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A solid is in the shape of a cone standi...

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of `pi`

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To find the volume of the solid that consists of a cone standing on a hemisphere, both with a radius of 1 cm and the height of the cone also being 1 cm, we can follow these steps: ### Step 1: Identify the dimensions - The radius \( r \) of both the cone and the hemisphere is given as \( 1 \, \text{cm} \). - The height \( h \) of the cone is also given as \( 1 \, \text{cm} \). ### Step 2: Calculate the volume of the cone The formula for the volume \( V \) of a cone is given by: \[ V = \frac{1}{3} \pi r^2 h \] Substituting the values of \( r \) and \( h \): \[ V_{\text{cone}} = \frac{1}{3} \pi (1)^2 (1) = \frac{1}{3} \pi \, \text{cm}^3 \] ### Step 3: Calculate the volume of the hemisphere The formula for the volume \( V \) of a hemisphere is given by: \[ V = \frac{2}{3} \pi r^3 \] Substituting the value of \( r \): \[ V_{\text{hemisphere}} = \frac{2}{3} \pi (1)^3 = \frac{2}{3} \pi \, \text{cm}^3 \] ### Step 4: Find the total volume of the solid The total volume of the solid is the sum of the volumes of the cone and the hemisphere: \[ V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} \] Substituting the volumes we calculated: \[ V_{\text{total}} = \frac{1}{3} \pi + \frac{2}{3} \pi = \frac{3}{3} \pi = \pi \, \text{cm}^3 \] ### Final Answer The volume of the solid is: \[ \boxed{\pi \, \text{cm}^3} \]

To find the volume of the solid that consists of a cone standing on a hemisphere, both with a radius of 1 cm and the height of the cone also being 1 cm, we can follow these steps: ### Step 1: Identify the dimensions - The radius \( r \) of both the cone and the hemisphere is given as \( 1 \, \text{cm} \). - The height \( h \) of the cone is also given as \( 1 \, \text{cm} \). ### Step 2: Calculate the volume of the cone The formula for the volume \( V \) of a cone is given by: ...
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