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What is the area, in square units, of the square with the four vertices at $(-2, 2), (2, -2), (-2, -6)$, and $(-6, -2)$?

 Feb 25, 2021
 #1
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Since $dist((-2,2), (2,-2))=\sqrt{4^2+4^2}=4\sqrt{2}$, the area is $(4\sqrt2)^2=\boxed{32}$.

 Feb 25, 2021

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