The solid shown has a square base of side length $s$. The upper edge is parallel to the base and has length $2s$. All other edges have length $s$. Given that $s=6\sqrt{2}$, what is the volume of the solid?
Solution 1
First, we find the height of the solid by dropping a perpendicular from the midpoint of $AD$ to $EF$. The hypotenuse of the triangle formed is the [median](https://artofproblemsolving.com/wiki/index.php/Median) of equilateral triangle $ADE$, and one of the legs is $3\sqrt{2}$. We apply the Pythagorean Theorem to deduce that the height is $6$.


Next, we complete t he figure into a triangular prism, and find its volume, which is $\frac{6\sqrt{2}\cdot 12\sqrt{2}\cdot 6}{2}=432$.
Now, we subtract off the two extra [pyramids](https://artofproblemsolving.com/wiki/index.php/Pyramid) that we included, whose combined volume is $2\cdot \left( \frac{6\sqrt{2}\cdot 3\sqrt{2} \cdot 6}{3} \right)=144$.
Thus, our answer is $432-144=\boxed{288}$.

Solution 2

Extend $EA$ and $FB$ to meet at $G$, and $ED$ and $FC$ to meet at $H$. Now, we have a regular tetrahedron $EFGH$, which by symmetry has twice the volume of our original solid. This tetrahedron has side length $2s = 12\sqrt{2}$. Using the formula for the volume of a regular tetrahedron, which is $V = \frac{\sqrt{2}S^3}{12}$, where S is the side length of the tetrahedron, the volume of our original solid is:
$V = \frac{1}{2} \cdot \frac{\sqrt{2} \cdot (12\sqrt{2})^3}{12} = \boxed{288}$.

Solution 3
We can also find the volume by considering horizontal cross-sections of the solid and using calculus. As in Solution 1, we can find that the height of the solid is $6$; thus, we will integrate with respect to height from $0$ to $6$, noting that each cross section of height $dh$ is a rectangle. The volume is then $\int_0^h(wl) \ \text{d}h$, where $w$ is the width of the rectangle and $l$ is the length. We can express $w$ in terms of $h$ as $w=6\sqrt{2}-\sqrt{2}h$ since it decreases linearly with respect to $h$, and $l=6\sqrt{2}+\sqrt{2}h$ since it similarly increases linearly with respect to $h$. Now we solve:\[\int_0^6(6\sqrt{2}-\sqrt{2}h)(6\sqrt{2}+\sqrt{2}h)\ \text{d}h =\int_0^6(72-2h^2)\ \text{d}h=72(6)-2\left(\frac{1}{3}\right)\left(6^3\right)=\boxed{288}\].

Solution 4
Draw an altitude from a vertex of the square base to the top edge. By using $30,60, 90$ triangle ratios, we obtain that the altitude has a length of $3 \sqrt{6}$, and that little portion that hangs out has a length of $3\sqrt2$. This is a triangular pyramid with a base of $3\sqrt6, 3\sqrt6, 3\sqrt2$, and a height of $3\sqrt{2}$. Since there are two of these, we can compute the sum of the volumes of these two to be $72$. Now we are left with a triangular prism with a base of dimensions $3\sqrt6, 3\sqrt6, 3\sqrt2$ and a height of $6\sqrt2$. We can compute the volume of this to be 216, and thus our answer is $\boxed{288}$.
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