A straight, symmetrical railway embankment has a lower width of $a$, an upper width of $b$, and a height of $m$. A road is being built that runs perpendicular to the embankment's axis. The embankment is cut down to its base. The removed earth is bounded by equally sloping planes, and the width of the cut will be $c$ at the bottom and $d$ at the top. What is the volume of the earth to be excavated?
It is evident that $a > b$ and $c < d$. The cut $K$ is symmetric with respect to the vertical plane passing through the axis of the road to be constructed, because the bounding planes have equal slopes. Let us cut $K$ with two vertical planes passing through the boundaries of the road. The prism that falls between them has a cross-section identical to its loading, and the distance between its end faces is $c$, so its volume is

$$
K_{1} = \frac{(a + b) m}{2} \cdot c
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_e7ca230d08a527f75335g-1.jpg?height=537&width=1027&top_left_y=390&top_left_x=538)

The two remaining wedge-shaped, tapering downward parts are identical to each other. Their upper boundary is a rectangle with dimensions $b$ and $(d - c)/2$. Let us cut them with vertical planes passing through the upper edges of the loading. A prism is formed between the planes, with a cross-section that is a right triangle with legs $m$ and $(d - c)/2$, and side length $b$. On the outer sides of the planes, two identical three-sided pyramids are formed, with the previously mentioned right triangles as their bases and height (in the lying position) $(a - b)/2$. Thus, the volume of one wedge is

$$
K_{2} = \frac{m(d - c) b}{4} + 2 \frac{m(d - c)}{4} \cdot \frac{a - b}{6}
$$

The volume of the earth to be excavated is therefore

$$
\begin{aligned}
K = K_{1} + 2 K_{2} = & \frac{m}{2}\left[(a + b) c + (d - c) b + \frac{(d - c)(a - b)}{3}\right] = \\
& = \frac{m}{6}(2 a c + 2 b d + a d + b c)
\end{aligned}
$$

Lajos Szlobodnyik (Fót, Vocational Secondary School, Class III. o. t.)