493. The height of a truncated cone is 3 meters, and the radii of its bases are 1 and 2 meters. Divide the volume into three parts, proportional to the numbers 2, 3, and 7, by two planes parallel to the bases.

## Problems from the course "Die Elemente der Mathematik" by R. Baltzer.
493. Let the volumes of the sought parts be $v_{1}, v_{2}, v_{3}$, so that

$$
v_{1}: v_{2}: v_{3}=2: 3: 7
$$

The total volume $v=7 \pi$, hence,

$$
v_{1}=\frac{7}{6} \pi ; v_{2}=\frac{7}{4} \pi ; v_{3}=\frac{49}{12} \pi
$$

Triangles $A C B$ and $D E B$ (Fig. 114) are similar, therefore:

![](https://cdn.mathpix.com/cropped/2024_05_21_926f0bb262c8f9569516g-212.jpg?height=300&width=468&top_left_y=1706&top_left_x=1225)

Fig. 114.

$$
A C: B C=D E: B E
$$

or

$$
1: 3=\left(r_{1}-1\right): h_{1}
$$

where $r_{1}=D K$ and $h_{1}=N K$. Hence, $h_{1}=3\left(r_{1}-1\right)$.

$14^{*}$

The volume

$$
v_{1}=\frac{7}{6} \pi=\frac{\pi h_{1}}{3}\left(r_{1}^{2}+1+r_{1}\right)
$$

from which

$$
\frac{7}{2}=h_{1}\left(r_{1}^{2}+1+r_{1}\right)=3\left(r_{1}-1\right)\left(r_{1}^{2}+1+r_{1}\right),
$$

or

$$
\frac{7}{6}=r_{1}^{3}-1
$$

and

$$
r_{1}=\sqrt[3]{\frac{13}{6}} ; h_{1}=3\left(\sqrt[3]{\frac{13}{6}}-1\right)
$$

Similarly, we find $r_{2}$ and $h_{2}$.

Baltzer, Richard, a German mathematician and educator, whose course "Die Elemente der Mathematik" enjoys enormous popularity in Germany.