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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.

 Jun 1, 2024
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First, from the first part of the problem, we know that

\(2x + 2y = 40 \\ x + y = 20 \\ y = 20 - x\)

 

Now, we use the fact that the diaganol is 10 sqrt2, we have

\(x^2 + y^2 = (10 sqrt 2) ^2 \\ x^2 + ( 20 - x)^2 = 200 \\ x^2 + x^2 - 40x + 400 = 200 \\ 2x^2 - 40x + 200 = 0 \\ x^2 - 20x + 100 = 0\)

\((x -10)^2 = 0 \\ x - 10 = 0 \\ x = 10 \\ y = 20 - 10 = 10\)

 

The area is just 10*10 = 20. 

 

Thanks! :)

 Jun 2, 2024

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