Question
The figure is made up of a cone and a hemisphere.
To the nearest whole number, what is the volume of this figure?
Use 3.14 to approximate π . Round only your final answer to the nearest whole number.
Enter your answer in the box.
cm³
The circle at the top is 8 cm and the cone is 12 cm
To the nearest whole number, what is the volume of this figure?
Use 3.14 to approximate π . Round only your final answer to the nearest whole number.
Enter your answer in the box.
cm³
The circle at the top is 8 cm and the cone is 12 cm
Answers
GPT-4o mini
To find the volume of the figure made up of a cone and a hemisphere, we can use the formulas for the volume of a cone and a hemisphere.
1. **Volume of the cone:**
\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h
\]
where:
- \( r \) is the radius of the base of the cone (which is half of the diameter),
- \( h \) is the height of the cone.
Given that the diameter of the circle is 8 cm, the radius \( r \) is:
\[
r = \frac{8}{2} = 4 \text{ cm}
\]
The height \( h \) of the cone is 12 cm.
Now we can calculate the volume of the cone:
\[
V_{\text{cone}} = \frac{1}{3} \cdot 3.14 \cdot (4^2) \cdot 12
\]
\[
= \frac{1}{3} \cdot 3.14 \cdot 16 \cdot 12
\]
\[
= \frac{1}{3} \cdot 3.14 \cdot 192
\]
\[
= \frac{602.88}{3} \approx 200.96 \text{ cm}^3
\]
2. **Volume of the hemisphere:**
\[
V_{\text{hemisphere}} = \frac{2}{3} \pi r^3
\]
Again, using the radius \( r = 4 \text{ cm} \):
\[
V_{\text{hemisphere}} = \frac{2}{3} \cdot 3.14 \cdot (4^3)
\]
\[
= \frac{2}{3} \cdot 3.14 \cdot 64
\]
\[
= \frac{2}{3} \cdot 201.76
\]
\[
\approx \frac{403.52}{3} \approx 134.51 \text{ cm}^3
\]
3. **Total volume of the figure:**
Now we can sum the volumes of the cone and the hemisphere:
\[
V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} = 200.96 + 134.51 \approx 335.47 \text{ cm}^3
\]
Rounding this to the nearest whole number,
\[
\text{Total Volume} \approx 335 \text{ cm}^3
\]
Thus, the final answer is:
\[
\boxed{335} \text{ cm}^3
\]
1. **Volume of the cone:**
\[
V_{\text{cone}} = \frac{1}{3} \pi r^2 h
\]
where:
- \( r \) is the radius of the base of the cone (which is half of the diameter),
- \( h \) is the height of the cone.
Given that the diameter of the circle is 8 cm, the radius \( r \) is:
\[
r = \frac{8}{2} = 4 \text{ cm}
\]
The height \( h \) of the cone is 12 cm.
Now we can calculate the volume of the cone:
\[
V_{\text{cone}} = \frac{1}{3} \cdot 3.14 \cdot (4^2) \cdot 12
\]
\[
= \frac{1}{3} \cdot 3.14 \cdot 16 \cdot 12
\]
\[
= \frac{1}{3} \cdot 3.14 \cdot 192
\]
\[
= \frac{602.88}{3} \approx 200.96 \text{ cm}^3
\]
2. **Volume of the hemisphere:**
\[
V_{\text{hemisphere}} = \frac{2}{3} \pi r^3
\]
Again, using the radius \( r = 4 \text{ cm} \):
\[
V_{\text{hemisphere}} = \frac{2}{3} \cdot 3.14 \cdot (4^3)
\]
\[
= \frac{2}{3} \cdot 3.14 \cdot 64
\]
\[
= \frac{2}{3} \cdot 201.76
\]
\[
\approx \frac{403.52}{3} \approx 134.51 \text{ cm}^3
\]
3. **Total volume of the figure:**
Now we can sum the volumes of the cone and the hemisphere:
\[
V_{\text{total}} = V_{\text{cone}} + V_{\text{hemisphere}} = 200.96 + 134.51 \approx 335.47 \text{ cm}^3
\]
Rounding this to the nearest whole number,
\[
\text{Total Volume} \approx 335 \text{ cm}^3
\]
Thus, the final answer is:
\[
\boxed{335} \text{ cm}^3
\]