Area of a polygon inscribed in a circle with 'n' sides

[Idealistic]

a) Let A[sub:2o6jy2vq]n[/sub:2o6jy2vq] be the area of a polygon with n sides inscribed in a circle with a radius of r. By dividing the polygon iinto n congruent triangles with central angle 2pi/n , show that:

A[sub:2o6jy2vq]n[/sub:2o6jy2vq] = (1/2)nr[sup:2o6jy2vq]2[/sup:2o6jy2vq]sin(2pi/n)

b) Show that the lim[sub:2o6jy2vq]n -> inf[/sub:2o6jy2vq] A[sub:2o6jy2vq]n[/sub:2o6jy2vq] = pir[sup:2o6jy2vq]2[/sup:2o6jy2vq]

Am I suposed to divide the top equation by (1/2)r[sup:2o6jy2vq]2[/sup:2o6jy2vq](2pi/n), the area of a sector? Im not really sure what to do in part b as n approaches infinity

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[Deleted member 4993]

a) Let A[sub:1tmpqutr]n[/sub:1tmpqutr] be the area of a polygon with n sides inscribed in a circle with a radius of r. By dividing the polygon iinto n congruent triangles with central angle 2pi/n , show that:

A[sub:1tmpqutr]n[/sub:1tmpqutr] = (1/2)nr[sup:1tmpqutr]2[/sup:1tmpqutr]sin(2pi/n)

b) Show that the lim[sub:1tmpqutr]n -> inf[/sub:1tmpqutr] A[sub:1tmpqutr]n[/sub:1tmpqutr] = pir[sup:1tmpqutr]2[/sup:1tmpqutr]

Am I suposed to divide the top equation by (1/2)r[sup:1tmpqutr]2[/sup:1tmpqutr](2pi/n), the area of a sector? Im not really sure what to do in part b as n approaches infinity

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Hint for (a)

It already says - divide it up into triangles - with vertex at the center of the circle - and find the height and abase oof the triangle - knowing two of the sides are = r .
Hintfor b)

lim[sub:1tmpqutr]n -> inf[/sub:1tmpqutr][sin(2?/n)/(2?/n)] = 1

\(\displaystyle Lim_{n \, to \, \infty}\frac{sin(\frac{2\pi}{n})}{\frac{2\pi}{n}} = 1\)

 

[soroban]

Hello, Idealistic!

\(\displaystyle \text{a) Let }A_n\text{ be the area of a polygon with }n\text{ sides inscribed in a circle with a radius of }r.\)

\(\displaystyle \text{By dividing the polygon into }n\text{ congruent triangles with central angle }\tfrac{2\pi}{n}\),
\(\displaystyle \text{show that: }\;A_n \; = \; \frac{1}{2}n r^2 \sin\tfrac{2\pi}{n}\)

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\(\displaystyle \text{There are }n\text{ congruent isosceles triangles.}\)
\(\displaystyle \text{The vertex angle is: }\:\theta \:=\:\tfrac{2\pi}{n}\text{, the equal sides are }r.\)

\(\displaystyle \text{The area of the triangle is: }\:A \:=\:\tfrac{1}{2}r^2\sin\tfrac{2\pi}{n}\)

\(\displaystyle \text{The area of the polygon is the sum of the areas of these }n\text{ triangles.}\)

. . \(\displaystyle \text{Therefore: }\;A_n \;=\;\tfrac{1}{2}nr^2\sin\tfrac{2\pi}{n}\)

\(\displaystyle \text{b) Show that: }\;\lim_{n\to\infty} A_n \:=\: \pi r^2\)

\(\displaystyle \text{We have: }\;\lim_{n\to\infty} A_n \;=\;\lim_{n\to\infty}\left(\tfrac{1}{2}nr^2\sin\tfrac{2\pi}{n}\right) \;=\;\tfrac{1}{2}r^2\!\cdot\! \lim_{n\to\infty}\left(n\sin\tfrac{2\pi}{n}\right)\)


\(\displaystyle \text{Multiply top and bottom by }\frac{2\pi}{n}\)

. . \(\displaystyle \lim_{n\to\infty}A_n \;=\;\tfrac{1}{2}r^2\cdot\lim_{n\to\infty}\left(n\sin\tfrac{2\pi}{n} \cdot \frac{\frac{2\pi}{n}}{\frac{2\pi}{n}}\right) \;=\;\tfrac{1}{2}r^2\cdot\lim_{n\to\infty}\left(2\pi\cdot\frac{\sin\frac{2\pi}{n}}{\frac{2\pi}{n}}\right) \;=\;\pi r^2\cdot\lim_{n\to\infty}\left(\frac{\sin\frac{2\pi}{n}}{\frac{2\pi}{n}}\right)\)


\(\displaystyle \text{Let }\,\theta \:=\:\frac{2\pi}{n}\)

\(\displaystyle \text{Note that, as }n \to\infty,\;\;\theta \to 0\)

\(\displaystyle \text{Hence, we have: }\;\lim_{n\to\infty}A_n \;=\; \pi r^2\cdot \underbrace{\lim_{\theta\to0}\frac{\sin\theta}{\theta}}_{\text{This is 1}} \;=\; \pi r^2\)

 

[Idealistic]

a) Let A[sub:15ehnsdh]n[/sub:15ehnsdh] be the area of a polygon with n sides inscribed in a circle with a radius of r. By dividing the polygon iinto n congruent triangles with central angle 2pi/n , show that:

A[sub:15ehnsdh]n[/sub:15ehnsdh] = (1/2)nr[sup:15ehnsdh]2[/sup:15ehnsdh]sin(2pi/n)

b) Show that the lim[sub:15ehnsdh]n -> inf[/sub:15ehnsdh] A[sub:15ehnsdh]n[/sub:15ehnsdh] = pir[sup:15ehnsdh]2[/sup:15ehnsdh]

Am I suposed to divide the top equation by (1/2)r[sup:15ehnsdh]2[/sup:15ehnsdh](2pi/n), the area of a sector? Im not really sure what to do in part b as n approaches infinity

.

Hint for (a)

It already says - divide it up into triangles - with vertex at the center of the circle - and find the height and abase oof the triangle - knowing two of the sides are = r .
Hintfor b)

lim[sub:15ehnsdh]n -> inf[/sub:15ehnsdh][sin(2?/n)/(2?/n)] = 1

\(\displaystyle Lim_{n \, to \, \infty}\frac{sin(\frac{2\pi}{n})}{\frac{2\pi}{n}} = 1\)

For part a I got this

A[sub:15ehnsdh]n[/sub:15ehnsdh] = n(0.5*bh) Where b = base of a triangle, h = height of a triangle, n = number of triangles.

Then I figured, b = sin(2pi/n)r, which brings me to here

A[sub:15ehnsdh]n[/sub:15ehnsdh] = n(0.5*rhsin(2pi/n).

I pretty much have the equation (A[sub:15ehnsdh]n[/sub:15ehnsdh] = n(0.5*r[sup:15ehnsdh]2[/sup:15ehnsdh]sin(2pi/n)), but how can I prove that h equals r when in a triangle it doesn't? I do know that as n approaches infinity, then h is approximately r

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