In the rectangular prism $A B C D E F G H$ - with the usual labeling - $A E=1, A D=2, A B=3$. What is the volume of the solid whose vertices are $A$ and $C$, as well as the midpoints of the edges of the $E F G H$ face?
Solution. Let the midpoints of the edges $E F, F G, G H$, and $H E$ of the rectangle $E F G H$ be $K, L, M$, and $N$ respectively. Instead of calculating the volume of the solid $A C K L M N=\Theta$, we will calculate the sum of the volumes of the parts obtained by removing the solid $\Theta$ from the rectangular prism.

![](https://cdn.mathpix.com/cropped/2024_05_02_bc41d159f6466878f00ag-1.jpg?height=331&width=553&top_left_y=309&top_left_x=775)

The remaining part consists of two congruent pyramids and two congruent frustums.

One of the pyramids is $A E K N$, with a right-angled triangle as its base, and its height is equal to the edge $A E$ of the rectangular prism. The other pyramid is the congruent $C G L M$. The congruence of the pyramids follows from the symmetry. By connecting the midpoints of the rectangle's sides, we get a rhombus. $\triangle E K N \cong \triangle G L M$. The area of $\triangle E K N$ is one-eighth of the area of the rectangle, which is $\frac{3}{4}$ square units.

![](https://cdn.mathpix.com/cropped/2024_05_02_bc41d159f6466878f00ag-1.jpg?height=337&width=417&top_left_y=978&top_left_x=843)

One of the frustums is the solid $A D C N H M$, with bases that are right-angled triangles, whose planes are parallel because they lie in opposite faces of the rectangular prism.

If we subtract the sum of the volumes of these pyramids (frustums) from the volume of the rectangular prism, $2 \times 3 \times 1=6$ cubic units, we get the volume of the desired solid. The sum of the volumes of the pyramids is:

$$
2 \cdot \frac{\frac{3}{4} \cdot 1}{3}=\frac{1}{2}
$$

The sum of the volumes of the frustums, using the known volume formula, is:

$$
2 \cdot \frac{1}{3} \cdot\left(\frac{3}{4}+3+\sqrt{\frac{9}{4}}\right)=2 \cdot \frac{7}{4}
$$

The volume of the desired solid is: $V_{\Theta}=6-\frac{1}{2}-\frac{7}{2}=2$ cubic units.