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Show that 4 rectangles of dimensions a * b can fit in a square with side length a + b.

 Jul 15, 2021

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 #1
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The area of 4 rectangles of dimensions a*b has an area of 4ab, while a square with side length a+b has an area of (a+b)^2=a^2+2ab+b^2. We want to show:

\(4ab\le a^2+2ab+b^2\\0\le a^2-2ab+b^2\\ 0 \le (a-b)^2\)

Since \((a-b)^2\) will always be nonnegative for real numbers a and b, it will always be greater or equal to zero. \(\square\)

 Jul 15, 2021
 #1
avatar+420 
+1
Best Answer

The area of 4 rectangles of dimensions a*b has an area of 4ab, while a square with side length a+b has an area of (a+b)^2=a^2+2ab+b^2. We want to show:

\(4ab\le a^2+2ab+b^2\\0\le a^2-2ab+b^2\\ 0 \le (a-b)^2\)

Since \((a-b)^2\) will always be nonnegative for real numbers a and b, it will always be greater or equal to zero. \(\square\)

textot Jul 15, 2021
 #2
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ahhh, i see now, thanks!

Guest Jul 15, 2021
 #3
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Very nice, textot   !!!!!!

 

 

cool cool cool

CPhill  Jul 15, 2021

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