A right pyramid has a square base with perimeter 24 inches. Its apex is 9 inches from each of the other vertices. What is the height of the pyramid from its peak to the center of its square base, in inches?
[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);
label("$F$",F,S);
triple M=(B+C)/2;
draw(P--F--B,dashed);
[/asy]

Let $F$ be the center of the square base.  Since the pyramid is a right pyramid, triangle $PFB$ is a right triangle.  The perimeter of the base of the pyramid is 24 inches, so the length of each side of the base is $6$ inches.  Since $F$ is the center of the base, $FB$ is half the diagonal of the base, or $(6\sqrt{2})/2 = 3\sqrt{2}$ inches.  Applying the Pythagorean Theorem to triangle $PFB$ gives  \[PF = \sqrt{PB^2 - FB^2} = \sqrt{81 - 18} = \sqrt{63} = \boxed{3\sqrt{7}} \text{ inches}.\]
3\sqrt{7}
Level 4
Geometry
Problem: A right pyramid has a square base with perimeter 24 inches. Its apex is 9 inches from each of the other vertices. What is the height of the pyramid from its peak to the center of its square base, in inches?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the height of the pyramid, we first determine the side length of the square base. The perimeter of the base is 24 inches, so each side length is 24 inches divided by 4, which equals 6 inches. 

Next, we identify the center of the square base. The coordinates of the vertices of the square can be represented as (3, 3), (3, -3), (-3, 3), and (-3, -3) if we place the center at the origin (0, 0). The apex of the pyramid is directly above the center of the base.

The distance from the apex to each vertex is given as 9 inches. We can use the distance formula to relate the height of the pyramid (h) to the distance from the apex to a vertex. The distance from the apex at (0, 0, h) to a vertex, say (3, 3, 0), is calculated as:

sqrt((3 - 0)² + (3 - 0)² + (0 - h)²) = 9.

This simplifies to:

sqrt(9 + 9 + h²) = 9.

Thus, we have:

sqrt(18 + h²) = 9.

Squaring both sides gives:

18 + h² = 81.

Subtracting 18 from both sides results in:

h² = 63.

Taking the square root yields:

h = sqrt(63) = sqrt(9 * 7) = 3sqrt{7}.

Therefore, the height of the pyramid from its peak to the center of its square base is 3sqrt{7} inches.<|END_OF_LATENT|> The final answer is $3\sqrt{7}$.