
#1
Oct604, 06:12 PM

P: 4

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=(1/2)nr^2sin(2pi/n).
B)Show that the limit as n approaches infinity = pir^2. Now, for part A, I don't understand how you can take the sin of the angle, because it is not a right triangle. And even if you divided the triangle in half, wouldn't it be more beneficial to take the tan, because then you would be getting the base and height of the triangle? I just need help on where to start this problem. 



#2
Oct604, 06:36 PM

Emeritus
Sci Advisor
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P: 11,154

Yes, you divide the triangle in half. But it is the sine you want  the hypotenuse of the triangle is the radius, r.




#3
Oct604, 06:53 PM

Emeritus
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P: 16,101

A good place to start is to derive the formula for the area of a triangle.
Notice that, for each of the triangles, you know three pieces of information: two of the sides, and their included angle. Your thoughts about splitting the triangle into two parts (each right triangles) is a good one... but you have to do it right. You're given one of the angles, so it wouldn't make sense for that angle to be the one you split up... 



#4
Oct604, 09:02 PM

P: 4

Areas and distances
Alright, thanks. I'm so used to integrating from highschool calculus, so now that we are not supposed to know how in my calc 125 class, I'm not really sure what I'm doing anymore. But I've worked it out by splitting the triangle in half, keeping the angle of 2pi/n, and I figured out how it works.



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