The sum of the lengths of all of the edges of rectangular prism $A B C D E F G H$ is 24 . If the total surface area of the prism is 11, determine the length of the diagonal $A H$.

![](https://cdn.mathpix.com/cropped/2024_04_30_22929074a9efc78b4168g-2.jpg?height=380&width=542&top_left_y=279&top_left_x=846)
Suppose that the prism has $A B=x, A D=y$, and $A F=z$.

Since the sum of all of the edge lengths is 24 , then $4 x+4 y+4 z=24$ or $x+y+z=6$.

(The prism has 4 edges of each length.)

Since the surface area is 11 , then $2 x y+2 x z+2 y z=11$.

(The prism has 2 faces that are $x$ by $y, 2$ faces that are $x$ by $z$, and two faces that are $y$ by $z$.) By the Pythagorean Theorem, $A H^{2}=A C^{2}+C H^{2}$.

![](https://cdn.mathpix.com/cropped/2024_04_30_75c1cf70fbe3735bfd92g-03.jpg?height=388&width=542&top_left_y=1595&top_left_x=846)

Again, by the Pythagorean Theorem, $A C^{2}=A B^{2}+B C^{2}$, and so $A H^{2}=A B^{2}+B C^{2}+C H^{2}$. But $A B=x, B C=A D=y$, and $C H=A F=z$.

Therefore, $A H^{2}=x^{2}+y^{2}+z^{2}$.

Finally,

$$
(x+y+z)^{2}=(x+(y+z))^{2}=x^{2}+2 x(y+z)+(y+z)^{2}=x^{2}+2 x y+2 x z+y^{2}+2 y z+z^{2}
$$

and so $(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2 x y+2 x z+2 y z$, which gives

$$
A H^{2}=(x+y+z)^{2}-(2 x y+2 x z+2 y z)=6^{2}-11=25
$$

Since $A H>0$, then $A H=\sqrt{25}=5$.