A $12 \times 12 \times 35 \mathrm{~cm}$ foam block, weighing $5 \mathrm{~kg}$, was cut diagonally such that the parallelogram-shaped cross-section has side lengths of $15$ and $20 \mathrm{~cm}$. What could be the mass of the two resulting pieces?
Solution. Let the vertices of one of the square faces of the column be denoted by $A, B, C$, and $D$. The volume of the cut-off part will be the smallest if we start the cut at one of the vertices, let this be $B$. Let the vertices of the parallelogram be $B$, $E, F$, and $G$ (as shown in the figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_027059a7ca6679f6d8e7g-1.jpg?height=273&width=471&top_left_y=308&top_left_x=816)

Draw a line through $E$ parallel to the edge $CD$, intersecting the opposite edge at point $M$, so $EC = MD$. From the right triangle $ECB$, we have $EC = \sqrt{15^2 - 12^2} = 9$. From the triangle $EMF$, we have $FM = \sqrt{20^2 - 12^2} = 16$, and therefore $FD = 25$.

The remaining part after the cut consists of a square frustum and a piece identical to the one cut off. The edges of the frustum are 10 and $12 \, \text{cm}$ long. The two identical pieces together form a $25 \times 12 \times 12$ frustum; the volume of the cut-off part is half of this, i.e., $1800 \, \text{cm}^3$.

We know that the volume of the column is $5040 \, \text{cm}^3$ and its mass is $5 \, \text{kg}$. The mass of the cut-off $1800 \, \text{cm}^3$ part is therefore

$$
x = \frac{1800 \cdot 5}{5040} \approx 1.7857 \, \text{kg}
$$

The mass of the cut-off piece can vary between $1.7857 \, \text{kg}$ and $3.2143 \, \text{kg}$, depending on where we start the cut.