A tent has been constructed which is in the form of a right circular cylinder surmounted by a right circular cone whose axis coincides with the axis of the cylinder. If the radius of the base is 50 m, the height of the cylinder is 10 m and the total height of the tent is 15 m, then what is the capacity of the tent in cubic meters?
87500π/3
This problem involves calculating the total volume (capacity) of a composite shape formed by a cylinder and a cone. The tent has a specific structure: a right circular cylinder with a right circular cone placed on top, sharing the same base and axis.
Understanding Tent Dimensions and Shape
We are given the following dimensions for the tent:
- Radius of the base (\(r\)): 50 m
- Height of the cylindrical part (\(h_{cyl}\)): 10 m
- Total height of the tent (\(H_{total}\)): 15 m
The tent is essentially made of two parts: a cylinder at the bottom and a cone on top.
Calculating the Cone's Height
The total height of the tent is the sum of the height of the cylinder and the height of the cone. We can find the height of the cone (\(h_{cone}\)) by subtracting the cylinder's height from the total height.
Using the formula:
\(H_{total} = h_{cyl} + h_{cone}\)
Rearranging to find \(h_{cone}\):
\(h_{cone} = H_{total} - h_{cyl}\)
Substituting the given values:
\(h_{cone} = 15 \text{ m} - 10 \text{ m} = 5 \text{ m}\)
So, the height of the conical part is 5 m.
Calculating the Volume of the Cylindrical Part
The volume of a right circular cylinder is given by the formula:
\(V_{cyl} = \pi r^2 h_{cyl}\)
Where:
- \(r\) is the radius of the base (50 m)
- \(h_{cyl}\) is the height of the cylinder (10 m)
Plugging in the values:
\(V_{cyl} = \pi \times (50 \text{ m})^2 \times 10 \text{ m}\)
\(V_{cyl} = \pi \times 2500 \text{ m}^2 \times 10 \text{ m}\)
\(V_{cyl} = 25000\pi \text{ m}^3\)
Calculating the Volume of the Conical Part
The volume of a right circular cone is given by the formula:
\(V_{cone} = \frac{1}{3} \pi r^2 h_{cone}\)
Where:
- \(r\) is the radius of the base (50 m)
- \(h_{cone}\) is the height of the cone (5 m)
Plugging in the values:
\(V_{cone} = \frac{1}{3} \pi \times (50 \text{ m})^2 \times 5 \text{ m}\)
\(V_{cone} = \frac{1}{3} \pi \times 2500 \text{ m}^2 \times 5 \text{ m}\)
\(V_{cone} = \frac{12500\pi}{3} \text{ m}^3\)
Determining the Total Capacity of the Tent
The total capacity (volume) of the tent is the sum of the volume of the cylindrical part and the volume of the conical part.
\(V_{total} = V_{cyl} + V_{cone}\)
\(V_{total} = 25000\pi \text{ m}^3 + \frac{12500\pi}{3} \text{ m}^3\)
To add these, we need a common denominator:
\(V_{total} = \frac{25000\pi \times 3}{3} \text{ m}^3 + \frac{12500\pi}{3} \text{ m}^3\)
\(V_{total} = \frac{75000\pi}{3} \text{ m}^3 + \frac{12500\pi}{3} \text{ m}^3\)
\(V_{total} = \frac{75000\pi + 12500\pi}{3} \text{ m}^3\)
\(V_{total} = \frac{87500\pi}{3} \text{ m}^3\)
Therefore, the capacity of the tent is \(\frac{87500\pi}{3}\) cubic meters.
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