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 30m3 30m cubed 6m3 6m cubed 1m3 1 m cubed 3 m3

Answer 1

To find the volume of the pyramid, we first need to recall the formulas for the volumes of both the cube and the square-based pyramid.

1. **Volume of the cube (V_cube)**:
The volume of a cube is given by:
\[
V_{\text{cube}} = s^3
\]
where \( s \) is the length of one side. We know from the problem that the volume of the cube is 9 cubic meters:
\[
s^3 = 9 \implies s = \sqrt[3]{9}
\]

2. **Volume of the pyramid (V_pyramid)**:
The volume of a square-based pyramid is given by:
\[
V_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]
We know from the problem that the pyramid has the same height and base area as the cube:
- The height \( h \) of the pyramid is equal to the side length \( s \) of the cube.
- The base area \( A \) of the pyramid is equal to \( s^2 \) (the area of the cube's base).

Since we already have the volume of the cube as 9 m³, we can deduce:
\[
s^3 = 9 \implies s = \sqrt[3]{9}
\]
Thus:
\[
\text{Height} = s = \sqrt[3]{9}
\]
\[
\text{Base Area} = s^2 = (\sqrt[3]{9})^2 = 9^{2/3}
\]

Now substituting into the volume formula for the pyramid:
\[
V_{\text{pyramid}} = \frac{1}{3} \times A \times h = \frac{1}{3} \times 9^{2/3} \times \sqrt[3]{9}
\]

By substituting \( \sqrt[3]{9} \) we simplify to:
\[
= \frac{1}{3} \times 9^{2/3} \times 9^{1/3} = \frac{1}{3} \times 9^{(2/3 + 1/3)} = \frac{1}{3} \times 9^{3/3} = \frac{1}{3} \times 9 = 3
\]

Thus, the volume of the pyramid is **3 cubic meters**, which is represented in the answer choice as **3 m³**.