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Sagot :
To solve for the volume of the frustum, let's start by understanding the structure of the frustum. The frustum is formed by removing a smaller cone from a larger cone.
The volume [tex]\(V\)[/tex] of a cone is given by the formula:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
Where:
- [tex]\(r\)[/tex] is the radius of the base of the cone
- [tex]\(h\)[/tex] is the height of the cone
Let's break down the problem into steps:
1. Volume of the Large Cone:
- Radius of the large cone, [tex]\(r_1 = 4\)[/tex] cm
- Height of the large cone, [tex]\(h_1 = 10\)[/tex] cm
The volume [tex]\(V_{large}\)[/tex] of the large cone is calculated as:
[tex]\[ V_{large} = \frac{1}{3} \pi r_1^2 h_1 \][/tex]
Substituting the values:
[tex]\[ V_{large} = \frac{1}{3} \pi (4)^2 (10) \][/tex]
[tex]\[ V_{large} = \frac{1}{3} \pi (16) (10) \][/tex]
[tex]\[ V_{large} = \frac{1}{3} \pi (160) \][/tex]
[tex]\[ V_{large} = \frac{160}{3} \pi \][/tex]
[tex]\[ V_{large} = 160\pi/3 \][/tex]
2. Volume of the Small Cone:
- Radius of the small cone, [tex]\(r_2 = 2\)[/tex] cm
- Height of the small cone, [tex]\(h_2 = 5\)[/tex] cm
The volume [tex]\(V_{small}\)[/tex] of the small cone is calculated as:
[tex]\[ V_{small} = \frac{1}{3} \pi r_2^2 h_2 \][/tex]
Substituting the values:
[tex]\[ V_{small} = \frac{1}{3} \pi (2)^2 (5) \][/tex]
[tex]\[ V_{small} = \frac{1}{3} \pi (4) (5) \][/tex]
[tex]\[ V_{small} = \frac{1}{3} \pi (20) \][/tex]
[tex]\[ V_{small} = \frac{20}{3} \pi \][/tex]
[tex]\[ V_{small} = 20\pi/3 \][/tex]
3. Volume of the Frustum:
The volume [tex]\(V_{frustum}\)[/tex] of the frustum is the volume of the large cone minus the volume of the small cone:
[tex]\[ V_{frustum} = V_{large} - V_{small} \][/tex]
Substituting the values:
[tex]\[ V_{frustum} = \left( \frac{160}{3} \pi \right) - \left( \frac{20}{3} \pi \right) \][/tex]
[tex]\[ V_{frustum} = \frac{160\pi - 20\pi}{3} \][/tex]
[tex]\[ V_{frustum} = \frac{140\pi}{3} \][/tex]
Therefore, the volume of the frustum is:
[tex]\[ V_{frustum} = \frac{140\pi}{3} \, \text{cm}^3 \][/tex]
In numerical form, this volume is approximately [tex]\(146.61 \, \text{cm}^3\)[/tex].
The volume [tex]\(V\)[/tex] of a cone is given by the formula:
[tex]\[ V = \frac{1}{3} \pi r^2 h \][/tex]
Where:
- [tex]\(r\)[/tex] is the radius of the base of the cone
- [tex]\(h\)[/tex] is the height of the cone
Let's break down the problem into steps:
1. Volume of the Large Cone:
- Radius of the large cone, [tex]\(r_1 = 4\)[/tex] cm
- Height of the large cone, [tex]\(h_1 = 10\)[/tex] cm
The volume [tex]\(V_{large}\)[/tex] of the large cone is calculated as:
[tex]\[ V_{large} = \frac{1}{3} \pi r_1^2 h_1 \][/tex]
Substituting the values:
[tex]\[ V_{large} = \frac{1}{3} \pi (4)^2 (10) \][/tex]
[tex]\[ V_{large} = \frac{1}{3} \pi (16) (10) \][/tex]
[tex]\[ V_{large} = \frac{1}{3} \pi (160) \][/tex]
[tex]\[ V_{large} = \frac{160}{3} \pi \][/tex]
[tex]\[ V_{large} = 160\pi/3 \][/tex]
2. Volume of the Small Cone:
- Radius of the small cone, [tex]\(r_2 = 2\)[/tex] cm
- Height of the small cone, [tex]\(h_2 = 5\)[/tex] cm
The volume [tex]\(V_{small}\)[/tex] of the small cone is calculated as:
[tex]\[ V_{small} = \frac{1}{3} \pi r_2^2 h_2 \][/tex]
Substituting the values:
[tex]\[ V_{small} = \frac{1}{3} \pi (2)^2 (5) \][/tex]
[tex]\[ V_{small} = \frac{1}{3} \pi (4) (5) \][/tex]
[tex]\[ V_{small} = \frac{1}{3} \pi (20) \][/tex]
[tex]\[ V_{small} = \frac{20}{3} \pi \][/tex]
[tex]\[ V_{small} = 20\pi/3 \][/tex]
3. Volume of the Frustum:
The volume [tex]\(V_{frustum}\)[/tex] of the frustum is the volume of the large cone minus the volume of the small cone:
[tex]\[ V_{frustum} = V_{large} - V_{small} \][/tex]
Substituting the values:
[tex]\[ V_{frustum} = \left( \frac{160}{3} \pi \right) - \left( \frac{20}{3} \pi \right) \][/tex]
[tex]\[ V_{frustum} = \frac{160\pi - 20\pi}{3} \][/tex]
[tex]\[ V_{frustum} = \frac{140\pi}{3} \][/tex]
Therefore, the volume of the frustum is:
[tex]\[ V_{frustum} = \frac{140\pi}{3} \, \text{cm}^3 \][/tex]
In numerical form, this volume is approximately [tex]\(146.61 \, \text{cm}^3\)[/tex].
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