(i) Since OBX is a right angle (90°), we use the cosine rule: \(\cos \theta = \frac{r}{2r} = \frac{1}{2}\). Therefore, \(\theta = \frac{1}{3}\pi\) radians.

(ii) The arc length AB is \(\frac{1}{3}r\pi\). The length BX is \(r\tan(\frac{1}{3}\pi) = r\sqrt{3}\). Thus, the perimeter P is \(r + \frac{1}{3}r\pi + r\sqrt{3}\).

(iii) The area of the shaded region is the area of triangle OBX minus the area of sector AOB. The area of triangle OBX is \(\frac{1}{2}r^2\sqrt{3}\) and the area of sector AOB is \(\frac{1}{6}r^2\pi\). Therefore, the area of the shaded region is \(\frac{1}{2}r^2\sqrt{3} - \frac{1}{6}r^2\pi\).