(i) The gradient of AC is \(-\frac{1}{2}\). The perpendicular gradient is 2. The equation of line BX is \(y - 2 = 2(x - 2)\), which simplifies to \(y = 2x - 2\). Solving the equations \(2y + x = 16\) and \(y = 2x - 2\) simultaneously gives the coordinates of X as \((4, 6)\).

(ii) Since X is the midpoint of BD, and using the coordinates of X \((4, 6)\), we find D as \((6, 10)\).

(iii) Using the distance formula, \(AB = \sqrt{(14)^2 + (2)^2} = \sqrt{200}\) and \(BC = \sqrt{(2)^2 + (6)^2} = \sqrt{40}\). The perimeter is \(2\sqrt{200} + 2\sqrt{40} = 40.9\).