24-a-let-a_n-be-the-area-of-a-polygon-with-n-equal-sides-inscribed-in-a-circle-with-radius-r-by-divi

Question

24. (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
$$\begin{array}{c}{A_{\kappa}=\frac{1}{2} n r^{2} \sin \left(\frac{2 \pi}{n}ight)} \ {	ext { (b) Show that } \lim _{x ightarrow \infty} A_{n}=\pi r^{2} \cdot[H i n t : 	ext { Use Equation } 2.4 .6} \ {	ext { on page } 133 .}\end{array}$$

Fuzail Shakir · Accepted Answer

Okay, So if we have any equal scientific radius are we have to show that the area is equal to 1/2 and our square times sign off to pi over and so this fairly easy to show. So let&#x27;s suppose we have a triangle with three sides. Okay, so therefore, the area of the triangle with three sides has the equal 1/2 1/2 times, three times the really square times sign off to high over three. Simply because if you said this is equal to the radius and then the sine of the angle is thehe mount of the circle that is being covered so we could set this The amount of circle that it&#x27;s being covered and we have pie are square to be the area of the whole circle. So you basically can get an area of the amount of the circle there is being covered multiplying by. Of course, you have to cut it off because if this is a circle, you&#x27;re not counting the red part. So therefore this theory must be approved

Frank Lin · Answer

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&#x27;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&#x27;re left with pi r squared.

Ayman Sherif · Answer

Okay. Hi. And this problem, we have a circulates Inside the circle is appointment of equal size, which can be by this. This is a center of the circle. This parliament is divided into a number of equal triangles. For example, this is this is one of this is one thing This is the next is the second language. This is the same triangle and support string this is this site is or And this site is also which is a sent a radius of the second. This angle is the central anger which is off. If you want to get the area off this triangle, we&#x27;ll be acquitted. Video, I hope deploy. But Roy Oh, off deploy swing is equal to boy over. And where n is the total number of triangles in sizable in this case with before so But we want to get the area of the overall volume. So you way will not deploy this area by end, which is a number off trains it so in which, which is a total area of the Taliban with the equal or squared Uh, sign Do boy deploy, which is the first requirement, is the second requirement. So we we knew that the area of the bargain will be equal and it will deploy or square deployed and oy, sign to over in. We want to take limit to this equation so limit from then turns to infinity. Half are square is a constant eyes is very circus constant So it will be equal hope squared limit when tends to infinity Oh, look it sighing to over in We will take this equation and multiply it mainly by one. But in our case it will be too boy over do for So this limit will become I will show in exploit to will deploy and over to boy must deploy sewing Cool boy boy over in if we talk If we take this turn on and make it a diminished denominator every year it would become equals to guard no deploy toe over and over to boy over in Let system be seated which is are some thrilling So in the next flight limit when end girls 24 in will be equal Oh squared deploy limit when sita zero Because what we&#x27;re increasing the number and the sita or the central angle will decrease gradually until it nearly become zero Flemington. See? It tends to zero toe boy Want deploy saying see you seek mint. We will take the two boy outside so exploit limit and well in turns to infinity will be acquitted. Pool off, squared, deploying to boy Look, Chloe, limit tens zero So I&#x27;m so in Saint Overseas. The tour Big gun with the two. So it will be equal. Sister is one. This time would be equipped to one because we see the term is 20 signs. It was one through This time is one. So this term will be equal next light limit and then turns to infinity. Boy R squared one. She could Boy. Oh, squish. This is a quote on This is the overall area of the city. This is the way as they used toe proof as overall area of the second as shown, uh, isn&#x27;t it you? Thank you.

Andrew Bassila · Answer

okay. And this problem, they&#x27;re talking about the area of a regular polygon, and they give us three task. But let&#x27;s just first name what things are and is the number of sides. All right, and then X is going to be the length upsides. I just want to clarify that before moving on. Why aren&#x27;t you writing for me? And in that length of sides? Here we go. Right. So the first part, they say, to rewrite in terms of one trig function, um, we can do that for sure. That&#x27;s quite easy to do. Co sign over Sign is co tangent. So I&#x27;m going to rewrite Well, I guess I should write a a is going to be equal to this whole thing and x squared over four times co tangent pi over four. Using a ratio identity that we&#x27;ve talked about in the section Um, the next part they want us to verify, um, when that we&#x27;re dealing with a square that has side length of eight each. Cool. We can do that. Um, well, we know in the area of a square with eight, um, side side length is eight times eight. So that&#x27;s going to be 64. So this is four square, right? And then we&#x27;re gonna use the polygon formula toe, help us find this. Okay, So, um, let&#x27;s plug in what we know, um, four square and is four and X is going to be eight. So this is part B. It&#x27;s just my little side over here. Um, I knows the answer should be 64 that&#x27;s what we&#x27;re doing. We&#x27;re verifying this. So a is going to be four times eight squared over four times co Tangent Group&#x27;s Co Tangent of Pi over four. All right. And the great thing about this is the force cancel and I&#x27;m left with Eight Squared, which is 64. And then the cool thing about co tangent that&#x27;s the reciprocal of tangent. Tangent of Higher before is going to be signed over. Co sign co Tangent is co sign over sign and we know co sign a pie before is a route to over to and sign a pyre before his Route 2/2. So this actually holding is one. And this does indeed equal 64. So it matches what? We expected it to you. So we&#x27;ve verified it or we&#x27;ve demonstrated, I guess, is a better word to show that this is indeed going toe work. And with that information, we&#x27;re then going to say, Well, you know what? Let&#x27;s find the area of a dodecahedron. I&#x27;m sorry I don&#x27;t deck a gun that has 12 sides, so that&#x27;s gonna be N is equal to 12. And the side links are going to be 10. So now, using this formula, we confined the area of these really cool polygon shapes. Um, that&#x27;s going to be a is equal to, um, 12 times 10 squared all over four times co tangent of pyre before. The great thing about this is this is just going to be one. It&#x27;s gonna be the same. All right, So, um, what we have here is 12 times 100 over four, and that&#x27;s going to be 12 times 100 over over four. That&#x27;s 300 and the square units meters meters squared. Okay. And that&#x27;s the three tasks for this problem.

