A tunnel with a circular cross-section of radius $r=2.3 \, \text{m}$ needs to be lined with a concrete layer to balance it. The surface of the concrete layer is horizontal, and its thickness, measured vertically (i.e., the height of the circular segment cross-section of the concrete layer) is $0.7 \, \text{m}$. How many sacks of cement are needed to concrete a $1 \, \text{km}$ long tunnel, if the finished concrete contains $150 \, \text{kg}$ of cement per cubic meter?
The diagram shows the cross-section of the tunnel and the reinforcement. The cross-section of the concrete layer is the segment of a circle that is cut off.

![](https://cdn.mathpix.com/cropped/2024_05_02_4df36f0a3f0b1d8f1bc1g-1.jpg?height=364&width=355&top_left_y=192&top_left_x=880)

The volume of the concrete layer is obtained by multiplying the area of the segment by the length of the tunnel. Denoting the area of the sector $O \widehat{A B}$ by $t_{c}$ and the area of the triangle $O A B$ by $t_{\Delta}$, the required weight of cement, in cubic meters and quintals, is

$$
\begin{gathered}
Q=1.5 \cdot 1000\left(t_{c}-t_{\triangle}\right)=1500\left(\frac{r^{2} \pi \alpha}{360}-O C \cdot A C\right) \\
\cos \frac{\alpha}{2}=\frac{1.6}{2.3}, \quad \frac{\alpha}{2}=45.92^{\circ}, \quad \alpha=91.84^{\circ} \\
A C=\sqrt{2.3^{2}-1.6^{2}}=\sqrt{3.9 \cdot 0.7}=\sqrt{2.73}
\end{gathered}
$$

Substituting the values under (2) and (3) into (1), using logarithm tables,

$$
\begin{gathered}
Q=1500\left(\frac{2.3^{2} \pi 91.84}{360}-1.6 \sqrt{2.73}\right)=1500(4.238-2.642)=1500 \cdot 1.596= \\
=2394 \mathrm{q}
\end{gathered}
$$

George Frank (Budapest, V. Eötvös József g. III. o. t.)