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?
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 $h$.
Let's denote the length of a side of the square base as $s$.
The area of the square base is $s^2$ square units.
Since the area of each triangular face is one-third the area of the square face, the area of each triangular face is $\frac{1}{3}(s^2)$ square units.
A right pyramid has 4 triangular faces and 1 square base. So, the total surface area of the pyramid is the sum of the areas of all 5 faces:
$432 = 4\left(\frac{1}{3}(s^2)\right) + s^2$
$432 = \frac{4}{3}s^2 + s^2$
$432 = \frac{7}{3}s^2$
To solve for $s^2$, we can multiply both sides of the equation by $\frac{3}{7}$:
$\frac{3}{7} \cdot 432 = s^2$
$s^2 = \frac{1296}{7}$
Now, let's find the volume of the pyramid.
Since the pyramid has a square base, the volume can be calculated using the formula:
$V = \frac{1}{3} \times \text{base area} \times \text{height}$
The height of the pyramid can be found using the Pythagorean theorem. Let's denote the height as $h$.
Using the property of right triangles, we can see that the height, the side length of the square base, and half the diagonal of the square base form a right triangle.
The diagonal of the square base can be found by using the Pythagorean theorem with the side length of the square base:
$ \text{diagonal} = \sqrt{2} \times \text{side length} = \sqrt{2} \times s$
The height, the side length of the square base, and the diagonal form a right triangle, so we can use the Pythagorean theorem:
$h^2 + \left(\frac{\sqrt{2}s}{2}\right)^2 = (\sqrt{2}s)^2 $
$h^2 + \frac{s^2}{2} = 2s^2$
$h^2 = \frac{s^2}{2}$
Substituting the value of $s^2$ from earlier:
$h^2 = \frac{\frac{1296}{7}}{2} = \frac{1296}{14} = \frac{648}{7}$
Now, we can calculate the volume of the pyramid:
$V = \frac{1}{3} \times s^2 \times h$
$V = \frac{1}{3} \times \frac{1296}{7} \times \frac{648}{7}$
$V = \frac{1296 \times 648}{3 \times 7 \times 7}$
$V = \frac{839,808}{147}$
Therefore, the volume of the pyramid is approximately $\boxed{5,712}$ cubic units.
To find the volume of the pyramid, we need to know the length of one side of the square base and the height of the pyramid. Let's say the length of one side of the square base is $x$, and the height of the pyramid is $h$.
Since the area of each triangular face is one-third the area of the square face, we can write the following equation:
Area of each triangular face = (1/3) * Area of square face
The area of the square face is $x^2$, so the area of each triangular face is $1/3$ * $x^2$. Since there are four triangular faces, the total surface area of the triangular faces is $4$ * $1/3$ * $x^2$ = $4/3$ * $x^2$.
The total surface area of the pyramid is given as $432$ square units, so we have the equation:
Total surface area = Area of square base + Total surface area of the triangular faces
$432$ = $x^2$ + $4/3$ * $x^2$
Combining like terms, we get:
$432$ = $7/3$ * $x^2$
To solve for $x$, we can multiply both sides of the equation by $3/7$:
$3/7$ * $432$ = $x^2$
Simplifying, we find that $x^2$ = $144$.
Taking the square root of both sides, we get $x$ = $12$.
Now that we know the length of one side of the square base is $12$, we can calculate the volume of the pyramid.
The volume of a pyramid is given by the formula:
Volume = (1/3) * Base Area * Height
In this case, the base is a square, so the base area is $12^2$ = $144$ square units.
The height of the pyramid is not given directly, but we can find it using the Pythagorean theorem. Since the height forms the hypotenuse, and the base and height of each triangular face form legs of a right triangle, the height is equal to $\sqrt{x^2 + x^2}$ = $\sqrt{2x^2}$.
Substituting the values, the height of the pyramid is $\sqrt{2} * 12$ = $12\sqrt{2}$.
Now, we can calculate the volume of the pyramid:
Volume = (1/3) * Base Area * Height
= (1/3) * 144 * 12\sqrt{2}
= $48\sqrt{2}$ cubic units.
Therefore, the volume of the pyramid is $48\sqrt{2}$ cubic units.