Task B-4.6. A container for liquid has the shape of a right circular cone with a height of $12 \mathrm{~cm}$ and a base radius of $5 \mathrm{~cm}$. If we pour liquid into the container and hold it (closed) so that its vertex is pointing downwards and the base is in a horizontal position, the depth of the poured liquid is $9 \mathrm{~cm}$. How many $\mathrm{cm}^{3}$ of liquid need to be added to the container so that the depth of the liquid is $9 \mathrm{~cm}$ and in the case where the container is inverted with the vertex pointing upwards (and the base in a horizontal position)?
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_9c6c142e457d4763ad17g-18.jpg?height=684&width=880&top_left_y=1801&top_left_x=596)

Let the vertex of the cone be pointing downwards as shown in the left cone in the image. The radius of the base of the cone and the radius $x$ (at a height of $9 \mathrm{~cm}$) are in the same ratio as the depth of the liquid and the height of the cone,

$$
k=\frac{x}{5}=\frac{9}{12}=\frac{3}{4}
$$

Then the volume of the liquid $V_{1}$ and the volume of the entire cone $V$ are in the ratio

$$
\frac{V_{1}}{V}=k^{3}=\left(\frac{3}{4}\right)^{3}=\frac{27}{64}
$$

It follows that the volume of the poured liquid is

$$
V_{1}=\frac{27}{64} V
$$

and the volume of the part of the cone without liquid is

$$
V_{2}=V-\frac{27}{64} V=\frac{37}{64} V
$$

Now consider the case when the vertex of the cone is pointing upwards as shown in the right cone in the image. The liquid we poured at the beginning occupies a volume $V_{1}=\frac{27}{64} V$, the volume of the liquid that needs to be added to reach a height of $9 \mathrm{~cm}$ will be denoted by $V_{t}$, and the volume of the part of the cone that is without liquid by $V_{3}$. As in the first case, we conclude that

$$
\frac{V_{3}}{V}=\left(\frac{y}{5}\right)^{3}=\left(\frac{3}{12}\right)^{3}=\frac{1}{64}
$$

so it follows that

$$
V_{3}=\frac{1}{64} V
$$

The desired volume can be calculated from

$$
V_{t}=V_{2}-V_{3}=\frac{36}{64} V=\frac{9}{16} V\left(\text { or } V_{t}=V-V_{1}-V_{3}\right)
$$

Since the volume of the given cone is

$$
V=\frac{r^{2} \pi}{3} \cdot h=100 \pi \mathrm{cm}^{3}
$$

it follows that

$$
V_{t}=\frac{225 \pi}{4} \mathrm{~cm}^{3}
$$