A right pyramid has a square base. The area of each triangular face is one-third the area of the square face. If the total surface area of the pyramid is $432$ square units, then what is the volume of the pyramid in cubic units?

Answer 1

Let the side length of the square base be $s$. Then if $A$ denotes area, we are given that \begin{align*}
A_{ABC} &= A_{BCD} = A_{CDA} = A_{DAB} =\frac13 s^2, \ \textrm{and} \\
A_{ABCD} &= s^2 + (s)(s) + (s)(s) + (s)(s) = 4s^2 = 432. %\, \textrm{square units}.
\end{align*}To find $s$, we solve $4s^2 = 432$, obtaining $s^2 = 108$, so $s = 6\sqrt{3}$.

The height of the pyramid is irrelevant. Its volume is thus given by \[
V = \frac13 Bh = \frac 13 \left (\frac13s^2\right) h = \frac{s^2 \cdot h}{27} = \frac{ (6\sqrt{3})^2 \cdot h}{27} = \frac{6^2 \cdot 3h}{27} = \boxed{12}h.
\] The problem statement does not contain enough information to find $h$.