Question 350235
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The perimeter of a rectangle is given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P\ =\ 2l\ +\ 2w]


So if the perimeter is 34, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2l +\ 2w\ =\ 34]


Which is to say


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ +\ w\ =\ 17]


and furthermore:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l\ =\ 17\ -\ w]


The diagonal of a rectangle forms the hypotenuse of a right triangle where the legs are the length and width of the rectangle.  So, Mr. Pythagoras says, if the hypotenuse is 13 and the legs are *[tex \Large l] and *[tex \Large w], then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ l^2\ +\ w^2\ =\ 13^2]


Substituting:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ (17\ -\ w)^2\ +\ w^2\ =\ 169]


Expand, collect like terms, and factor to solve for *[tex \Large w].  One of the roots will be the length and the other will be the width.


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