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

36 m3
36 m cubed

18 m3
18 m cubed

There is not enough information to say.
There is not enough information to say.

24 m3
24 m cubed
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1 answer

To determine the volume of the object composed of a hemisphere and a cone, we will first need to establish the relationship between the dimensions given.

  1. Volume of the Cone: We know the volume of the cone is \( V_{cone} = 12 , m^3 \).

  2. Relationship between the cone and the hemisphere:

    • The height of the cone (h) is equal to the diameter of the hemisphere (d), which is twice the radius (r) of the hemisphere.
    • Therefore, we have:
      • \( h = d = 2r \)
  3. Volume of the Hemisphere:

    • The volume of a hemisphere is given by the formula: \[ V_{hemisphere} = \frac{2}{3} \pi r^3 \]
  4. Volume of the Cone Formula:

    • The volume of a cone is given by the formula: \[ V_{cone} = \frac{1}{3} \pi r^2 h \]
    • Since \( h = 2r \), we can replace \( h \) in the cone's volume formula: \[ V_{cone} = \frac{1}{3} \pi r^2 (2r) = \frac{2}{3} \pi r^3 \]
  5. Equating Volumes:

    • We know from the problem statement that \( V_{cone} = 12 , m^3 \): \[ \frac{2}{3} \pi r^3 = 12 \]
    • Solving for \( r^3 \): \[ r^3 = \frac{12 \cdot 3}{2\pi} = \frac{36}{2\pi} = \frac{18}{\pi} \]
  6. Calculating Volume of the Hemisphere:

    • Plugging \( r^3 \) back into the hemisphere volume formula: \[ V_{hemisphere} = \frac{2}{3} \pi r^3 = \frac{2}{3} \pi \left(\frac{18}{\pi}\right) = \frac{2 \cdot 18}{3} = 12 \]
  7. Total Volume of the Object:

    • Now, we can find the total volume of the object (hemisphere + cone): \[ V_{total} = V_{hemisphere} + V_{cone} = 12 , m^3 + 12 , m^3 = 24 , m^3 \]

Thus, the volume of the whole object is 24 m³.