Answer to Question #5329 in Calculus for Jessica

Question

(a) Let A sub n be the area of a polygon with n equal sides inscribed in a circle with radius r. By dividing the polygon into n congruent triangles with central angle 2*(pi)/n, show that A sub n=1/2*(n/r^2)*sin*(2pi/n)

Answer 1

Let S be the area of each of n triangles, then A_n = n*S.
So we have to find
the area of S.
Notice that S is an isosceles trianlge with two sides of
length r.
The angle betwenn these sides is 2*pi/n.
It is known that the
area of a triangle with sides a, b, and angle phi between them is
a*b*sin(phi)/2.
In our case
a=b=r
phi=2*pi/n.
Hence
S = 1/2
* r^2 * sin(2*pi/n)

Therefore
A_n = n*S = 1/2 * n * r^2 *
sin(2*pi/n)