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In the $ABCD-PQRS$ parallelepiped shown, $AH=\sqrt{39}$, $HS=\sqrt{57}$. Calculate its volume ($H$ belongs to base).
(Answer:$V=220,5\sqrt{13}$)
My progress:
I can only discover AS
$SAS^2 =39+57 \implies AS = \sqrt{96} \therefore AS = 4\sqrt6$
but i can't find the base area..
Missing any informa...