(i) In triangle OAB, using the cosine rule, we have:

\(AB = 2r \cos x\)

(ii) The area of the semicircle is \(\frac{1}{2} \pi r^2\). The area of the sector with angle x is \(\frac{1}{2} (2r \cos x)^2 x\).

Equating half the area of the semicircle to the area of the sector:

\(\frac{1}{2} \pi r^2 = \frac{1}{2} (2r \cos x)^2 x\)

\(\pi r^2 = 4r^2 \cos^2 x \cdot x\)

\(\cos^2 x = \frac{\pi}{16x}\)

\(x = \cos^{-1}\left(\sqrt{\frac{\pi}{16x}}\right)\)

(iii) Calculate values at \(x = 1\) and \(x = 1.5\):

For \(x = 1\), \(f(1) = 1.11\)

For \(x = 1.5\), \(f(1.5) = 1.20\)

Thus, \(x\) lies between 1 and 1.5.

(iv) Using the iterative formula:

\(x_{n+1} = \cos^{-1}\left(\sqrt{\frac{\pi}{16x_n}}\right)\)

Iterations: 1.11173, 1.13707, 1.14225, 1.14329, 1.14349, 1.14354, 1.14354

Final answer: 1.144

(i) The area of the semicircle with diameter AB is \(\frac{\pi a^2}{8}\) and similarly for AC. The total area of the semicircles is \(\frac{\pi a^2}{4}\).

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

Equating the areas: \(\frac{1}{2}a^2\theta - \frac{1}{2}a^2\sin \theta = \frac{\pi a^2}{4}\).

Dividing through by \(\frac{1}{2}a^2\): \(\theta - \sin \theta = \frac{\pi}{2}\).

Rearranging gives \(\theta = \frac{1}{2}\pi + \sin \theta\).

(ii) Calculate \(f(\theta) = \frac{\pi}{2} + \sin \theta\) at \(\theta = 2.2\) and \(\theta = 2.4\):

\(f(2.2) = 2.3793\) and \(f(2.4) = 2.2463\).

Since \(2.2 < f(2.2)\) and \(f(2.4) < 2.4\), \(\theta\) lies between 2.2 and 2.4.

(iii) Using the iterative formula \(\theta_{n+1} = \frac{1}{2}\pi + \sin \theta_n\):

Start with \(\theta_0 = 2.2\):

\(\theta_1 = 2.3793\)

\(\theta_2 = 2.2614\)

\(\theta_3 = 2.3417\)

\(\theta_4 = 2.2881\)

\(\theta_5 = 2.3244\)

\(\theta_6 = 2.3000\)

\(\theta_7 = 2.3165\)

\(\theta_8 = 2.3054\)

\(\theta_9 = 2.3129\)

\(\theta_{10} = 2.3072\)

The value converges to \(\theta = 2.31\) to 2 decimal places.

(i) The area of the segment on AP is given by \(\frac{1}{2} r^2 (x - \sin x)\) and the area of the segment on BP is \(\frac{1}{2} r^2 (\pi - x + \sin x)\). Given that the area of the segment on AP is half of the area of the segment on BP, we have:

\(\frac{1}{2} r^2 (x - \sin x) = \frac{1}{2} \times \frac{1}{2} r^2 (\pi - x + \sin x)\)

\(x - \sin x = \frac{1}{2} (\pi - x + \sin x)\)

\(2(x - \sin x) = \pi - x + \sin x\)

\(3x = \pi + \sin x\)

\(x = \frac{1}{3}(\pi + \sin x)\)

(ii) To verify that x lies between 1 and 1.5, calculate:

For \(x = 1\), \(\frac{1}{3}(\pi + \sin 1) \approx 1.380\)

For \(x = 1.5\), \(\frac{1}{3}(\pi + \sin 1.5) \approx 1.497\)

Since \(1.380 < 1.5\) and \(1.497 > 1\), x lies between 1 and 1.5.

(iii) Using the iterative formula \(x_{n+1} = \frac{1}{3}(\pi + \sin x_n)\):

Start with \(x_0 = 1\).

\(x_1 = \frac{1}{3}(\pi + \sin 1) = 1.38052\)

\(x_2 = \frac{1}{3}(\pi + \sin 1.38052) = 1.37492\)

\(x_3 = \frac{1}{3}(\pi + \sin 1.37492) = 1.37417\)

\(x_4 = \frac{1}{3}(\pi + \sin 1.37417) = 1.37405\)

\(x_5 = \frac{1}{3}(\pi + \sin 1.37405) = 1.37403\)

The value of x correct to 3 decimal places is 1.374.

(i) The tangents AT and BT are equal in length, and each is equal to \(r \tan x\). The arc length of AB is \(2xr\). The perimeter of the shaded region is \(2r \tan x + 2xr\). This is equal to the circumference of the circle, \(2\pi r\). Therefore, \(2r \tan x + 2xr = 2\pi r\). Dividing by \(2r\), we get \(\tan x + x = \pi\), which rearranges to \(\tan x = \pi - x\).

(ii) Calculate \(\tan 1 \approx 1.5574\) and \(\tan 1.3 \approx 3.6021\). For \(x = 1\), \(\tan 1 - (\pi - 1) \approx 1.5574 - 2.1416 = -0.5842\). For \(x = 1.3\), \(\tan 1.3 - (\pi - 1.3) \approx 3.6021 - 1.8416 = 1.7605\). The sign change indicates a root between 1 and 1.3.

(iii) Using the iterative formula \(x_{n+1} = \arctan(\pi - x_n)\):

Start with \(x_0 = 1\).

\(x_1 = \arctan(\pi - 1) \approx \arctan(2.1416) \approx 1.1366\).

\(x_2 = \arctan(\pi - 1.1366) \approx \arctan(2.0044) \approx 1.1071\).

\(x_3 = \arctan(\pi - 1.1071) \approx \arctan(2.0345) \approx 1.1102\).

\(x_4 = \arctan(\pi - 1.1102) \approx \arctan(2.0314) \approx 1.1098\).

The root correct to 2 decimal places is 1.11.

(i) The perimeter of the shaded region is given by the arc length \(r \cdot x\) plus the length of the arc \(2r \cdot \theta\), where \(\theta = \frac{\pi}{2} - x\). The total perimeter is \(r \cdot x + 2r \cdot \left( \frac{\pi}{2} - x \right) = \frac{1}{2} \cdot 2\pi r\). Simplifying gives \(x = \cos^{-1} \left( \frac{\pi}{4 + 4x} \right)\).

(ii) Calculate \(\cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1} \right) \approx 1.318\) and \(\cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.5} \right) \approx 1.047\). Since \(1.047 < x < 1.318\), x lies between 1 and 1.5.

(iii) Using the iterative formula:

\(x_1 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1} \right) = 1.3183\)

\(x_2 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.3183} \right) = 1.2146\)

\(x_3 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.2146} \right) = 1.2099\)

\(x_4 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.2099} \right) = 1.2109\)

\(x_5 = \cos^{-1} \left( \frac{\pi}{4 + 4 \cdot 1.2109} \right) = 1.2106\)

The value of x correct to 2 decimal places is 1.21.