Question 438030
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Yep.  30-60-90 triangle.  So if the short side of the rectangle is *[tex \Large x], the hypotenuse of the triangle (that is to say the diagonal of the rectangle) has to be *[tex \Large 2x] and the long side of the rectangle has to be *[tex \Large x\sqrt{3}].


Since the area of the rectangle has to be length times width or *[tex \Large x^2\sqrt{3}], we can say:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2\sqrt{3}\ =\ 64\sqrt{3}]


Just solve for *[tex \Large x]


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