for his correct solution to this week's POTW, which was Problem 307 in the MAA Challenges. His solution follows:
By the converse of Pythagoras’ theorem, the triangle $BCD$ is a right triangle with hypotenuse $DB$; the area of that triangle is equal to 6.
From $A$, we draw the perpendicular to the plane $BCD$, intersecting the plane at $O$; $AO$ is the altitude of the pyramid relative to the base $BCD$.
The right triangles $AOB$, $AOC$, and $AOD$ are congruent, since they share the side $AO$ and their hypotenuses are equal. This shows that $BO=CO=DO$, and $O$ is the center of the circumcircle of the triangle $BCD$. As that triangle is a right triangle, $O$ is the midpoint of the hypotenuse $DB$, and $BO=\dfrac52$. This gives:
and the volume of the pyramid is equal to: