Area, approximating triangles?

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  • #1
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Let [tex]A_n[/tex] 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 [tex]\frac{2\pi}{n}[/tex], show that [tex]A_n=\frac 1 2 \pi r^2\sin{\frac{2\pi}{n}}.[/tex]

Ok, I drew a circle with congruent triangles inscribed in it. I assumed that it was an equilateral triangle, so it has height [tex]\frac{\sqrt{3}}{2}r[/tex].

So far I have

[tex]A_{triangle}=\frac 1 2 \cdot r \cdot \frac{\sqrt{3}}{2}r[/tex]

[tex]\sin{\frac{2\pi}{n}}=\frac{\frac{\sqrt{3}}{2}r}{r}=\frac{\sqrt{3}}{2}[/tex]

[tex]A_{triangle}=\frac 1 2 \cdot r^2\cdot \sin{\frac{2\pi}{n}}[/tex]

Now I'm stuck, maybe my assumption was incorrect, and I also do not know how to incorporate the fact that it is inscribed in the circle. I know I need to take it into consideration noticing that it wants me to express the answer with the area of a circle as part of the answer. Or perhaps [tex]\pi r^2[/tex] appears through substitutions?
 
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Answers and Replies

  • #2
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I remember learning a variation of this years ago in high school, this is a nifty little formula

The first thing I see is that I'm not sure you even understood what you were being asked to prove

Do it with a simple shape, like a hexagon(I tried an octagon myself but couldn't draw a circle worth a darn that circumscribed it :( )

Draw the circle around it that touches every intersection on the hexagon. Now from the center of the circle, draw a line to every intersection and behold six triangles!

Note that they won't necessarily be equilateral triangles since two sides are the radius of the circle and one's a chord(I think that's the term >_>) Isosceles always though, I think

So what's the area of that triangle? The base is r, you need 1/2*base*height, the height you have to drop a perpendicular and find that, you need the sine of that angle...well you have the full circle broken into 6 things, so...

Anyways that's a better way to start
 
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  • #3
Dick
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1) I don't know what an equilateral triangle has to do with anything if you have n sides. 2) Your A_n approaches 0 as n approaches infinity, hence A_n is NOT the area of a polygon with n equal sides inscribed in a circle of radius r. Look, what's the area of an isosceles triangle with apex angle 2pi/n? Multiply that by n to get the total area.
 
  • #4
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Note that they won't necessarily be equilateral triangles since two sides are the radius of the circle and one's a chord(I think that's the term >_>) Isosceles always though, I think
LOL, I assumed the chord was length r, hence the equilateral triangle. Ok, let me continue reading you and Dick's post. Must solve this!!!
 
  • #5
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Right, which is basically where I went

I googled, and as feared, his equation is wrong, I think you misread pi for n
 
  • #6
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Right, which is basically where I went

I googled, and as feared, his equation is wrong, I think you misread pi for n
Confused pi for n? That is the final equation it wants though.

[tex]A_n=\frac 1 2 \pi r^2\sin{\frac{2\pi}{n}}[/tex]

Stewart 5th edition, page 326
 
  • #7
Dick
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Confused pi for n? That is the final equation it wants though.

[tex]A_n=\frac 1 2 \pi r^2\sin{\frac{2\pi}{n}}[/tex]

Stewart 5th edition, page 326
If that's supposed to be the area of an n sided inscribed polygon, it's wrong. There must be a typo in "Stewart 5th edition, page 326".
 
  • #8
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My calc 3 professor was a proofreader for math textbook solutions

I doubt he was very good >_>

Of course every time I've been so certain I'm right and the book's wrong I've been just ludicrously wrong, still it's not too surprising.
 
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