For a circle of radius $r$, if I draw a regular polygon with $n \geq 3$ sides (equilateral triangle, square, pentagon, etc.) contained inside the circle, such that each vertex intersects the circle...
What is the area of the polygon? (both in terms of $r$, and as a ratio of the circle's area $\pi r^2$) 
I assume that ratio $\to 1$ as $n \to \infty$
On
Consider this picture. I took yours and modified it a little by adding some constructions that will be useful for proving this statement. Keep in mind that this picture should only be used as a reference, because the proof has to be valid for any regular polygon with n-sides inscribed into the circle.
Consider a regular polygon with any number of sides ( $n$ sides ) inscribed into the circle.
We can now draw some lines starting from the center and ending at each of the vertices of the polygon. These lines have the same length as the radius of the circumference, which I can call $R$
You can see that the original polygon is now divided into some triangles. The idea is that I can express the area of the original polygon as the sum of the areas of the triangles.
If we're considering a square, we have four of these triangles, but how many do we have if we consider a general regular polygon with $n$ sides?
Let's do the same process again, but in a more general way. Each triangle is formed by two of these lines starting from the center and one side of the inscribed polygon. Consider the sides of the inscribed polygon as the bases of these triangles, this means that we can have at most $n$ possible bases and so at most $n$ possible triangles for a polygon with $n$ sides.
Now I want you to consider each triangle separately. What are the measures of its sides?
Each triangle is composed of two equal sides equivalent to the radius of the circumference $R$ and the third side is one of the sides of the inscribed polygon, which I can call $l$. Since sides of the same regular polygon are equal, all of the $n$ triangles are congruent because they have 3 equal sides.
Now ask yourself, what does congruence mean? A congruence relationship between two polygons means that the polygons can be overlapped. We have $n$ triangles and they are congruent, so they can be overlapped. If you can overlap all of these six triangles, then each one must occupy the same area, so all of these $n$ triangles have the same area.
Now let's analyze one of these triangles separately and let's try to calculate its Area.
the Area of a triangle is $\frac{1}{2}bh$ and,since the base is the side of the regular polygon, I can write it as $\frac{1}{2}lh$.
I will begin by drawing the height $h$ of the triangle , which divides the original triangle into two right-angle triangles. The original triangle has two sides equal to the radius of the circumference, so its height is also the mean point of the base. This means that each of the two right-angle triangles has $\frac{l}{2}$ as its base and $h$ as its height.
Now consider the height $h$ of the triangle. You can apply the Pythagorean Theorem to one of the two right-angle triangles and you get:
$ipotenuse^2 = side1^2 + side2^2$
By substituting the values:
$R^2 = h^2 + (\frac{l}{2})^2$
Now let's solve for h
$h^2 = R^2 - (\frac{l}{2})^2$
This implies that $h^2 = R^2 - \frac{l^2}{4} = \frac{4R^2-l^2}{4}$
Let's take the square root of both sides:
$h = \sqrt{\frac{4R^2-l^2}{4}} = \frac{\sqrt{4R^2-l^2}}{2}$
Now we can calculate the area of the triangle by substituting the value that I found for $h$ into the formula:
$A = \frac{1}{2}l\frac{\sqrt{4R^2-l^2}}{2} = \frac{l\sqrt{4R^2-l^2}}{4}$
We can finally calculate the area of the regular inscribed polygon. We have $n$ triangles with equal area, so the total area will be n multiplied by the area of a single triangle
$Atot=\frac{nl\sqrt{4R^2-l^2}}{4}$
Let's make some observations to simplify the formula. First of all a regular polygon has n equal sides, so its perimeter is n multiplied by l, $nl = p$, so the formula becomes
$Atot = \frac{p\sqrt{4R^2-l^2}}{4}$
This is the formula expressed with the radius, but it's pretty long. You can use the fact that the height of each triangle is the radius of the circle inscribed into the regular polygon, so the formula becomes
$Atot = \frac{1}{2}pr$ where r is the radius of the circle inscribed into the polygon.
Since you also asked for the ratio in terms of the circle circumscribed around the regular polygon:
$Acircle = \pi R^2$
$\frac{Acircle}{Atot} = \frac{\pi R^2}{\frac{1}{2}pr} = \frac{2\pi R^2}{pr} = \frac{2Acircle}{pr}$
Use the standard formula: the area of an elementary triangle is half the apothem ($a$) times the length of a side ($s$).
If the regular polygon has $n$ sides, the apothem and half the side length are $$a=r\cos\frac\pi n,\qquad s=r\sin\frac\pi n.$$ Can you proceed?