Elizabeth Xu · Answer

so here, regular in that and as you go to three and R is four inches, that means our area is equal to 1/2 times, three times four squared times. Sign of two pi over three, which if we put into a car litters, you&#x27;ll get this is equal to 20.7 eight inches squared.

Andrew Bassila · Answer

all right, This problem, They&#x27;re talking about a polygon that&#x27;s has a circle inscribed in it. And I just want the only thing I want to add is talking about what n and r. R n is. The number of sides and are is the radius of the inscribe circle. All right, so the first part asks us to rewrite it so that we have only one trig function, and that&#x27;s kind of actually easy. I&#x27;m gonna rewrite this and recognize that sign of something over coastline of that same something is actually a fancy way of saying tangent of that something. So now we&#x27;ve simplified that. The second part says we&#x27;ll do it with the square with that has a radius of four meters. So circles being inscribed into a square. The radius of that circle is four meters. And, um so let&#x27;s just talk about us. Um, the area of this square, we know the area of squares. The area of a square is side squared, right? So if the this case, it tells us that the radius is four. If R equals four, then that means the side is the entire diameter, right? So it&#x27;s gonna be eight. So in this case, hate squared is going to be 64. So that&#x27;s the answer that we&#x27;re expecting to get. Now, let&#x27;s verify this because the square is a polygon. Let&#x27;s verify it using this new formula, which is a little extra. But, you know, it works for, uh, polygons with more than four sites. So here we have foresighted polygon. The radius is four that&#x27;s going to be squared and then we&#x27;re gonna being tangent of pi over four. All right, all ends become four. And the cool thing here is that tangent of pyre before is just one, and then this is actually four cubed or 16 times for which is 64. So that kind of agrees with what we expected to see with a square What they&#x27;re asking in The last part is they want us to use a 12 sided object. Um, that&#x27;s uses the same circle of radius four. Okay, so a 12 sided object will be area of, uh, FIS. We will have 12 sides. Um, the theater is the number of side. I&#x27;m sorry. The radius is going to be four, and that will be squared. And then this is the interesting part. It&#x27;s going to tangent of Pi over 12. Okay, so this this number&#x27;s gonna not be a nice, neat number. That&#x27;s okay. Um, looking at this, we have 12 times. 16 So 12 times 16 that&#x27;s four square. That&#x27;s gonna be 1 92 times Tangent of pi over 12 and we&#x27;re gonna use our calculator for this. But please, please make sure pre calculus is that one course where you&#x27;re gonna be switching between degrees and radiance a lot. So make sure you are in radiance when you type it in this way. So I&#x27;m doing that now. Have tangent a pi over 12 and I&#x27;m just gonna use a decimal approximation. It&#x27;s It&#x27;s about, um, 0.268 rounding to three decimal places. So I have 1 92 times that number, and I get that. The area is about 51.446 square meters. All right? And that&#x27;s part C of this problem.

Problem 24 Medium Difficulty

Answer

a) $$
A_{n}=\frac{1}{2} n r^{2} \sin \left(\frac{2 \pi}{n}\right)
$$
b) $$
\lim _{n \rightarrow \infty} A_{n}=\pi r^{2}
$$

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James Stewart

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Grace H.

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a) $$A_{n}=\frac{1}{2} n r^{2} \sin \left(\frac{2 \pi}{n}\right)$$b) $$\lim _{n \rightarrow \infty} A_{n}=\pi r^{2}$$

Biocalculus Calculus for the Life Sciences

Grace H.

Catherine R.

Michael J.

Joseph L.

Integration Review - Intro

Definite vs Indefinite

James StewartBiocalculus Calculus for the Life Sciences

Grace H.

Catherine R.

Michael J.

Joseph L.

a) $$
A_{n}=\frac{1}{2} n r^{2} \sin \left(\frac{2 \pi}{n}\right)
$$
b) $$
\lim _{n \rightarrow \infty} A_{n}=\pi r^{2}
$$

James Stewart

Biocalculus Calculus for the Life Sciences