A cuboid with dimensions $20 \mathrm{~cm} \times 30 \mathrm{~cm} \times 40 \mathrm{~cm}$ is placed such that the edge of length $20 \mathrm{~cm}$ lies on the table and the edge of length $40 \mathrm{~cm}$ forms an angle of $30^{\circ}$ with the table. The cuboid is partially filled with water, which wets one quarter of the top face with dimensions $20 \mathrm{~cm} \times 40 \mathrm{~cm}$.

Determine the volume of water in the cuboid.

(M. Krejčová)

![](https://cdn.mathpix.com/cropped/2024_04_17_f2fcb56db57152892a4ag-5.jpg?height=579&width=1082&top_left_y=453&top_left_x=493)
The volume of water in the cuboid can be determined as the difference between the volume of the cuboid and the volume of its empty part. The volume of the cuboid is $2 \cdot 3 \cdot 4 \mathrm{dm}^{3}$, i.e., 24 liters. The empty part of the cuboid forms a triangular prism, whose base is a right triangle and whose height measures $2 \mathrm{dm}$. It is sufficient to determine the area of the right triangle $A B H$:

![](https://cdn.mathpix.com/cropped/2024_04_17_f2fcb56db57152892a4ag-5.jpg?height=534&width=625&top_left_y=1338&top_left_x=721)

The upper face of the cuboid is wetted by water from one quarter. This means that the edge $D H$ of length $4 \mathrm{dm}$ is divided by the water level at point $A$, which is $1 \mathrm{dm}$ from point $D$ and $3 \mathrm{dm}$ from point $H$. The water level is parallel to the table surface, and the opposite edges of the cuboid are parallel to each other, so the angle $B A H$ is $30^{\circ}$. The right angle is at vertex $H$, leaving the angle at vertex $B$ to be $60^{\circ}$.

Triangle $A B H$ can be understood as half of an equilateral triangle divided by the height $A H$. The length of side $B H$ is therefore half the length of side $A B$. If we denote the length of side $B H$ in $\mathrm{dm}$ by $x$, then according to the Pythagorean theorem in triangle $A B H$ we have

$$
3^{2}+x^{2}=(2 x)^{2}
$$

From this, it follows that $3 x^{2}=9$ or $x=\sqrt{3}$. The area of the right triangle $A B H$ is thus $\frac{3 \sqrt{3}}{2} \mathrm{dm}^{2}$.

The volume of the triangular prism representing the empty part of the cuboid is $2 \cdot \frac{3 \sqrt{3}}{2}=3 \sqrt{3} \mathrm{dm}^{3}$, which is approximately 5.2 liters. The volume of water in the cuboid is $24-3 \sqrt{3} \mathrm{dm}^{3}$, which is approximately 18.8 liters.

Evaluation. 2 points for the initial consideration and the length of segment $A H$, or $A D$; 3 points for the length of segment $B H$ and the area of triangle $A B H$; 1 point for the calculation and possible rounding.