I'm trying to apply the method of ehxaustion to have an approximation of the area of the circle:
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 $h$.
Best Answer
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
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 $$a=b\cdot\dfrac{\sin(2\pi/n)}{\sin(\frac{\pi-(2\pi/n)}{2})}.$$