3. A wooden block $\begin{aligned} & E F G H \\ & A B C D\end{aligned}$ is cut into 8 smaller blocks with three saw cuts. Each saw cut is parallel to one of the three pairs of opposite faces. Each pair of saw cuts is perpendicular to each other. The smaller blocks at the vertices $A, C$, $F$ and $H$ have volumes of 9, 12, 8, 24, respectively.

(The proportions in the picture are not accurate!!)

Calculate the volume of the entire block.

![](https://cdn.mathpix.com/cropped/2024_04_17_1d41116fda1ec097cf4ag-1.jpg?height=531&width=440&top_left_y=1593&top_left_x=1433)
3. Let the lengths of the edges of the large block be $x, y, z$ and the segments into which the edges are divided by the three cutting planes be $x_{1}, x_{2}$ and $y_{1}, y_{2}$ and $z_{1}, z_{2}$ with $x_{1}+x_{2}=x$ and $y_{1}+y_{2}=y$ and $z_{1}+z_{2}=z$. Then it holds that $x_{1} y_{1} z_{1}=9, x_{2} y_{2} z_{1}=12, x_{2} y_{1} z_{2}=8$ and $x_{1} y_{2} z_{2}=24$. Furthermore, it holds that:

$\frac{x_{1} y_{1} z_{1} \times x_{2} y_{2} z_{1} \times x_{2} y_{1} z_{2}}{x_{1} y_{2} z_{2}}=x_{2}^{2} y_{1}^{2} z_{1}^{2}$ so $x_{2}^{2} y_{1}^{2} z_{1}^{2}=\frac{9 \times 12 \times 8}{24}=36$,

thus $x_{2} y_{1} z_{1}=6$. Therefore, the volume of the small block at vertex $B$ is

![](https://cdn.mathpix.com/cropped/2024_04_17_29d173b1c3c8d37aa9abg-1.jpg?height=491&width=419&top_left_y=1819&top_left_x=1481)
equal to 6. By multiplying three of the given volumes and dividing the product by the fourth volume, you find the squares of the volumes of the four small blocks whose volumes are not given. This way, you find the volumes of the other small blocks at D, E, and G to be 18, 12, and 16, respectively. Adding all the volumes together gives the volume of the entire block: 105.