To find the volume of the square-based pyramid, we first need to determine the dimensions of the cube. The volume \( V \) of a cube is given by the formula:
\[ V = s^3 \]
where \( s \) is the length of a side of the cube. Given that the volume of the cube is 54 cubic units, we can solve for \( s \):
\[ s^3 = 54 \]
Taking the cube root of both sides, we find:
\[ s = \sqrt[3]{54} = 3\sqrt[3]{2} \]
Next, we need to find the height of the cube, which is equal to \( s \). Since the cube has height equal to its side length \( s \), the height of the cube (and therefore the height of the pyramid) is:
\[ h = s = 3\sqrt[3]{2} \]
Now, since the cube and the pyramid have the same base area and the same height, we need to find the base area of the cube. The base area \( A \) of the cube is:
\[ A = s^2 \]
Calculating the base area:
\[ A = (3\sqrt[3]{2})^2 = 9 \cdot \sqrt[3]{4} = 9 \cdot 2^{2/3} \]
Now, for the pyramid, the volume \( V_p \) is given by the formula:
\[ V_p = \frac{1}{3} \cdot \text{Base Area} \cdot \text{Height} \] Since the base area of the pyramid is the same as that of the cube, we have:
\[ V_p = \frac{1}{3} \cdot (9 \cdot 2^{2/3}) \cdot (3\sqrt[3]{2}) \]
Calculating the volume:
\[ V_p = \frac{1}{3} \cdot 9 \cdot 2^{2/3} \cdot 3 \cdot 2^{1/3} \]
Combine the powers of 2:
\[ V_p = \frac{1}{3} \cdot 9 \cdot 3 \cdot (2^{2/3} \cdot 2^{1/3}) = \frac{1}{3} \cdot 27 \cdot 2^{(2/3 + 1/3)} = \frac{1}{3} \cdot 27 \cdot 2^{1} = \frac{27 \cdot 2}{3} \]
\[ V_p = 9 \cdot 2 = 18 \text{ cubic units} \]
Thus, the volume of the pyramid is 18 cubic units. The correct response is:
18 cubic units.