Question 620559
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If the perimeter of the square is *[tex \LARGE a], then the measure of a side has to be *[tex \LARGE \frac{a}{4}], so the area must be *[tex \LARGE \left(\frac{a}{4}\right)^2], but the area is *[tex \LARGE b], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{a}{4}\right)^2\ =\ b]


But since *[tex \LARGE a\ =\ b],


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \left(\frac{a}{4}\right)^2\ =\ a]


Solve for *[tex \LARGE a], then calculate *[tex \LARGE \frac{a}{4}] so that you have the measure of one of the sides of the square.  Then use Pythagoras to calculate the measure of a diagonal or use the fact that the diagonal of a square measuring *[tex \LARGE s] on a side measures *[tex \LARGE s\sqrt{2}].


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