Maria filled a rectangular parallelepiped box with $160 \mathrm{ml}$ of water and placed it on one of its faces, as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_05_01_d1e5bbfbdffc7558a598g-16.jpg?height=428&width=719&top_left_y=505&top_left_x=592)

Maria then measured the height the water reached and obtained $2 \mathrm{~cm}$. After that, she repeated the experiment by placing the box on other faces and obtained heights of $4 \mathrm{~cm}$ and $5 \mathrm{~cm}$. What are the dimensions (width, height, and length) of the box?
Solution

Let $x, y$ and $z$ be the dimensions of the boxes in centimeters, as shown in the figure below. First, we remember that 160 milliliters are $160 \mathrm{~cm}^{3}$. Then, if we place the box on the face with dimensions $y$ and $z$, the water reaches a height of $2 \mathrm{~cm}$, as shown in the figure below:

![](https://cdn.mathpix.com/cropped/2024_01_01_d1e5bbfbdffc7558a598g-44.jpg?height=560&width=912&top_left_y=1684&top_left_x=675)

The region occupied by the water forms a parallelepiped with dimensions $y, z$ and $2 \mathrm{~cm}$, $\log 0$

$$
\begin{array}{cc}
y \times z \times 2 & =160 \\
y \times z & =80
\end{array}
$$

Similarly, when we place the box on the face with dimensions $x$ and $z$, we get

$$
\begin{aligned}
x \times z \times 4 & =160 \\
x \times z & =40
\end{aligned}
$$

And finally, we have

$$
\begin{gathered}
x \times y \times 5=160 \\
x \times y=32 .
\end{gathered}
$$

Now we multiply the first two equations, and we get

$$
x \times y \times z^{2}=3200
$$

Since $x \times y=32$, then

$$
32 z^{2}=3200
$$

Thus $z^{2}=100$ and, therefore $z=10 \mathrm{~cm}$. Now, knowing the value of $z$, we can find the values of $x$ and $y$. From the first equation, we have $10 y=80$, so $y=8 \mathrm{~cm}$, and from the second equation, we have $10 x=40$, so $x=4 \mathrm{~cm}$. Therefore, the dimensions of the box are $4 \mathrm{~cm}, 8 \mathrm{~cm}$ and $10 \mathrm{~cm}$.