Question 1121378
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Perimeter is *[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l\ +\ 2w\ =\ 100]


Half the perimeter is *[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ +\ w\ =\ 50]


Hence *[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 50\ - w]


Area is *[tex \LARGE \ \ \ \ \ \ \ \ \ \ lw\ =\ 25]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (50\ -\ w)w\ =\ 25]


You can put it into whatever form you like.


By the way, Jogsarithmetic is right, you don't have a square, you have a rectangle.  A square with a perimeter of 100 has an area of 625, not 25.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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