To find the volume of the square-based pyramid, we can use the formula for the volume of a pyramid, which is given by:
\[ V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \]
We know the following:
- The base area of the pyramid is equal to the base area of the cube.
- The height of the pyramid is equal to the height of the cube.
- The volume of the cube is 9 cubic meters.
Let's find the side length of the cube first. The volume \( V \) of a cube is given by:
\[ V = \text{side length}^3 \]
If the volume of the cube is 9 m³, then:
\[ \text{side length}^3 = 9 \quad \Rightarrow \quad \text{side length} = \sqrt[3]{9} \]
Now, the volume of a cube also gives us the base area (which is side length squared). Therefore, the base area of the cube (and the pyramid) is:
\[ \text{Base Area} = \text{side length}^2 = (\sqrt[3]{9})^2 = 9^{2/3} = 3^{4/3} \]
The height of the cube would be equal to the side length:
\[ \text{Height} = \sqrt[3]{9} \]
Now, substituting these into the volume formula for the pyramid:
\[ V_{\text{pyramid}} = \frac{1}{3} \times (3^{4/3}) \times (\sqrt[3]{9}) \]
Let's simplify:
\[ = \frac{1}{3} \times 3^{4/3} \times 3^{2/3} = \frac{1}{3} \times 3^{(4/3) + (2/3)} = \frac{1}{3} \times 3^{6/3} = \frac{1}{3} \times 3^2 = \frac{1}{3} \times 9 = 3 \]
So, the volume of the square-based pyramid is \( 3 , \text{m}^3 \).
Thus, the correct response is:
3 m³
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