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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.

 May 23, 2024
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Let's let the two sides of the rectange x and y. 

 

We can use the problem to write the equation

 

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

 

Isolating y, we have \(y = 20 - x\)

 

Subsituting this value for y into the second part of the qeustion and using the pythaogrean theorem, we have

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

 

Finally, we find that we have x = 10. From the first equation, we find that y = 10 as well. 

 

The area is just 10*10=100!

 

Thanks! :)

 May 23, 2024

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