An engineer wants to know the volume of the following object composed of a hemisphere and a cone. She knows the height of the cone is equal to the diameter of the hemisphere, and the volume of the cone is 12 m3. What is the volume of the whole object?(1 point)
Responses
18 m3
18 m cubed
36 m3
36 m cubed
24 m3
24 m cubed
There is not enough information to say.
To find the volume of the whole object, we need to sum up the volumes of the cone and the hemisphere.
Given:
Volume of the cone (V_cone) = 12 m³
The volume of a cone is calculated using the formula:
\[ V_{cone} = \frac{1}{3} \pi r^2 h \]
where:
\( r \) = radius of the cone's base
\( h \) = height of the cone
The volume of a hemisphere (V_hemisphere) is half the volume of a full sphere. The volume of a full sphere is calculated using the formula:
\[ V_{sphere} = \frac{4}{3} \pi r^3 \]
So the volume of the hemisphere is:
\[ V_{hemisphere} = \frac{1}{2} \times \frac{4}{3} \pi r^3 = \frac{2}{3} \pi r^3 \]
From the information given, we know that the height of the cone is equal to the diameter of the hemisphere, which means:
\( h = 2r \)
Now, since we already are given the volume of the cone, we can solve for the radius by rearranging the cone volume formula:
\[ 12 = \frac{1}{3} \pi r^2 h \]
\[ 12 = \frac{1}{3} \pi r^2 (2r) \]
\[ 12 = \frac{2}{3} \pi r^3 \]
Now we want to find the radius:
\[ r^3 = \frac{12}{\frac{2}{3} \pi} \]
\[ r^3 = \frac{12 \times 3}{2 \pi} \]
\[ r^3 = \frac{36}{2 \pi} \]
\[ r^3 = \frac{18}{\pi} \]
But we don't actually need to find the exact value of \( r \). Because we found that the cone's volume in terms of \( r \) perfectly matches the formula for the hemisphere's volume, we can deduce that:
\[ V_{hemisphere} = V_{cone} = 12 m^3 \]
Thus, the volume of the whole object is:
\[ V_{total} = V_{cone} + V_{hemisphere} \]
\[ V_{total} = 12 m^3 + 12 m^3 \]
\[ V_{total} = 24 m^3 \]
So the volume of the whole object is 24 cubic meters.
The correct answer is:
24 m³
24 m cubed