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November 27, 2014

November 27, 2014

Posted by **kaykay** on Sunday, January 9, 2011 at 12:51pm.

- math -
**Ms. Sue**, Sunday, January 9, 2011 at 1:11pm - math -
**tchrwill**, Sunday, January 9, 2011 at 3:47pmConsidering all rectangles with the same perimeter, the square encloses the greatest area.

Proof: Consider a square of dimensions x by x, the area of which is x^2. Adjusting the dimensions by adding a to one side and subtracting a from the other side results in an area of (x + a)(x - a) = x^2 - a^2. Thus, however small the dimension "a" is, the area of the modified rectangle is always less than the square of area x^2.

- math -
**Anonymous**, Wednesday, May 4, 2011 at 8:25pm20x15

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