What is the area, in square units, of the square with the four vertices at $(-2, 2), (2, -2), (-2, -6)$, and $(-6, -2)$?
Since $dist((-2,2), (2,-2))=\sqrt{4^2+4^2}=4\sqrt{2}$, the area is $(4\sqrt2)^2=\boxed{32}$.
+205 
+600 
