Question 978105
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The perimeter is two times the length plus 2 times the width.  Therefore the sum of the length and the width is half of the perimeter.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ +\ w\ =\ 31]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 31\ -\ w]


The area is the length times the width, so in terms of just the width, the area is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 31w\ -\ w^2\ =\ 220]


or, in standard quadratic form:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ -\ 31w\ +\ 220\ =\ 0]


When you solve this factorable quadratic, you will get two positive real roots.  One value is the length and the other value is the width.  


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it

*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \