Volume of a given parallelepiped

Question

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

$AS^2 =39+57 \implies AS = \sqrt{96} \therefore AS = 4\sqrt6$

but I can't find the base area..

Missing any information?

Answer 1

Orthogonally project $PA$ onto the base plane $PAsin(a)=PH$ then project it orthogonally onto $PS$ ($PAsin(a)cos(30°)$) which is the rectangular component of $AP$ in the direction of $PS$. Now that same component is achieved if you project $PA$ orthogonally giving $PAcos(53°)$.

Let $\angle HAP = a$

$PA\cdot sin(a)cos(30°)=PA\cdot cos(53^o)\\ sin(a)=\frac{cos(53^o)}{cos(30°)}=\frac{\frac{3}{5}}{\frac{\sqrt3}{2}}=\frac{2\sqrt3}{5}\\ \therefore sec(a)=\frac{5}{\sqrt13}\\ PA =AH sec(a)=5\sqrt3\\ \triangle PAS (PA =5\sqrt{3}, AS=\sqrt{96})\implies PS=\sqrt{73}\\ V =(\sqrt{73})\cdot2sen(60^o)\cdot\sqrt{39}=220,5\sqrt{13}$

(By hMendez)