The cross-section of a tunnel is a square of side 7 m surmounted by a semicircle. The tunnel is 80 m long ,Calculate : its volume

Answer

To calculate the volume of the tunnel, we need to consider the cross-section, which consists of a square and a semicircle. Here’s a step-by-step solution:

### Step 1: Understand the dimensions of the cross-section
The cross-section of the tunnel is a square with a side length of 7 m, and it is surmounted by a semicircle.

### Step 2: Calculate the area of the square
The area \( A \) of a square is given by the formula:
\[ A = \text{side}^2 \]
For our square:
\[ A = 7^2 = 49 \, \text{m}^2 \]

### Step 3: Calculate the radius of the semicircle
The semicircle is based on the top side of the square. The diameter of the semicircle is equal to the side of the square, which is 7 m. Thus, the radius \( r \) of the semicircle is:
\[ r = \frac{7}{2} = 3.5 \, \text{m} \]

### Step 4: Calculate the area of the semicircle
The area \( A \) of a semicircle is given by the formula:
\[ A = \frac{1}{2} \pi r^2 \]
Substituting the radius:
\[ A = \frac{1}{2} \times \pi \times (3.5)^2 \]
Calculating \( (3.5)^2 \):
\[ (3.5)^2 = 12.25 \]
Thus, the area becomes:
\[ A = \frac{1}{2} \times \pi \times 12.25 = \frac{12.25\pi}{2} = 6.125\pi \, \text{m}^2 \]
Using \( \pi \approx \frac{22}{7} \):
\[ A \approx 6.125 \times \frac{22}{7} = \frac{135.75}{7} \approx 19.39 \, \text{m}^2 \]

### Step 5: Calculate the total cross-sectional area
Now, we add the area of the square and the area of the semicircle:
\[ \text{Total Area} = \text{Area of Square} + \text{Area of Semicircle} \]
\[ \text{Total Area} = 49 + 19.39 \approx 68.39 \, \text{m}^2 \]

### Step 6: Calculate the volume of the tunnel
The volume \( V \) of the tunnel is given by the formula:
\[ V = \text{Total Area} \times \text{Length} \]
Given that the length of the tunnel is 80 m:
\[ V = 68.39 \times 80 \]
Calculating this gives:
\[ V \approx 5471.2 \, \text{m}^3 \]

### Final Answer
The volume of the tunnel is approximately \( 5471.2 \, \text{m}^3 \).

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