jmc

Let the dimensions of the rectangle be $a$ and $b.$ Since $a$ and $b$ are the legs of a triangle with hypotenuse 2, $a^2+b^2=4.$



Then by AM-GM, $$4=a^2+b^2\ge 2ab,$$so $ab\le 2.$

Equality occurs when $a=b=\sqrt{2},$ so the largest possible area is $\boxed{2}.$