The perimeter of a rectangle is $40,$ and the length of one of its diagonals is $10 \sqrt{2}.$ Find the area of the rectangle.
\(\sqrt{200} = 10\sqrt2\), so \(a^2 + b^2 = 200\)
Assuming that a and b are both integer side lengths, \(10^2+10^2=200\), so the sides are 10 and 10.
10 times 10 is 100, so the area of the rectangle is \(\bf100\)
+908 
