Question 987371
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When the wire is cut, the length of one piece will be the circumference of a circle of radius *[tex \Large r], which is to say *[tex \Large 2\pi{r}].  The length of the other piece will be the perimeter of a square of side length *[tex \Large s], which is to say *[tex \Large 4s].  The measures of these two pieces sum to 360 inches according to the problem statement.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2\pi{r}\ +\ 4s\ =\ 360\ \ \ \ \ ] (1)


The area of the circle of radius *[tex \Large r] is *[tex \Large \pi{r}^2] and the area of the square of side *[tex \Large s] is *[tex \Large s^2].  These two areas are equal according to the problem statement.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \pi{r}^2\ =\ s^2\ \ \ \ \ \ ] (2)


Solve the system for *[tex \Large s] and *[tex \Large r].  Then calculate *[tex \Large 4s] and *[tex \Large 360\ -\ 4s]


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

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