What i thought was simple, a circle can be formed by increasing the number of sides of regular polygon( like pentagons, hexagons, etc ) up to infinity by keeping the distance between the center and the side constant.
it is an another way to say that a circle is a polygon with infinite number of sides.
the figure is below
let $AO=BO=r$
since it is a regular polygon, hence $$ar(AOB)= \frac 12 AB\times OC= r^2\times \frac {AC}r \times \frac{OC}r$$
since $sin \theta= AC/r, cos \theta = OC/r$ $$\therefore ar(AOB) = r^2sin\theta cos\theta= \frac{r^2}2sin\phi$$
since it is an n sided regular polygon hence the total area of this polygon is $$ar(polygon)= n\frac{r^2}2 sin \phi$$ and we know that because it is a regular polygon hence $\phi = 360^0/n$ $$ar(polygon)= \frac{nr^2}2sin{\frac {360^0}n}$$
now i told you earlier that the number of sides is tending towards infinity which will tends to make the polygon a circle, hence
$$ar(circle) = \lim_{n\to \infty}\frac{nr^2}2sin{\frac {360^0}n} $$
and we know that area of a circle is $\pi r^2$
$$\therefore \pi r^2= \lim_{n\to \infty}\frac{nr^2}2sin{\frac {360^0}n} \implies \pi= \lim_{n\to \infty}\frac{n}2sin{\frac {360^0}n}\ldots (1)$$
so the value of $\pi$ is above.
the equation 1 says that when n will become larger and larger the value of the function will become closer to the value of $\pi$ i.e. 3.14 (approx.).
Do you feel it is wrong? Can this thing be improved?
and yes the Graph
thanks in advance.....
As you wrote, the area of the polygon is given by $$A_n=\frac{nr^2}2\sin{(\frac {2\pi}n})$$ Now, for large values of $n$, consider Taylor series around $x=0$ $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ and replace $x=\frac {2\pi}n$ $$\sin({\frac {2\pi}n})=\frac{2 \pi }{n}-\frac{4 \pi ^3}{3 n^3}+O\left(\frac{1}{n^4}\right)$$ which makes $$A_n=\pi r^2-\frac{2 \pi ^3 }{3 n^2}r^2+O\left(\frac{1}{n^3}\right)$$ and, as expected, the limit is approached from below.