Question 991997
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If the length is half of the width and the area is 25, then *[tex \Large w^2\ =\ 50].  Solve for *[tex \Large w], calculate the length, then use the width and the length in the perimeter formula.


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

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