2008 AMC 12B Problem 18

The square has side $\sqrt{196} = 14.$ Let $F$ and $G$ be the feet of the perpendiculars from $E$ to $AB$ and $CD.$ Then $FG = 14,$ $EF = \tfrac{2 \cdot 105}{14} = 15,$ and $EG = \tfrac{2 \cdot 91}{14} = 13.$ Triangle $EFG$ lies in a plane perpendicular to the base, so its altitude to $FG$ is the pyramid’s height. By Heron’s formula with $s = 21,$ its area is $\sqrt{21 \cdot 6 \cdot 8 \cdot 7} = 84,$ so the altitude to $FG$ is $\tfrac{2 \cdot 84}{14} = 12.$ The volume is $\tfrac13 \cdot 196 \cdot 12 = 784.$ Thus, the correct answer is E.