A right pyramid has a square base. The area of each triangular face is half the area of the square face. If the total surface area of the pyramid is $432$ square units, then what is the area of the square base?
Let $s$ be the side length of the square base. If the area of each triangular face is half the area of the square, then each triangular face has area $\frac{s^2}{2}$. Since there are four triangular faces, their total area is $2s^2$. The surface area of a pyramid is the sum of the areas of the base and the triangular faces, so \[2s^2 + s^2 = 432.\] Simplifying the left side gives $3s^2 = 432$, so $s^2 = 144$. Since $s$ is the length of a side, $s$ must be a positive number, so $s = 12$. The area of the square base is then $12^2 = \boxed{144}$ square units.
Let's denote the side length of the square base as $s$.
We know that the area of a square is given by $A_s = s^2$, and the area of a triangle is given by $A_t = \frac{1}{2}bh$, where $b$ is the base length and $h$ is the height.
Since the area of each triangular face is half the area of the square face, we have $A_t = \frac{1}{2} A_s$.
The total surface area of the pyramid is given by $A_p = A_s + 4A_t$. Substituting the values, we have $432 = s^2 + 4(\frac{1}{2} s^2)$.
Simplifying further, we get $432 = s^2 + 2s^2$.
Combining like terms, we have $432 = 3s^2$.
Dividing both sides by 3, we get $s^2 = 144$.
Taking the square root of both sides, we find $s = 12$.
Therefore, the area of the square base is $s^2 = 12^2 = \boxed{144}$ square units.