Question 534439
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The perimeter of a rectangle is given by:


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


From that we can conclude for this rectangle that:


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


Then, since the area, given by the length times the width, is 60,


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ w^2\ -\ 17w\ +\ 60\ =\ 0]


Solve the quadratic to get the length and width.  Then use Pythagoras to get the measure of the diagonal -- or note that the length and width are 2 values in a Pythagorean triple.


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