A vessel in the shape of an inverted right pyramid has a square base of side 12 cm and a height of 30 cm. Water is leaking from the vessel at a constant rate of 5 cm$^3$/s.

1. Find the side length of the square water surface in terms of $h$ using similar triangles.

Let the side length of the square water surface be $x$ cm.
By similar triangles,
$\frac{x}{12} = \frac{h}{30}$
$x = \frac{12h}{30} = \frac{2h}{5}$

2. Express the volume of the water in terms of $h$ and simplify.

The volume of the water, $V$, is given by the volume of a pyramid:
$V = \frac{1}{3} \times \text{base area} \times \text{height}$
$V = \frac{1}{3} x^2 h$
Substitute $x = \frac{2h}{5}$:
$V = \frac{1}{3} \left(\frac{2h}{5}\right)^2 h$
$V = \frac{1}{3} \left(\frac{4h^2}{25}\right) h$
$V = \frac{4h^3}{75} \text{ (Shown)}$

1. Identify the given rate of change and the required rate of change.

Given $\frac{dV}{dt} = -5$ cm$^3$/s (since water is leaking).
We need to find $\frac{dh}{dt}$ when $h = 6$.

2. Differentiate the volume formula with respect to $h$.

From (i), $V = \frac{4h^3}{75}$.
$\frac{dV}{dh} = \frac{d}{dh}\left(\frac{4h^3}{75}\right) = \frac{12h^2}{75} = \frac{4h^2}{25}$

3. Apply the chain rule to find $\frac{dh}{dt}$ when $h = 6$.

By the chain rule, $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt}$.
$-5 = \frac{4(6)^2}{25} \times \frac{dh}{dt}$
$-5 = \frac{144}{25} \times \frac{dh}{dt}$
$\frac{dh}{dt} = -5 \times \frac{25}{144} = -\frac{125}{144}$ cm/s