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.
Find the area of the rectangle.
2(a+b)=40a=402−b102⋅√22=a2+b2200=400−40b+2b2b2−20b+100=0b=10±√100−100b=10a=10A=ab=10⋅10A=100
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