Question
Mei Li measures a cube and a square-based pyramid and finds that the pyramid has the same height and base area as the cube. She also calculates that the volume of the cube is 9 cubic meters. What is the volume of the pyramid?(1 point) Responses 1m3 1 m cubed 30m3 30m cubed 6m3 6m cubed 3 m3
Answers
GPT-4o mini
To find the volume of the square-based pyramid, we need to first determine the relationship between the volumes of the cube and the pyramid.
We know that:
1. The volume of the cube (V_cube) is given as 9 cubic meters.
2. The formula for the volume of a cube is:
\[
V_{\text{cube}} = s^3
\]
where \(s\) is the side length of the cube.
Since \(V_{\text{cube}} = 9 \, \text{m}^3\), we can find \(s\):
\[
s^3 = 9 \implies s = \sqrt[3]{9} \approx 2.08 \, \text{m}
\]
3. The volume of a square-based pyramid (V_pyramid) is given by the formula:
\[
V_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
4. Since Mei Li found that the pyramid has the same base area and height as the cube, let’s calculate:
- The base area of the cube (which is also a square) is:
\[
\text{Base Area}_{\text{cube}} = s^2 = (\sqrt[3]{9})^2 = 9^{2/3} \approx 4.24 \, \text{m}^2
\]
- The height of the cube is the same as its side length, \(s = \sqrt[3]{9}\).
5. Now we can plug the base area and height into the volume formula for the pyramid:
\[
V_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 9^{2/3} \times \sqrt[3]{9}
\]
\[
V_{\text{pyramid}} = \frac{1}{3} \times 9^{2/3} \times 9^{1/3} = \frac{1}{3} \times 9^{(2/3) + (1/3)} = \frac{1}{3} \times 9^1 = \frac{9}{3} = 3 \, \text{m}^3
\]
Therefore, the volume of the pyramid is **3 cubic meters**.
The correct response is: **3 m³** (or **3m3**).
We know that:
1. The volume of the cube (V_cube) is given as 9 cubic meters.
2. The formula for the volume of a cube is:
\[
V_{\text{cube}} = s^3
\]
where \(s\) is the side length of the cube.
Since \(V_{\text{cube}} = 9 \, \text{m}^3\), we can find \(s\):
\[
s^3 = 9 \implies s = \sqrt[3]{9} \approx 2.08 \, \text{m}
\]
3. The volume of a square-based pyramid (V_pyramid) is given by the formula:
\[
V_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
4. Since Mei Li found that the pyramid has the same base area and height as the cube, let’s calculate:
- The base area of the cube (which is also a square) is:
\[
\text{Base Area}_{\text{cube}} = s^2 = (\sqrt[3]{9})^2 = 9^{2/3} \approx 4.24 \, \text{m}^2
\]
- The height of the cube is the same as its side length, \(s = \sqrt[3]{9}\).
5. Now we can plug the base area and height into the volume formula for the pyramid:
\[
V_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height} = \frac{1}{3} \times 9^{2/3} \times \sqrt[3]{9}
\]
\[
V_{\text{pyramid}} = \frac{1}{3} \times 9^{2/3} \times 9^{1/3} = \frac{1}{3} \times 9^{(2/3) + (1/3)} = \frac{1}{3} \times 9^1 = \frac{9}{3} = 3 \, \text{m}^3
\]
Therefore, the volume of the pyramid is **3 cubic meters**.
The correct response is: **3 m³** (or **3m3**).