Answer
(a)$ \frac{2\pi}{n}$
(b) $A=\frac{1}{2}r^2sin(T)$
(c)$ \pi r^2$
Work Step by Step
(a)
Total angle around a point = 360 deg = $2\pi$
Therefore the central angle $$T = \frac{2\pi}{n}$$
(b)
Area of a triangle = $\frac{1}{2}\times base \times height = \frac{1}{2} \times r \times rsin(T ) = \frac{1}{2}r^2sin(T)$
(c)
Total Area of the polygon = $n\times$ area of a triangle
Therefore,
$A_n = n \times \frac{1}{2}r^2sin(\frac{2\pi}{n})$
$\lim\limits_{n \to \infty}A_n = \frac{1}{2}r^2 \lim\limits_{n \to \infty} n\times sin(\frac{2\pi}{n}) = 2\pi \times \frac{1}{2}r^2 \lim\limits_{n \to \infty} \frac{n}{2\pi}\times sin(\frac{2\pi}{n}) $
Now, $ \lim\limits_{n \to \infty} \frac{n}{2\pi}\times sin(\frac{2\pi}{n}) = \lim\limits_{n \to \infty}\frac{sin(\frac{2\pi}{n}) }{ \frac{2\pi}{n}} = \lim\limits_{x \to 0} \frac{sin(x)}{x} = 1$
So, $$\lim\limits_{n \to \infty}A_n = 2\pi \times \frac{1}{2}r^2 \times 1 = \pi r^2$$
You might note that this is equal to the area of the circle.
