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.

Quote:

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Originally Posted by Recklessid

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.

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You want to find,
lim n--> oo (1/2)n(r^2)*sin(2pi/n)

Instead do,

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.

Thus,
pi*r^2