Express the area of a regular n-gon in terms of its perimeter P, using trigonometry.

Question

Express the area of a regular n-gon in terms of its perimeter P, using trigonometry. What happens to this area as n tends to infinity?

It might be helpful to explore an example regular polygon or two first before generalizing to the area of any regular n-gon in terms of its perimeter and using trigonometry. Specifically, use trigonometry to find the area of a regular hexagon or octagon (using trigonometry and getting your final answer to be in terms of the regular hexagon or octagon's perimeter). Then, use your example to help you generalize the process and solution for any regular n-gon.

Answer 1

draw one of the triangles of the n-gon; they are all congruent and all are isosceles triangles

draw the altitude from the center of the n-gon perpendicular to one side

call the altitude "h"

the altitude bisects the central angle from where it was drawn

call one of the angles (half of the bisected central angle) θ

θ=(1/2)(360/n)=(180/n) where n=# of sides

find the altitude "h"

tan(θ)=(s/2)/h where "s" is the length of one of the sides of the n-gon (the altitude bisects the side)

h=(s/2)/tan(θ)

h=s/2tan(θ)

A=(1/2)(b)(h)

A=(1/2)(s) (s/2tan(θ))

A=(s^2)/4tan(θ)

since there are n number of triangles, multiply this area by "n"

A=(n)(s^2)/4tan(θ)

A=(ns)(s)/4tan(180/n) because θ is (1/2)(360/n)

Perimeter=sn where "s" is the length of any one of the sides and "n" is the number of sides

substitute P(perimeter) for sn

A=Ps/4tan(180/n)

Answer 2

P = na; where a is side. So a = P/n; r is aphothem, r = a/2cot(180°/n); r = Pcot(180°/n)/(2n)

Area A = nar/2 = nP/nPcot(180°/2)/(4n) = P²cot(180°/n)/(4n)