(A) Find and prove a formula for the area of the unit $n$-gon. An $n$-gon has $n$ evenly spaced points on the unit circle. EDIT: Solution to part (a): The area of a triangle is $\frac{1}{2}ab\sin{C}$, and if we let $a,b$ be radii and $C$ be the angle $\frac{360^\circ}{n}$, we get: Area of one triangle = $\frac{1}{2}\times 1\times 1\times\sin{\frac{360^\circ}{n}}=\frac{1}{2}\sin{\frac{360^\circ}{n}}$
Because we have $n$ triangles, the area of the unit $n$-gon is $n\times\left(\frac{1}{2}\sin{\frac{360^\circ}{n}}\right)=\boxed{\frac{n}{2}\sin{\frac{360^\circ}{n}}}$
(B) Prove that has $n$ goes to $\infty$, the area approaches the area of the unit circle which is $\pi$.
For part (B), you may use calculus if necessary.
Your answer for part A is correct, but how you got it is incorrect. You don't need Calculus for part B, just an understanding of limits. Indeed, this is the example I like to use, except using the perimeter approaching the circumference to show
$$ \lim_{\theta \to 0} \frac{\sin(\theta)}{\theta} = 1 $$
Where $\theta$ is measured in radians. You are using degrees, so I will stick with that.
Using the center of the circle as common vertex, your polygon can be partitioned into $n$ triangles. Each is an isosceles triangle with the two equal sides having a length of one. For a sample triangle, draw a line from the common vertex to halfway on the edge, call the length of this line $h$. Let the length of the base be $b$. The $h$ line is perpendicular to the $b$ line so the area of the triangle is $ \frac{1}{2} b h $. The angle between the $h$ line and the long side is $ 180 / n $. Thus the length of $b$ is $ 2 \sin(180/n) $. Similarly, the length of $h$ is $ \cos(180/n) $. Thus the area of the triangle:
$$ A = \frac{1}{2} b h = \frac{1}{2} 2 \sin(180/n) \cos(180/n) = \frac{1}{2} \sin(360/n) $$
The last step comes from the double angle formula for sines.
The perimeter, call it $p$, is the sum of all the bases of the isosceles triangles.
$$ p = n b $$
The total area for all the triangles is:
$$ A_{total} = n \frac{1}{2} b h $$
Now come the limits. The first is easy to see.
$$ lim_{ n \to \infty } h = 1 $$
The second is just as intuitive. The length of the perimeter of the polygon as the number of sides approaches infinity is the circumference of the circle.
$$ \lim_{ n \to \infty } p = 2 \pi $$
$$ \lim_{ n \to \infty } A_{total} = \lim_{ n \to \infty } n \frac{1}{2} b h = \lim_{ n \to \infty } \frac{1}{2} p h = \frac{1}{2} \cdot 2 \pi \cdot 1 = \pi $$
Hope this helps.
Ced