(i) The base of the pyramid is a square with side \(\frac{1}{2}h\), so the area \(A\) of the base is \(\left(\frac{1}{2}h\right)^2 = \frac{1}{4}h^2\).

The volume \(V\) of the pyramid is given by \(V = \frac{1}{3}Ah\).

Substituting for \(A\), we have \(V = \frac{1}{3} \times \frac{1}{4}h^2 \times h = \frac{1}{12}h^3\).

(ii) We know \(\frac{dV}{dt} = 20 \text{ cm}^3/\text{min}\).

From (i), \(V = \frac{1}{12}h^3\), so \(\frac{dV}{dh} = \frac{1}{4}h^2\).

Using the chain rule, \(\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt} = \frac{4}{h^2} \times 20\).

When \(h = 10\), \(\frac{dh}{dt} = \frac{4}{10^2} \times 20 = 0.8 \text{ cm/min}\).