Area of a polygon
Hello, I was wondering if anyone can help me with this problem.
A) Let A_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 of (2pi)/n, show that A_n = [n(r^2)sin(2pi/n)]/2.
B) Show that limit as n goes to infinity A_n = (pi)(r^2).
For this one, I think that you can use the fact that (sin x)/x = 1
Appreciate it if anyone can contribute anything.
You want to find,
Originally Posted by Recklessid
lim n--> oo (1/2)n(r^2)*sin(2pi/n)
r^2*lim n--> oo (1/2) (1/n)(r^2)*sin(2pi n)
Multiply numerator and denominator by pi to get,
r^2 *lim n--> oo pi*(1/2pi*n)*sin(2pi n)
But lim n--> oo sin(2pi n)/(2pi n) = 1.