A body bounded by 6 planes has one face that is a square with a side of 1 meter, another face that is a rectangle 40 cm long and 20 cm wide, and four other faces that are trapezoids, the sum of the lengths of whose legs is 130 cm. What is the volume of the body?
A test $A B C D=N$ square plate has no common edge with the rectangular $E F G H=T$ plate $(E F=40 \mathrm{~cm}, E H=$ $20 \mathrm{~cm}$ ), because the lengths of their sides are different. For the same reason, the trapezoidal plates can only connect to $N$ and $T$ with their parallel sides, and to each other along their legs. Thus, the planes of $T$ and $N$ are parallel, because they contain two pairs of parallel edges in different directions: $A B \| E F$ and $B C \| F G$ (Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_36b1d1d8203b11b2efe3g-1.jpg?height=840&width=756&top_left_y=380&top_left_x=663)

Figure 1

Due to the equality of the legs, the trapezoidal plates are axially symmetric. Thus, the perpendicular bisector plane of the $E F$ edge is a common symmetry plane of the $T$, $E F B A, N$, and $C D H G$ plates, and therefore of the entire body, and the same applies to the perpendicular bisector plane of the $F G$ edge.

Cut the body into three parts with planes passing through the $E H, F G$ edges and perpendicular to the $N$ plane, so the middle part is a prism with a trapezoidal base (lying on one of its side faces). Cut the other two parts with planes passing through the $E F, G H$ lines and perpendicular to the $N$ plane, so the middle parts are (lying) triangular prisms, and on the edges, four congruent quadrilateral pyramids are formed.

Let the projection of $F$ onto the $N$ plane be $F^{\prime}$, and the point on the $A B$ edge be $F_{1}$, so the $A B F E$ plate, and then the height of the body are

$$
\begin{aligned}
& F F_{1}=\sqrt{F B^{2}-\left(\frac{A B-E F}{2}\right)^{2}} \\
& F F^{\prime}=\sqrt{F F_{1}^{2}-\left(\frac{B C-F G}{2}\right)^{2}}=120 \mathrm{~cm}
\end{aligned}
$$

The volume of the prism with a trapezoidal base is

$$
\frac{B C+F G}{2} \cdot F F^{\prime} \cdot E F=288 \mathrm{dm}^{3}
$$

The combined volume of the two triangular prisms is

$$
2 \cdot \frac{A B-E F}{4} \cdot F F^{\prime} \cdot F G=72 \mathrm{dm}^{3}
$$

The combined volume of the four pyramids is

$$
\frac{4}{3} \cdot \frac{A B-E F}{2} \cdot \frac{B C-F G}{2} \cdot F F^{\prime}=192 \mathrm{dm}^{3}
$$

Based on these, the volume of the body is: $V=552 \mathrm{dm}^{3}$.

Zoltán Ábrahám (Nagykőrös, Arany J. Gymnasium, 2nd grade)

Remark. The body is not a truncated pyramid, because the intersection points of the $B F, C G$, and $A E, D H$ edge pairs are different. We call a convex body with planar faces, all of whose vertices lie in two parallel planes, a prizmoid (each of its side faces is a triangle, trapezoid, or parallelogram). It can be shown that the volume of every such body is

$$
V=m \cdot \frac{A_{1}+4 K+A_{2}}{6}
$$

where $A_{1}$ and $A_{2}$ are the areas of the parallel planes, $m$ is the distance between the two planes (height), and $K$ is the area of the section cut by the perpendicular bisector plane of the height. In our case, $K$ is a $7 \times 6 \mathrm{dm}$ rectangle (Figure 2).

![](https://cdn.mathpix.com/cropped/2024_05_02_36b1d1d8203b11b2efe3g-2.jpg?height=556&width=501&top_left_y=329&top_left_x=801)

[^0]:    ${ }^{1}$ See, for example, György Pólya: The School of Problem Solving, Volume I, Tankönyvkiadó, Budapest, 1967, pp. 123-124, problems 4.22-4.29.