Question 988671
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The sum of the length and the width of a rectangle is half of the perimeter.  Proof of this assertion is left as an exercise for the student.


For this rectangle, length plus width equals 28, so we can say *[tex \Large l\ =\ 28\ -\ w].


Then we know that the area, which is the product of the length times the width, is 187.  Substituting from above:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ -w^2\ +\ 28w\ =\ 187]


Solve the quadratic.  One of the roots is the width, the other is the length.


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

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