The vertices all satisfy $(x+y)^{2}=x^{2}+y^{2}+2 x y=73+2 \cdot 24=121$, so $x+y= \pm 11$. Similarly, $(x-y)^{2}=x^{2}+y^{2}-2 x y=73-2 \cdot 24=25$, so $x-y= \pm 5$. Thus, there are four solutions: $(x, y)=(8,3),(3,8),(-3,-8),(-8,-3)$. All four of these solutions satisfy the original equations. The quadrilateral is therefore a rectangle with side lengths of $5 \sqrt{2}$ and $11 \sqrt{2}$, so its area is 110.