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Question

If the number of sides of two regular polygons having the same perimeter be n and 2n respectively, prove that their areas are in the ratio 
2 cos ($$\pi$$/n) : [1 + cos ($$\pi$$/n)].


Solution

Let p be the perimeter of both the polygons . Then each sides of first polygon is of length (p / n ) and that of the second polygon is (p / 2n) . If $$A_1 , A_2 $$ demote their areas then by 
$$A_1 \, = \, \dfrac {1}{4} n . \dfrac {p^2}{n^2} \, cot \dfrac {\pi}{n}$$
and $$A_2 \, = \dfrac {1}{4} \, \cdot \, 2n \, \cdot \, \dfrac{p^2}{4n^{2} }\, \cdot  \, cot \dfrac {\pi }{2n }$$
$$\therefore \, \dfrac {A_1}{A_2}\, = \, \dfrac {2 cot \dfrac {\pi}{n}}{cot {\dfrac {\pi}{2n}}} \, = \, \dfrac {2 cos \dfrac {\pi}{n}sin \dfrac {\pi}{2n}} {sin \dfrac {\pi}{n}cos \dfrac {\pi}{2n}}$$
$$= \, \dfrac {2 cos \dfrac {\pi}{n}sin \dfrac {\pi}{2n}} {2sin \dfrac {\pi}{2n}cos \dfrac {\pi}{2n}cos \dfrac {\pi}{2n}}$$
$$= \, \dfrac {2 cos \dfrac {\pi}{n}} {2cos^2 \dfrac {\pi}{2n}} \, = \, \dfrac {2 cos \dfrac {\pi}{n}} {1 \, + \, cos \dfrac {\pi}{2n}} $$

Maths

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