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(Continuation of Exercise $21 . )$a. Inscribe a regular $n$ -sided polygon inside a circle of radius1 and compute the area of one of the $n$ congruent trianglesformed by drawing radii to the vertices of the polygon.b. Compute the limit of the area of the inscribed polygon as$n \rightarrow \infty .$c. Repeat the computations in parts (a) and (b) for a circle ofradius $r .$

a) $\frac{1}{2} \sin \frac{2 \pi}{n}$b) $\pi$c) $\frac{1}{2} r^{2} \sin \frac{2 \pi}{n}$ and $\pi r^{2}$

Calculus 1 / AB

Chapter 5

Integrals

Section 1

Area and Estimating with Finite Sums

Campbell University

Oregon State University

Harvey Mudd College

University of Nottingham

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in case this is another three part question. So it says that we have this inside polygon inside a circle of radius one, and we want to compete. The area of one of the N congruent triangles first get individual. So we have this circle with a radius one, and we're gonna have all of these triangles apologize from picture and keeps going. And there's a me end of these. So burn and just look at one of these triangles here. I'm gonna pull it out here just so we could have a closer look at it. No. Okay, let's say that's this one are here. So we have this, I softly strangle we known both of these. Sir Radius is going to be one, so we can actually just split this into two triangles too. Right? Triangles? Where's my pot? News is going to be equal to one, and then we can find the base in the heights. But if you remember, you're so Catella Rules. So we have So Kito asshole sign gonna be opposite over iPod news and so then are outside to this one. It's just gonna be our pot News times The sign of this angle, which we know from before that that acute angle is just going to equal to pi over two times and cause and is the sides of our polygon. So it's just gonna equal pie over and and we're in our ages, just one. So now we know that our right this sign of pie over and our base, it's co sign of pie over end. And so now we know that the area of our triangle up here, it's just gonna be We have to the zone and put it to you out here since we've again month divide that triangle too, right, triangles. And then we really know that the former and left for your triangle is one half times base times height. So no one half here my height again. Is that sign of pie over in. And then times co sign a pie over in which, if you remember, your trig identities sign Chief data is equal. Teo to sign if they tako science data. Well, we see that we have out here, so this can be simplified, Tio, what we have here. So one half sign we're doing Tuesday buses going to sign of to pop over. And so this is gonna be the answer to the first part of that question. Okay, Snow for part B, we want to compute the limit of the area of the inscribe polygon as an approaches infinity. So before you found area, just triangle. Now we're going to find of the whole pa Leon. So that's just going to be n times the area, the triangle since we have in triangles. So that's just gonna equal and over, too inside of two pi over. And it's not gonna take the limit as an approaches infinity of that. So I know this If I plug in and his poaching Fendi right now here I end up with infinity here. If I put in and hear this whole thing's been go, zero sign of zero goes a zero saiget times here here, which is an indeterminate form. So we can't have that. So we're gonna have to manipulate this equation. I see how the sine of to pry over end right here and if you remember also from you're right tricks with limits, the limit as Expro jizz Mindy, A sign of X over X is just equal to once I want to try and create that on this side someone and go ahead and multiply this by what I have inside signed right now, so to pry over end over t pire end. So when I do that when I did two prior end times this I see my end cancels out, my two cancels out some just left with the pie. And then I have exactly what I wanted over here. So sign of Sioux pyre and divided by two pie or n So then I take that limit, this whole thing, it's just going to go to one side. Then I just end up with pie. Then for the last question is asking us to repeat the computations in part and b for a circle of radius are. So if we notice over here, that's just gonna change. Where are partners is, so this would have been our instead. And since I have that for both of sign and co sign for my base in height, that just means I'm going to be multiplying each area by r squared so that in my area of the triangle, then become one half r squared times. Sign of two pi over end in the area of the polygon. Well, then be exactly that. T just multiplying by r squared. And when I take the limit, it's an approaches infinity of that using the same trick just before we're going to find that that's just going to be pi r squared, which, if you notice that is exactly the area of the circle.

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