To solve the problem, we need to find the total volume of both the hemisphere and the cone.
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Volume of the hemisphere: The volume \( V \) of a hemisphere is given by the formula: \[ V = \frac{2}{3} \pi r^3 \] Given that the volume of the hemisphere is 12, we have: \[ \frac{2}{3} \pi r^3 = 12 \] Rearranging this gives: \[ r^3 = \frac{12 \cdot 3}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \]
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Finding the radius: Ignoring the factors of \(\pi\), we continue with \( r^3 = \frac{18}{\pi} \).
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Volume of the cone: The volume \( V \) of a cone is given by the formula: \[ V = \frac{1}{3} \pi r^2 h \] From the problem, we know that the diameter of the cone is equal to its height. Therefore, if the diameter is \( d \), then the height \( h = d \) and \( r = \frac{d}{2} = \frac{h}{2} \).
If we let the height \( h = d \), then: \[ r = \frac{h}{2} \]
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Substituting into the volume of the cone: The volume of the cone becomes: \[ V = \frac{1}{3} \pi \left(\frac{h}{2}\right)^2 h = \frac{1}{3} \pi \frac{h^2}{4} h = \frac{1}{12} \pi h^3 \]
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Relating \( h \) back to \( r \): We can use \( r^3 = \frac{18}{\pi} \) to express \( h \): Since \( r = \frac{h}{2} \): \[ \left(\frac{h}{2}\right)^3 = \frac{18}{\pi} \] This means: \[ \frac{h^3}{8} = \frac{18}{\pi} \] Therefore, rearranging gives: \[ h^3 = \frac{18 \cdot 8}{\pi} = \frac{144}{\pi} \]
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Substituting back into the cone's volume equation: The volume of the cone now becomes: \[ V = \frac{1}{12} \pi h^3 = \frac{1}{12} \pi \left(\frac{144}{\pi}\right) = \frac{144}{12} = 12 \]
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Calculating the total volume of ice cream: The total volume of ice cream in and on top of the cone is the sum of the volumes of the hemisphere and the cone: \[ V_{\text{total}} = V_{\text{hemisphere}} + V_{\text{cone}} = 12 + 12 = 24 \]
Thus, the total volume of the ice cream in and on top of the cone is: \[ \boxed{24} \]