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Answer :
Sure! Let's find the total volume of the grain silo step-by-step.
The grain silo consists of two parts: a cylindrical part and a hemispherical part. The diameter of both parts is given as 4.4 meters.
### Step 1: Calculate the radius
The radius [tex]\( r \)[/tex] is half of the diameter:
[tex]\[ r = \frac{d}{2} = \frac{4.4}{2} = 2.2 \text{ meters} \][/tex]
### Step 2: Calculate the volume of the cylindrical part
The formula for the volume of a cylinder is:
[tex]\[ V_{\text{cylinder}} = \pi r^2 h \][/tex]
where:
- [tex]\( r \)[/tex] is the radius
- [tex]\( h \)[/tex] is the height of the cylindrical part, which is 6.2 meters
- [tex]\( \pi \)[/tex] is given as 3.14
Substitute the values:
[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]\[ V_{\text{cylinder}} = 3.14 \times 30.008 \][/tex]
[tex]\[ V_{\text{cylinder}} \approx 94.2 \text{ cubic meters} \][/tex]
### Step 3: Calculate the volume of the hemispherical part
The formula for the volume of a sphere is:
[tex]\[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \][/tex]
Since we have a hemisphere, we take half of the sphere's volume:
[tex]\[ V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 \][/tex]
[tex]\[ V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 \][/tex]
Substitute the values:
[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]\[ V_{\text{hemisphere}} = \frac{2}{3} \times 33.43672 \][/tex]
[tex]\[ V_{\text{hemisphere}} \approx 22.3 \text{ cubic meters} \][/tex]
### Step 4: Calculate the total volume of the silo
Add the volumes of the cylindrical part and the hemispherical part:
[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]
The approximate total volume of the silo is:
[tex]\[ 116.5 \text{ cubic meters} \][/tex]
So, the correct answer is:
[tex]\[ 116.5 \, m^3 \][/tex]
The grain silo consists of two parts: a cylindrical part and a hemispherical part. The diameter of both parts is given as 4.4 meters.
### Step 1: Calculate the radius
The radius [tex]\( r \)[/tex] is half of the diameter:
[tex]\[ r = \frac{d}{2} = \frac{4.4}{2} = 2.2 \text{ meters} \][/tex]
### Step 2: Calculate the volume of the cylindrical part
The formula for the volume of a cylinder is:
[tex]\[ V_{\text{cylinder}} = \pi r^2 h \][/tex]
where:
- [tex]\( r \)[/tex] is the radius
- [tex]\( h \)[/tex] is the height of the cylindrical part, which is 6.2 meters
- [tex]\( \pi \)[/tex] is given as 3.14
Substitute the values:
[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]\[ V_{\text{cylinder}} = 3.14 \times 30.008 \][/tex]
[tex]\[ V_{\text{cylinder}} \approx 94.2 \text{ cubic meters} \][/tex]
### Step 3: Calculate the volume of the hemispherical part
The formula for the volume of a sphere is:
[tex]\[ V_{\text{sphere}} = \frac{4}{3} \pi r^3 \][/tex]
Since we have a hemisphere, we take half of the sphere's volume:
[tex]\[ V_{\text{hemisphere}} = \frac{1}{2} \times \frac{4}{3} \pi r^3 \][/tex]
[tex]\[ V_{\text{hemisphere}} = \frac{2}{3} \pi r^3 \][/tex]
Substitute the values:
[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]\[ V_{\text{hemisphere}} = \frac{2}{3} \times 33.43672 \][/tex]
[tex]\[ V_{\text{hemisphere}} \approx 22.3 \text{ cubic meters} \][/tex]
### Step 4: Calculate the total volume of the silo
Add the volumes of the cylindrical part and the hemispherical part:
[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]
The approximate total volume of the silo is:
[tex]\[ 116.5 \text{ cubic meters} \][/tex]
So, the correct answer is:
[tex]\[ 116.5 \, m^3 \][/tex]
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