To find the total volume of the object composed of a hemisphere on top of a cone, we can follow these steps:
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Volume of the Cone: The problem states that the volume of the cone is 12 m³.
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Height of the Cone and Diameter of the Hemisphere: According to the given information, the height of the cone is equal to the diameter of the hemisphere.
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Radius of the Hemisphere: Since the height of the cone is equal to the diameter of the hemisphere, if we denote the radius of the hemisphere as \( r \), then the height of the cone \( h \) is \( 2r \).
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Volume of the Hemisphere: The volume \( V_h \) of a hemisphere is calculated using the formula: \[ V_h = \frac{2}{3} \pi r^3 \]
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Total Volume: The total volume \( V_{total} \) of the object that includes both the cone and the hemisphere is: \[ V_{total} = V_{cone} + V_{hemisphere} \] Substituting in the values: \[ V_{total} = 12 + \frac{2}{3} \pi r^3 \]
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Finding \( r \): Since we know the relationship between the cone's height and hemisphere's diameter, and we know the volume of the cone, we can solve for \( r \) as follows. The height of the cone, which is \( 2r \), can be related to the volume of the cone using the cone volume formula: \[ V_{cone} = \frac{1}{3} \pi r^2 h \] Since \( h = 2r \): \[ V_{cone} = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 \] Setting up the equation with the known volume (12 m³): \[ \frac{2}{3} \pi r^3 = 12 \] Solving for \( r^3 \): \[ r^3 = \frac{12 \cdot 3}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \]
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Calculating Volume of Hemisphere: Now substituting \( r^3 \) back into the volume of the hemisphere: \[ V_{hemisphere} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi \left(\frac{18}{\pi}\right) = \frac{2 \cdot 18}{3} = 12 \]
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Final Total Volume Calculation: Thus, the total volume of the whole object is: \[ V_{total} = V_{cone} + V_{hemisphere} = 12 + 12 = 24 \text{ m}^3 \]
So, the correct response is:
24 m³
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