(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