1. **State the problem:** We need to find the volume of a tunnel with a cross section shaped as a major segment of a circle with radius $2.5$ m and height $3$ m, and length $800$ m. 2. **Understand the geometry:** The cross section is a major segment of a circle. The radius $r=2.5$ m, and the height (sagitta) $h=3$ m. The length of the tunnel $L=800$ m. 3. **Formula for segment area:** The area $A$ of a circular segment with radius $r$ and height $h$ is given by $$A = r^2 \arccos\left(\frac{r-h}{r}\right) - (r-h)\sqrt{2rh - h^2}$$ 4. **Calculate the area of the minor segment:** - Compute $r - h = 2.5 - 3 = -0.5$ m (negative because height is greater than radius, indicating a major segment). 5. **Since $h > r$, the segment is a major segment.** The major segment area is the area of the circle minus the minor segment area. 6. **Calculate minor segment height:** The minor segment height is $h_{minor} = r - (h - r) = 2r - h = 2 \times 2.5 - 3 = 2$ m. 7. **Calculate minor segment area:** $$A_{minor} = r^2 \arccos\left(\frac{r - h_{minor}}{r}\right) - (r - h_{minor}) \sqrt{2 r h_{minor} - h_{minor}^2}$$ Calculate each term: - $\frac{r - h_{minor}}{r} = \frac{2.5 - 2}{2.5} = 0.2$ - $\arccos(0.2) \approx 1.369$ radians - $\sqrt{2 \times 2.5 \times 2 - 2^2} = \sqrt{10 - 4} = \sqrt{6} \approx 2.449$ So, $$A_{minor} = 2.5^2 \times 1.369 - 0.5 \times 2.449 = 6.25 \times 1.369 - 1.2245 = 8.556 - 1.2245 = 7.3315 \text{ m}^2$$ 8. **Calculate total circle area:** $$A_{circle} = \pi r^2 = \pi \times 2.5^2 = \pi \times 6.25 = 19.635 \text{ m}^2$$ 9. **Calculate major segment area:** $$A_{major} = A_{circle} - A_{minor} = 19.635 - 7.3315 = 12.3035 \text{ m}^2$$ 10. **Calculate volume of tunnel:** $$V = A_{major} \times L = 12.3035 \times 800 = 9842.8 \text{ m}^3$$ **Final answer:** The volume of the tunnel is approximately $9843$ m³.