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I'm trying to apply the method of ehxaustion to have an approximation of the area of the circle:

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I know that the task is about decomposing the circle into a great number of triangles and then summing the area. In the first figure, there are two regular polygons, the inner one and the outer one. If I use the outer polygon, I'll have the length of $h$, if I do it with the inner polygon, I'll have only the length of $b$ - how to find the other lengths? I guess there must be a connection between the angle and length of $h$ or $b$ but until now I couldn't find it. All methods for finding the area of the isoceles triangle I've found until now need $b$ and $a$ or $b$ and $h$.

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Right. There's a connection with the angle. And for a regular $n$-gon, the angle at the centre of the circle is $\frac{2\pi}{n}$. –  Daniel Fischer Aug 3 '13 at 20:38
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If you were to search Wikipedia, then you would find formula $\frac{1}{2}bc\sin\alpha$. –  dtldarek Aug 3 '13 at 20:47
    
The area of a triangle is fount as half the product of two side times the sine of the angle between them. This is the same as half base times height, as the height is side times the sine of the relevant angle. –  Mark Bennet Aug 3 '13 at 20:49
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The answer to the question in the title is "it is not possible". –  zyx Aug 3 '13 at 23:33
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"only the length of the two equal sides" does not determine the angle between them. With length $L$ the area could be anything from $0$ to $(L^2)/2$. Really you are asking a different question than the one in the title. –  zyx Aug 4 '13 at 3:12

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You have a regular $n$-gon inscribed in a circle of radius $b$, broken into $n$ isoceles triangles. In one of these triangles, the angle formed by the two radii (i.e., the two triangle sides of length $b$) is $1/n$th of the angle of the entire circle, so that this angle is $2\pi/n$. Note that

  • the altitude $h$ bisects this angle, and
  • the other two angles will be $\frac{\pi-(2\pi/n)}{2}$ since all three angles must sum to $\pi$.

Thus, you can either use the definition of $\cos$ and the first observation to see that $$\cos\Bigl(\frac{2\pi/n}{2}\Bigr)=\cos(\pi/n)=\frac{h}{b}$$ and hence $$h=b\cdot\cos(\pi/n),$$ or you can use the law of sines and the second bullet to see that $$\frac{\sin(2\pi/n)}{a}=\frac{\sin(\frac{\pi-(2\pi/n)}{2})}{b}$$ and hence $$a=b\cdot\dfrac{\sin(2\pi/n)}{\sin(\frac{\pi-(2\pi/n)}{2})}.$$

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