Question 982718
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If the rectangle is inscribed in a circle, a diagonal of the rectangle is a diameter.  So if the sides are *[tex \Large x] and *[tex \Large 2x], then the diagonal is *[tex \Large \sqrt{x^2\ +\ (2x)^2}\ =\ x\sqrt{5}], so the circumference, being *[tex \Large \pi d], is *[tex \Large x\pi\sqrt{5}]


However, since the sides are *[tex \Large x] and *[tex \Large 2x] and the area is 32,


*[tex \LARGE \ \ \ \ \ \ \ \ \ 2x^2\ =\ 32\ \Leftrightarrow\ x\ =\ 4]


So substitute into *[tex \Large x\pi\sqrt{5}] and the circumference is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 4\pi\sqrt{5}]


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

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