A right pyramid with a square base has total surface area 432 square units.  The area of each triangular face is half the area of the square face. What is the volume of the pyramid in cubic units?
Let $ABCD$ be the base of the pyramid and let $P$ be the pyramid's apex.

[asy]

import three;

triple A = (0,0,0);

triple B = (1,0,0);

triple C = (1,1,0);

triple D = (0,1,0);

triple P = (0.5,0.5,1);

draw(B--C--D--P--B);

draw(P--C);

draw(B--A--D,dashed);

draw(P--A,dashed);

label("$A$",A,NW);

label("$B$",B,W);

label("$C$",C,S);

label("$D$",D,E);

label("$P$",P,N);

triple F= (0.5,0.5,0);

triple M=(B+C)/2;

draw(P--F--M,dashed);

draw(P--M);

label("$F$",F,S);

label("$M$",M,SW);

[/asy]

Let $F$ be the center of the square base and $M$ be the midpoint of an edge of the square, as shown.  There are four triangular faces, each with area half the area of the square face.  So, the total surface area of the pyramid is 3 times the area of the square face.  Therefore, the area of the square face is $432/3=144$ square units, which means that each side of the square has length 12.

Since the area of the triangle is half the area of the square, we have $(BC)(PM)/2 = 72$, so $(BC)(PM) = 144$, which means $PM = 144/12 = 12$.  Since $F$ is the center of the square base, we have $FM = 6$, so $PF = \sqrt{12^2 - 6^2} = 6\sqrt{3}$.  Finally, the volume of the pyramid is \[\frac{[ABCD]\cdot PF}{3} = \frac{144\cdot 6\sqrt{3}}{3} = \boxed{288\sqrt{3}}.\]
288\sqrt{3}
Level 5
Geometry
Problem: A right pyramid with a square base has total surface area 432 square units.  The area of each triangular face is half the area of the square face. What is the volume of the pyramid in cubic units?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the volume of the pyramid, we first need to determine the dimensions of the square base and the height of the pyramid. The total surface area of the pyramid is given as 432 square units. 

Let the side length of the square base be denoted as s. The area of the square base is s^2. Each triangular face has an area that is half the area of the square face, which means the area of each triangular face is (1/2)s^2. Since a pyramid with a square base has four triangular faces, the total area of the triangular faces is 4 * (1/2)s^2 = 2s^2.

The total surface area of the pyramid can be expressed as the sum of the area of the base and the area of the triangular faces:
Total Surface Area = Area of Base + Area of Triangular Faces
432 = s^2 + 2s^2
432 = 3s^2

To find s^2, we solve for s:
s^2 = 432 / 3
s^2 = 144
s = √144
s = 12

Now that we have the side length of the base, we can find the height of the pyramid. The area of each triangular face is (1/2)s^2, which we already established as 72 square units. The area of a triangular face can also be expressed in terms of the base and height of the triangle. The base of each triangular face is s, and let h_t be the height of the triangular face. Thus, we have:
(1/2) * s * h_t = 72
(1/2) * 12 * h_t = 72
6h_t = 72
h_t = 12

The height of the triangular face is the slant height of the pyramid. To find the vertical height (h) of the pyramid, we can use the Pythagorean theorem. The height of the pyramid, the height of the triangular face, and half the base form a right triangle:
h_t^2 = h^2 + (s/2)^2
12^2 = h^2 + (12/2)^2
144 = h^2 + 36
h^2 = 144 - 36
h^2 = 108
h = √108
h = 6√3

Now we can calculate the volume (V) of the pyramid using the formula:
V = (1/3) * Base Area * Height
V = (1/3) * s^2 * h
V = (1/3) * 144 * 6√3
V = (1/3) * 864√3
V = 288√3

Thus, the final answer is 288√3.<|END_OF_LATENT|> The final answer is $288\sqrt{3}$.