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?

Answer 1

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.