College

A grain silo is composed of a cylinder and a hemisphere. The diameter is 4.4 meters, and the height of its cylindrical portion is 6.2 meters.

What is the approximate total volume of the silo? Use 3.14 for [tex]\pi[/tex] and round the answer to the nearest tenth of a cubic meter.

A. [tex]37.1 \, \text{m}^3[/tex]
B. [tex]71.9 \, \text{m}^3[/tex]
C. [tex]116.5 \, \text{m}^3[/tex]
D. [tex]130.8 \, \text{m}^3[/tex]

Answer :

To find the total volume of the grain silo, which consists of a cylindrical portion and a hemispherical top, we can break down the problem into two parts: calculating the volume of the cylinder and the volume of the hemisphere, and then adding them together.

1. Identify and Use the Given Dimensions:
- The diameter of the silo is 4.4 meters. This means the radius ([tex]\(r\)[/tex]) is half of the diameter:
[tex]\[ \text{Radius} = \frac{4.4}{2} = 2.2 \text{ meters} \][/tex]
- The height of the cylindrical portion is 6.2 meters.
- We will use [tex]\(\pi \approx 3.14\)[/tex].

2. Calculate the Volume of the Cylinder:
- The formula for the volume of a cylinder is:
[tex]\[ V_{\text{cylinder}} = \pi r^2 h \][/tex]
- Substitute the known values into the formula:
[tex]\[ V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2 \][/tex]
- After performing the calculations, the volume of the cylinder is approximately 94.2 cubic meters.

3. Calculate the Volume of the Hemisphere:
- The formula for the volume of a hemisphere is half of that for a sphere:
[tex]\[ V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 \][/tex]
- Substitute the radius into the formula:
[tex]\[ V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3 \][/tex]
- After performing the calculations, the volume of the hemisphere is approximately 22.3 cubic meters.

4. Add the Volumes to Get the Total Volume of the Silo:
- Total volume of the silo:
[tex]\[ V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemisphere}} = 94.2 + 22.3 \][/tex]
- The total volume is approximately 116.5 cubic meters.

5. Round to the Nearest Tenth:
- The final calculated volume, when rounded to the nearest tenth, results in 116.5 cubic meters.

Therefore, the approximate total volume of the silo is [tex]\( \boxed{116.5} \)[/tex] cubic meters.