Question 591961
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In order to be certain that your quadrilateral actually has an apothem, then it must be a regular quadrilateral, in other words, a square.  I'm also presuming that *[tex \LARGE 8\sqrt{2}] is the circumradius, i.e. the distance from the center of the square to any vertex.  Given those assumptions, the apothem of a regular *[tex \LARGE n]-gon is found by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ r\,\cdot\,\cos\left(\frac{180}{n}\right)]


where *[tex \LARGE r] is the circumradius and *[tex \LARGE n] is the number of sides.


Now that you know both the circumradius and the inradius (same thing as the apothem) you have two methods to compute the area of a regular *[tex \LARGE n]-gon:


Using the circumradius, *[tex \LARGE R]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Area\ =\ \frac{R^2n\sin\left(\frac{360}{n}\right)}{2}]


Using the inradius (apothem), *[tex \LARGE A]:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ Area\ =\ A^2n\tan\left(\frac{360}{n}\right)]


Just plug in your values and do the arithmetic.


Hint:  *[tex \LARGE \cos(45^\circ)\ =\ \sin(45^\circ)\ =\ \frac{\sqrt{2}}{2}], *[tex \LARGE \sin(90^\circ)\ =\ 1], and *[tex \LARGE \tan(45^\circ)\ =\ 1]



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