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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 \, m^3[/tex]
B. [tex]71.9 \, m^3[/tex]
C. [tex]116.5 \, m^3[/tex]
D. [tex]130.8 \, m^3[/tex]

Answer :

To find the total volume of the grain silo, we need to calculate the volumes of the cylindrical and hemispherical portions separately and then add them together.

1. Calculate the Volume of the Cylinder:
- The formula for the volume of a cylinder is:
[tex]\[
V_{\text{cylinder}} = \pi r^2 h
\][/tex]
- Given:
- Diameter of the cylinder is 4.4 meters, so the radius [tex]\( r \)[/tex] is half of that:
[tex]\[
r = \frac{4.4}{2} = 2.2 \text{ meters}
\][/tex]
- Height of the cylinder [tex]\( h \)[/tex] is 6.2 meters.
- Plug these values into the formula:
[tex]\[
V_{\text{cylinder}} = 3.14 \times (2.2)^2 \times 6.2
\][/tex]
[tex]\[
V_{\text{cylinder}} = 3.14 \times 4.84 \times 6.2
\][/tex]
[tex]\[
V_{\text{cylinder}} \approx 94.2 \text{ cubic meters}
\][/tex]

2. Calculate the Volume of the Hemisphere:
- The formula for the volume of a hemisphere is:
[tex]\[
V_{\text{hemisphere}} = \frac{2}{3} \pi r^3
\][/tex]
- Use the radius [tex]\( r = 2.2 \)[/tex] meters:
[tex]\[
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times (2.2)^3
\][/tex]
[tex]\[
V_{\text{hemisphere}} = \frac{2}{3} \times 3.14 \times 10.648
\][/tex]
[tex]\[
V_{\text{hemisphere}} \approx 22.3 \text{ cubic meters}
\][/tex]

3. Calculate the Total Volume of the Silo:
- Add the volumes of the cylindrical and hemispherical portions:
[tex]\[
V_{\text{total}} = V_{\text{cylinder}} + V_{\text{hemisphere}}
\][/tex]
[tex]\[
V_{\text{total}} = 94.2 + 22.3
\][/tex]
[tex]\[
V_{\text{total}} \approx 116.5 \text{ cubic meters}
\][/tex]

After calculating both volumes and adding them, we find that the total volume of the grain silo is approximately 116.5 cubic meters. The correct answer, rounded to the nearest tenth, is [tex]\(\boxed{116.5 \, \text{m}^3}\)[/tex].

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