HELP!!!!! This is an integration (calculus) problem; will you please help?

Question

The problem reads: a) Let An 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 2pi/n, show that An = 1/2nr^2(sin 2pi/n). b) Show that lim as n approaches infinity = An = r^2 pi.

Thank you so much for your help. I have no idea how to work this problem. I am totally stumped!!!

Okay, now I can work part b, but I'm still very stumped on how to work part a. I still don't understand.

You people rock!!! THANK YOU SO MUCH!!!!!!!!

Answer 1

area of an isoceles triangle is 2 * right triangle half
area = h*b/2

let the symmetric sides be on the left and right converging at the top. Each with length = r. Inside angle at tip = w (in units of radians)

let the different side be the base at the bottom. Length = b

h = the central height = r*cos(w/2)
b = base length = 2*r*sin(w/2)

area = h*b/2 = (r^2)*cos(w/2)*sin(w/2)

using the following identity:
sin(2x) = 2 sin x cos x

hence
area = h*b/2 = (1/2)*(r^2)*sin(w)

now w = 2pi/n

area of n triangles = (n/2)*(r^2)*sin(2pi/n)

take the limit as n goes to infinity

area = (r^2) * infinity * zero ----- uh oh!

use the series expansion of sin(x)

sin(x) = x - (1/6)*x^3 + (1/120)*x^5 -...

(n/2)*sin(2pi/n) = pi - (pi/6)*(2pi/n)^2 + (pi/120)*(2pi/n)^4 -....

limit as n goes to infinity, all the terms past the first one are zero, only the first term is left which equals "pi"

so total area = (r^2)*pi

Answer 2

For the first part, just use the normal triangle area = (1/2) base times altitude formula. Draw the altitude (a line from the apex of the triangle to the midpoint of the base.) That divides the triangle into two right triangles; you know the hypotenuse (it's the radius of the circle) and the opposite angle (it's half of the 2 pi / n) so basic trig lets you figure out the opposite side using sine (double that to get the base of the original triangle) and cosine gets you the adjacent side, which is the altitude of the original triangle.

For the second part, the limit, you just have to figure out lim n->infinity of n sin(2 pi/n). The only way I'm seeing to do that is via power series (just multiply n through the expansion of sin(2 pi/n), all the terms after the first will clearly go to zero) but you probably haven't gotten to power series yet...

Answer 3

Looking at 1 triangle, it is an isosceles triangle with the congruent sides of length r. The central angle of the triangle is 2pi/n. Using trigonometry, we find that 1/2 the base of the isoceles triangle is rsin(pi/n). The height of the triangle is rcos(pi/n). Thus the area of one triangle is r^2sin(pi/n)cos(pi/n). There are n of these triangles, so n of them have an area An = nr^2sin(pi/n)cos(pi/n)
Using the trig identity sin2x = 2sinx cosx we get

An = 0.5n r^2 sin(2pi/n)

Now take the limit as n approaches infinity.

This is the same thing as the limit as n approaches zero of

0.5 r^2 sin(2pi/n)
------------------------
(1/n)

= pi r^2 * (sin (2pi/n)) / (2pi/n)

The (sin (2pi/n)) / (2pi/n) as n approaches zero is 1

= pi r^2

Answer 4

1st of just think, even those are congruent trianles but each triangle itself is isosceles.
find the other two similar angles of the triangle.
Calculate one angle of the polygon.
Apply polygon formula i. e: one interior angle of a regular polygon ={(n-2)180}/n