A cube-shaped container with an edge length of $12 \mathrm{~cm}$ was filled to $\frac{5}{8}$ of its capacity with a liquid, and then it was slightly tilted along one of its edges. The diagram shows the cross-section of the container with the horizontal level of the liquid inside. We know that the length of segment $L C$ is exactly twice the length of segment $K B$. Determine the length of segment $L C$.

![](https://cdn.mathpix.com/cropped/2024_05_02_2ef57db81a3ef472104fg-1.jpg?height=265&width=271&top_left_y=331&top_left_x=922)
Solution. Before the impact, the fluid level was at $\frac{5}{8}$ of the 12 cm height, i.e., at $7.5 \mathrm{~cm}$. The area of this rectangle's cross-section is $12 \cdot 7.5=90 \mathrm{~cm}^{2}$. After the impact, the area of the trapezoid $B C L K$ will be $90 \mathrm{~cm}^{2}$. Let $L C=2 x$, then according to the text, $K B=x$, so we can write the area of the trapezoid as:

$$
\frac{(2 x+x) \cdot 12}{2}=90
$$

from which $x=5$. The length of the segment $L C$ is $10 \mathrm{~cm}$.