A driver is transporting milk in a cylindrical tank. The diameter of the base is $180 \mathrm{~cm}$, and the length of the tank is $4 \mathrm{~m}$. How many hectoliters of milk are in the tank if it is filled to three-quarters of the diameter?

![](https://cdn.mathpix.com/cropped/2024_04_17_2a0500418bf8c778174dg-5.jpg?height=337&width=920&top_left_y=1051&top_left_x=608)

(M. Krejčová)
The part of the base that is below the milk level in the tank will be divided into a non-convex circular sector and an isosceles triangle $X Y S$.

![](https://cdn.mathpix.com/cropped/2024_04_17_2a0500418bf8c778174dg-5.jpg?height=414&width=406&top_left_y=1666&top_left_x=859)

The length of segment $S X$, as well as $S Z$, is equal to the radius of the base, i.e., $|S X|=|S Z|=r=90 \text{ cm}$. According to the problem, point $R$ is the midpoint of segment $S Z$, i.e., $|S R|=|R Z|=\frac{1}{2} r$. Therefore, (right-angled) triangles $S X R$ and $Z X R$ are congruent, so $|Z X|=|S X|$ and triangle $S X Z$ is equilateral. Thus, the measure of the interior angle $R S X$ is $60^{\circ}$, the measure of the interior angle $X S Y$ is $120^{\circ}$, and the measure of the exterior angle $X S Y$ is $240^{\circ}$. The non-convex circular sector therefore has $\frac{2}{3}$ of the area of the entire circle, i.e.,

$$
S_{v}=\frac{2}{3} \pi r^{2} \doteq 16965 \text{ cm}^{2} \doteq 170 \text{ dm}^{2}
$$

The area of triangle $X Y S$ is equal to $S_{t}=|R X| \cdot|R S|$. The length of segment $R S$ is $\frac{1}{2} r$ and the length of $|R X|$ can be expressed using the Pythagorean theorem in triangle $S X R:|R X|^{2}=$ $=|S X|^{2}-|R S|^{2}=r^{2}-\left(\frac{1}{2} r\right)^{2}=\frac{3}{4} r^{2}$, thus $|R X|=\frac{\sqrt{3}}{2} r$. After substitution, we get

$$
S_{t}=\frac{\sqrt{3}}{4} r^{2} \doteq 3507 \text{ cm}^{2} \doteq 35 \text{ dm}^{2}
$$

The part of the base that is below the milk level in the tank therefore has an area of

$$
S=S_{v}+S_{t} \doteq 205 \text{ dm}^{2} .
$$

The tank is $d=4 \text{ m}=40 \text{ dm}$ long, so the volume of the milk being transported is approximately equal to

$$
V=S \cdot d \doteq 8200 \text{ dm}^{3}=82 \text{ hl} .
$$