Determine depth of a partially filled hemisphere

Question

Recently came across a question in a Year 9 math book of which there was no "working out" supplied and offers now description on how they obtained the answer.

The question goes like this:

A bowl is in the shape of a hemisphere with radius 10cm. The surface of the water in the container has a radius of 7cm. How deeps is the water?

The supplied answer is 2.86 cm but I am lost for how they go it. Does anyone have any pointers or an answer?

Answer 1

Draw a vertical cross-section of the bowl, passing through the centre $O$ of the hemisphere.

Let $N$ be the nearest point to $O$ on the surface of the water, and let $F$ be the point on the surface of the water which is furthest from $O$.

Then $NOF$ is a right triangle, with right angle at $N$.

By the Pythagorean Theorem, we have $(ON)^2+7^2=10^2$, so $ON=\sqrt{51}\approx 7.14$. The depth of the water is therefore $10-\sqrt{51}$, approximately $2.85857$.

Answer 2

The equation for the sphere is $\rho^2+z^2=R^2$. Here, you are given $\rho=7$ and $R=10$, so that

$$z=\sqrt{10^2-7^2} = \sqrt{51}$$

and the depth in the bowl is $R-z = 10-\sqrt{51} \approx 2.85857$ cm.