A clear plastic cone is 9 inches tall, with some liquid sealed inside. 
When the cone is held point down, the liquid is 8 inches deep. When
  the cone is inverted and held point up, the liquid is d inches deep.
  Find d, to the nearest hundredth of an inch.

However, I don't know if I should be finding the volume of the cone or not? I don't know how to find the height d, with the information that was given to me.
 A: Refer to the figure:
$\hspace{4cm}$
The volumes of the blue cone on the left and truncated blue cone on the right are equal:
$$\frac83\pi r_1^2=\frac93\pi R^2-\frac{9-d}{3}\pi r_2^2.$$
Also from the similarity of triangles we get:
$$\frac{r_1}{8}=\frac{r_2}{9-d}=\frac{R}{9}.$$
Can you interpret and continue yourself?
A: We don't know the radius of the cone.  But is that going to matter?
One way to do this is to assign a radius, and see if it drops away.
And to make numbers easier, lets assume the radius of the cone at the top of the cone is $9r$
At the surface of the water the radius is $8r.$ And, at any given height $hr$
The volume of a cone is $\frac 13$ the area of the base times the height.  The area of the base is $\pi (hr)^2$
The volume of the cone from vertex to $h$ is $\frac 13 \pi h^3 r^2$
Then the volume of water is $\frac 13 8^3 \pi r^2$
And inverting the cone the volume filled is the volume of the full cone less the volume of the gap at the top.
$\frac 13 9^3 \pi r^2 - \frac 13 h^3 \pi r^2$
$\frac 13 \pi 8^3 r^2 = \frac 13 \pi 9^3 r^2 - \frac 13 \pi h^3 r^2$
It looks like those r's are going to be dropping away...
$8^3 = 9^3 - h^3\\
h^3 = 9^3 - 8^3\\
h = \sqrt[3] {9^3-8^3}$
and in the figure $d = 9-h$
