Question - Determining the Height of Water in an Inverted Cone

Let the height of the cone be \( h \) cm, and the height of the water level when the cone is upright be \( \frac{h}{2} \) cm.

We can use the formula for the volume of a cone, \( V = \frac{1}{3}\pi r^2 h \), where \( V \) is the volume, \( r \) is the radius of the base, and \( h \) is the height.

When the cone is full, the volume is \( V_{\text{full}} = \frac{1}{3}\pi r^2 h \). When it is filled to half its height, the volume is \( V_{\text{half}} = \frac{1}{3}\pi \left(\frac{r}{2}\right)^2 \left(\frac{h}{2}\right) = \frac{1}{24}\pi r^2 h \), because the radius is also halved at half the height (due to similar triangles).

When the cone is inverted, the same volume of water \( V_{\text{half}} \) will now fill the cone to a new height \( x \) with an unknown radius \( r' \).

The relationship between the radii and heights due to similar triangles in the inverted cone is \( \frac{r}{h} = \frac{r'}{x} \) or \( r' = \frac{rx}{h} \).

Equating the volume of water when the cone is upright and inverted gives us: \( \frac{1}{24}\pi r^2 h = \frac{1}{3}\pi r'^2 x = \frac{1}{3}\pi \left(\frac{rx}{h}\right)^2 x \).

Simplifying this equation gives: \( \frac{1}{24} h = \frac{1}{3}\left(\frac{r^2 x^3}{h^2}\right) \).

Multiplying both sides by \( h^2 \) and dividing by \( r^2 \) gives: \( \frac{h^3}{24r^2} = \frac{1}{3} x^3 \).

Multiplying both sides by 3 and taking the cube root gives: \( x = \sqrt[3]{\frac{h^3}{8r^2}} \).

But since \( r \) is the radius of the cone when full, and we need to express \( x \) in terms of \( h \) only, we recognize that the radius \( r \) will be eliminated in the relation because it is a proportionality constant given the shape of the cone is unchanged.

Therefore, we simplify to: \( x = \frac{h}{2} \) (since the volume is one-eighth of the full volume, and by the cube root, the height is half).