Question 198939
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A regular *[tex \Large n]-gon has an area given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A = \frac{n}{4}s^2\cot\left(\frac{\pi}{n}\right)]


So if two similar regular polygons have a scale factor of *[tex \Large 5:2] then  for a unit length of *[tex \Large k], the measure of the side of the smaller polygon is *[tex \Large 2k] and the measure of the side of the larger is *[tex \Large 5k] and the areas of the two *[tex \Large n]-gons are:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_s = \frac{n}{4}(2k)^2\cot\left(\frac{\pi}{n}\right)=4\left(\frac{n}{4}k^2\cot\left(\frac{\pi}{n}\right)\right)]


and


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_l = \frac{n}{4}(5k)^2\cot\left(\frac{\pi}{n}\right)=25\left(\frac{n}{4}k^2\cot\left(\frac{\pi}{n}\right)\right)]


So now we can say that:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{A_s}{4} = \frac{A_l}{25}]


But we are given that *[tex \Large A_l = 100], so:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{A_s}{4} = \frac{100}{25}=4]


So:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_s = 16]


In general, two similar *[tex \Large n]-gons whose scale factor proportion is *[tex \Large a:b] have areas that are in proportion *[tex \Large a^2:b^2].  Using this general rule, the area proportion is *[tex \Large 25:4] which is equivalent to *[tex \Large 6.25:1] and *[tex \Large 100 \div 6.25 = 16].


Sorry, but *[tex \Large \pi] has nothing to do with this.  Shrug it off and go have a piece of cake.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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