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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 10, 2024
 #1
avatar+1334 
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Let's let the two sides of the rectange be a and b. 

 

From the problem, we have

\(2a + 2b =40 \\ a + b =20 \\ \)

Now, we just sqaure both sides to get

\(a^2 + 2ab + b^2 = 400 \\ a^2 + b^2 + 2ab = 400\)

 

From here, we have

\(a^2 + 2ab + b^2 = 400 \\ a^2 + b^2 + 2ab = 400 \\ a^2 + b^2 = (10\sqrt{ 2})^2 \\ a^2 + b^2 = 200\\ 200 + 2ab = 400 \\ 100 +ab = 200 \\ ab = 200 - 100\)

 

So our answer is 100. 

 

Thanks! :)

 Jun 10, 2024

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