract the area of triangle \\ ADC \\ from it.

The area of sector \\ ACD \\ is:

\\ \frac{1}{2} \times 5^2 \times 2.214 = 27.7 \\ cm².

The area of triangle \\ ADC \\ is:

\\ \frac{1}{2} \times 5 \times 4 = 10 \\ cm².

Thus, the area of the shaded region is:

\\ 27.7 - 10 = 17.7 \\ cm².