Answer
a. $\frac{1}{2}sin(\frac{2\pi}{n})$
b. $\pi$,
c. $\pi r^2$
Work Step by Step
a. One of the n congruent triangles is shown in the figure. The angle from the center of the circle is $\theta=\frac{2\pi}{n}$. With a unit circle $r=1$, the area of the triangle is given by $A_n=\frac{1}{2}(base)(height)=\frac{1}{2}(2sin\frac{\theta}{2})(cos\frac{\theta}{2})=\frac{1}{2}sin\theta=\frac{1}{2}sin(\frac{2\pi}{n})$
b. The total area will be the sum of all small triangles $A=\sum A_n =\frac{n}{2}sin(\frac{2\pi}{n})$. Thus we have $\lim_{n\to\infty}A=\lim_{n\to\infty}\frac{n}{2}sin(\frac{2\pi}{n})=\lim_{n\to\infty}\pi \frac{sin(2\pi/n)}{2\pi/n}=\pi$, (as $n\to\infty, \frac{2\pi}{n}\to 0$)
which is exactly the area of the unit circle.
c. Repeating the above steps for a radius $r$, we have $A_n=\frac{1}{2}(base)(height)=\frac{1}{2}(2r\ sin\frac{\theta}{2})(r\ cos\frac{\theta}{2})=\frac{r^2}{2}sin\theta=\frac{r^2}{2}sin(\frac{2\pi}{n})$ . And $\lim_{n\to\infty}A=\lim_{n\to\infty}\frac{nr^2}{2}sin(\frac{2\pi}{n})=\lim_{n\to\infty}\pi r^2\frac{sin(2\pi/n)}{2\pi/n}=\pi r^2$
which is exactly the area of the circle with a radius $r$.
