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.
\(2(a+b)=40\\ a=\frac{40}{2}-b\\ 10^2\cdot \sqrt{2}^2=a^2+b^2\\ 200=400-40b+2b^2\\ b^2-20b+100=0\\ b=10\pm \sqrt{100-100}\\ b=10\\ a=10\\ A=ab=10\cdot 10\\ \color{blue} A=100\)
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+70 
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