Example

A large pond in the shape of a hemisphere has a radius of 5 m and it is planned to fill the pond so that the water in the centre of the pond is 2 m deep. The pond's builders then want to attach a metal ring around the waterline which will be used to hold a cover for the pond in the winter months.

They want to know how long to make the metal ring, correct to the nearest mm, and what area of cover material will be required in square metres correct to two decimal places.

Solution

To find the length of the waterline and the area of the surface, we need to know the radius of the small circle as shown with the dotted line.

If ${AD}$ represents the depth of water in the pond,
then the air space above.

$\begin{aligned}[t] (OA)&= OD - AD\\ &= 5 - 2 \\ &= 3 \end{aligned}$

${OB}$ is 5 m since that is the radius of the hemisphere.

We can find the radius of the small circle using Pythagoras' theorem.

${{AB} = \sqrt{5^2 ? 3^2} = \sqrt{25 ? 9} = \sqrt{16} = 4}$

We find the length of the waterline (and hence the length of the metal ring) using ${C = 2\pi r}$.

${{C} = 2 \times \pi \times 4 = 8 \pi = 25.13274\ldots \approx 25.133 m}$

The metal ring will need to be 25.133 m in circumference.

The area of the water surface can be found using ${A = \pi r^2}$

${{A} = \pi \times 4^2 = 16 \pi = 50.2654\ldots \approx 50.27 m^2}$

The area of the cover will be 50.27 square metres.