Question 438554
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I'll solve the problem in general and then you can apply the solution to your specific case.


If a circle is inscribed in a square, that is the circle is tangent to all 4 sides of the square, then the diameter of the circle and the side of the square are congruent.


Let *[tex \Large d] represent the diameter of the circle.


1.  Since the diameter of the circle is congruent with the side of the square, the area of the square is represented by *[tex \Large d^2].


2.  Since the radius of a circle is half of the diameter, the radius of this circle is *[tex \Large \frac{d}{2}].  And the area of the circle is then *[tex \Large \frac{\pi{d^2}}{4}] (*[tex \Large \pi] times the radius squared)


3.  The desired area is the area of the square minus the area of the circle, so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d^2\ -\ \frac{\pi{d^2}}{4}]


A little algebra:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ d^2\left(1\ -\ \frac{\pi}{4}\right)]


And a little calculator work:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 1\ -\ \frac{\pi}{4}\ \approx\ 0.215]


Hence, roughly 21.5% of the square remains uncovered.  All you need now is the area of your given square and multiply by 0.215]


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