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Area, approximating triangles?

  1. Feb 21, 2008 #1
    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]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?
    Last edited: Feb 21, 2008
  2. jcsd
  3. Feb 21, 2008 #2
    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
    Last edited: Feb 21, 2008
  4. Feb 21, 2008 #3


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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.
  5. Feb 21, 2008 #4
    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!!!
  6. Feb 21, 2008 #5
    Right, which is basically where I went

    I googled, and as feared, his equation is wrong, I think you misread pi for n
  7. Feb 21, 2008 #6
    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
  8. Feb 21, 2008 #7


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    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".
  9. Feb 21, 2008 #8
    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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