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Problem 32

(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 $ 2\pi/n $, show that
$$ A_n = \frac{1}{2} nr^2 \sin \biggl( \frac{2 \pi}{n} \biggr) $$

(b) Show that $ \displaystyle \lim_{n \to \infty} A_n = \pi r^2 $. [$ Hint: $ Use Equation 3.3.2 on page 191.]

Answer

(A). $A_{n}=\frac{1}{2} n r^{2} \sin \left(\frac{2 \pi}{n}\right)$
(B). $\pi r^{2}$



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Video Transcript

So here we use untrue I and goes to approximate a polygon to approximate area of the circle. So here, so it should eyes try. Go on, look at this. This one has radius. Are this wise radius? Are this is one over the the horse Sir Ho sang Go So the whole circle goes to part over Wei have learned from the high school that the area of the triangle this one over two length of this one times Lance off this one time sigh of this and go Oh, so it's not yet also distressed area of the triangle is one over two times dance of this side and lands of this side Time side off the angle to pi over and the alien is simply trying goes at it together Therefore he gives us up This formula and computer are limit on fraud on course the infinity here that is this so here we still have to take limit here We put on over to pie here that we put pie are square here and we have side you pie over this just below there How to brought shake the limit here on this term This part it is. Actually, we actually use a famous result from calculus. What unconsciously infinity? This argument this too high over and goes to zero. And this can be view us divided by two pi over. So we used the famous result that science over hacks the limit of ab support zero give us one So this part of the limit will goes to one and we're left with pi r squared.

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