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.
Let a,b be the sides
2 ( a + b) = 40
a + b = 20
b = 20 - a
a^2 + b^2 = (10 sqrt 2)^2
a^2 + (20 - a)^2 = 200
a^2 + a^2 -40a + 400 = 200
2a^2 -40a + 200 = 0
a^2 -20a + 100 = 0
(a - 10*2 = 0
a =10
b =20 -10 =10
Area = a*b = 10 * 10 = 100
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