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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.

 Mar 16, 2024

Best Answer 

 #1
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\(\sqrt{200} = 10\sqrt2\), so \(a^2 + b^2 = 200\) 

 

Assuming that a and b are both integer side lengths, \(10^2+10^2=200\), so the sides are 10 and 10.

 

10 times 10 is 100, so the area of the rectangle is \(\bf100\)

 Mar 16, 2024
 #1
avatar+34 
+1
Best Answer

 

\(\sqrt{200} = 10\sqrt2\), so \(a^2 + b^2 = 200\) 

 

Assuming that a and b are both integer side lengths, \(10^2+10^2=200\), so the sides are 10 and 10.

 

10 times 10 is 100, so the area of the rectangle is \(\bf100\)

jilin73 Mar 16, 2024

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