Proving the area of the circle using sticks.

Question

I was just trying to prove the area of the circle but couldn't reach any conclusion.so here i went----- I know For 2 identical sticks making a regular polygon we have the regular polygon as square i.e with 4 sides. For 3 identical sticks which are able to make aregular polygon we have hexagon similarly for n identical sticks we have 2n sided polygon. Now if n tends to infinity we get a circle. Now i am stuck in connecting the length of the stick and the equal angle between them with the side of the regular polygon. I cant generalise. Please help. Please note- all the sticks intersect each other at a single point i.e the centre of the geometrical figure. We get the polygons by joining the terminal end points of the sticks.

Answer 1

I suppose the diameter is the length of the stick, let it be $d$, then we need to prove that the area of the circle with infinite sticks is $\pi\, \frac{d^2}{4}$.

The angle of the internal triangle formed with $n$ sticks is $\frac{n-1}{n}\times \frac{\pi}{2}$

Hence,

\begin{align*} \text{base} &= d \, \cos{\left(\frac{n-1}{n}\cdot \frac{\pi}{2}\right)}\\ \text{height} &= \frac{d}{2}\, \sin{\left(\frac{n-1}{n}\cdot \frac{\pi}{2}\right)}\\ \end{align*}

Hence, the total area of the polygon is

\begin{align*} A &= \frac{1}{2}\cdot d \, \cos{\left(\frac{n-1}{n}\cdot \frac{\pi}{2}\right)} \cdot \frac{d}{2}\, \sin{\left(\frac{n-1}{n}\cdot \frac{\pi}{2}\right)} \times 2\, n\\ \end{align*}

Then, take the limit

\begin{align*} A &= \lim_{n\to\infty} \frac{d^2}{4} \sin{\left(\frac{n-1}{n}\cdot \pi\right)} \times n\\ &= \frac{d^2}{4}\cdot \pi \end{align*}