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Homework Help: Integral calculus question, using limits.

  1. May 16, 2010 #1
    1. The problem statement, all variables and given/known data

    Let A 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 2pi/n, show that A = [tex](\frac{1}{2})nr^{2}sin(2(\pi)/n)[/tex]

    2. Relevant equations

    Area of a circle = pi x r^2
    Area of triangle = (base x height)/2
    sin x = opposite/hypotenuse

    3. The attempt at a solution

    My attempt is in the attached word document. I spent over an hour trying to use the latex/equation stuff, and normally I know how to use it, but it kept putting in stuff that I wasn't actually typing. Every line would just show up as something random.

    I apologize in advance.

    Attached Files:

  2. jcsd
  3. May 16, 2010 #2


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    Science Advisor

    Don't use the Pythagorean theorem to get h in terms of r and n. Instead, note that you have a right triangle with angle [itex]\theta/2[/itex], "near side" h and "hypotenuse" r. [itex]cos(\theta/2)= h/r[/itex] so [itex]h= r cos(\theta/2)[/itex]. Similarly, the base of the triangle is [itex]r sin(\theta/2)[/itex] so the area of each right triangle is [itex](1/2)r^2 sin(\theta/2)cos(\theta/2)[/itex].

    Since sin(2x)= 2sin(x)cos(x) for any x, [itex]sin(\theta/2)cos(\theta/2)= (1/2)sin(\theta)[/itex]. And since each of the triangles having a side of the polygon as base is two of those right triangles, the area of each such triangle is [itex](1/2)r^2 sin(\theta)[/tex].

    Now use the fact that there are n such triangles and that [itex]\theta= 2\pi/n[/itex]
  4. May 16, 2010 #3
    makes perfect sense! Then all I need to do is multiply [itex](1/2)r^2 sin(\theta)[/itex] by [itex]n[/itex] and substitute [itex]\theta= 2\pi/n[/itex].

    But, why wasn't I able to use Pythagorean theorem to find h?
  5. May 17, 2010 #4


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    Science Advisor

    Well, you could, but you clearly want everything in terms of [itex]\theta[/itex]. Writing h in terms of other things you don't know doesn't help!
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