Notice the volume of water in the container also takes the shape of an inverted right pyramid. It is geometrically similar to the container.

Let volume be V = ⅓a²h, where a is the side of the square base and h is the height of the pyramid (i.e depth of water)

Now, for the container, a/h = 18cm/30cm
= 3/5

Cross multiply, 5a = 3h
a = 3/5 h

So V = ⅓(3/5 h)²h = ⅓(9/25 h²)h
= 3/25 h³

(The a/h ratio will be the same for the water since the shape is geometrically similar to the container)


Differentiate with respect to h,
dV/dh = 9/25 h²



dV/dt = 5cm³/s

what you want to find is the rate of change of water level over time (dh/dt)

So,

dh/dt = dV/dt x dh/dV

= dV/dt ÷ dV/dh

= 5cm³/s ÷ (9/25 h²)

= 5cm³/s ÷ (9/25 x (10cm)²)

= 5cm³/s ÷ 36cm²

= 5/36 cm/s