What is the area of the polygon whose vertices are the points of intersection of the curves $x^2+y^2=25$ and $xy=12$?

1. From $x^2+y^2=25$ and $xy=12$: $(x+y)^2=25+24=49$ and $(x-y)^2=25-24=1$, so $x+y=\pm7$ and $x-y=\pm1$.
2. The four solutions are $(4,3),(3,4),(-3,-4),(-4,-3)$.
3. The polygon is a rectangle. Its sides are $\sqrt{(4-3)^2+(3-4)^2}=\sqrt2$ and $\sqrt{(3+4)^2+(4+3)^2}=7\sqrt2$.
4. The area is $\sqrt2\cdot7\sqrt2=\boxed{14}$.