The base areas of two geometrically similar cones are in the ratio of 9 : 49.

1. Find the linear ratio from the base area ratio.

Base areas of smaller cone : bigger cone $= 9:49$.

Since area ratio $=$ square of linear ratio,
$\text{linear ratio} = \sqrt{9}:\sqrt{49} = 3:7.$

2. Use the cube of the linear ratio to find the smaller cone's volume.

Volume ratio of smaller cone : bigger cone
$= 3^3:7^3 = 27:343.$

Given volume of bigger cone $=1200\text{ cm}^3$,
$\text{Volume of smaller cone} = 1200 \times \frac{27}{343} = 94.46\ldots$
$\approx 94.4\text{ cm}^3 \quad \text{to 3 s.f.}$

1. Find the full height of the smaller cone using its volume and radius.

From part (a), the volume of the smaller cone is $94.4\text{ cm}^3$.

For the smaller cone,
$V=\frac{1}{3}\pi r^2H$
$94.4=\frac{1}{3}\pi(3)^2H$
$94.4=3\pi H$
$H=\frac{94.4}{3\pi}$

2. Use the volume ratio to find the height of the water cone.

The water occupies $80\%$ of the smaller cone's volume.

Since the water cone is similar to the smaller cone,
$\left(\frac{h}{H}\right)^3=0.8$
$\frac{h}{H}=\sqrt[3]{0.8}$
$h=\frac{94.4}{3\pi}\times \sqrt[3]{0.8}$
$h=9.30\text{ cm} \quad \text{(3 s.f.)}$