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?

Answer 1

To find the volume of the pyramid, we can start by determining the side length of the cube based on its volume.

The formula for the volume \( V \) of a cube with side length \( s \) is:
\[
V = s^3
\]

Given that the volume of the cube is \( 9 \) cubic meters, we have:
\[
s^3 = 9
\]

To find \( s \), we take the cube root:
\[
s = \sqrt[3]{9}
\]

Next, we calculate the base area of the cube. Since the base of the cube is a square with side length \( s \), the base area \( A \) of the cube (and hence the base area of the pyramid) is:
\[
A = s^2
\]

Now we compute \( s^2 \):
\[
s^2 = (\sqrt[3]{9})^2 = \sqrt[3]{9^2} = \sqrt[3]{81}
\]

We can calculate the volume of the pyramid. The volume \( V_p \) of a square-based pyramid is given by the formula:
\[
V_p = \frac{1}{3} \times \text{Base Area} \times \text{Height}
\]

Here, the base area of the pyramid is \( s^2 \) (which we calculated to be \( \sqrt[3]{81} \)), and the height of the pyramid is the same as the height of the cube, which is \( s \):
\[
\text{Height} = s = \sqrt[3]{9}
\]

Now, substituting these values into the volume formula for the pyramid:
\[
V_p = \frac{1}{3} \times s^2 \times s
\]
\[
= \frac{1}{3} \times (\sqrt[3]{81}) \times (\sqrt[3]{9})
\]
\[
= \frac{1}{3} \times \sqrt[3]{81 \cdot 9}
\]

Calculating \( 81 \cdot 9 \):
\[
81 \cdot 9 = 729
\]

Therefore,
\[
V_p = \frac{1}{3} \times \sqrt[3]{729}
\]
Since \( 729 = 9^3 \):
\[
\sqrt[3]{729} = 9
\]

Thus:
\[
V_p = \frac{1}{3} \times 9 = 3
\]

Finally, the volume of the pyramid is:
\[
\boxed{3} \text{ cubic meters.}
\]