The first statement is true. Reasoning below.
= = =
We want to find the area of a fixed circle, so we can throw out the last option. If
![r](https://tex.z-dn.net/?f=r)
changes at all, then so does the area of the circle.
For
![s](https://tex.z-dn.net/?f=s)
to increase would require using a circumscribed polygon with less sides. Again, the circle is fixed, so only a certain length
![s](https://tex.z-dn.net/?f=s)
can fit inside the circle. This eliminates the third option.
Note that if we use a regular hexagon, then
![s=r](https://tex.z-dn.net/?f=s%3Dr)
automatically, because the component triangles that make up the hexagon are equilateral. Increasing
![h](https://tex.z-dn.net/?f=h)
would require that we use a polygon with more sides, which would simultaneously make
![s](https://tex.z-dn.net/?f=s)
stray away from
![r](https://tex.z-dn.net/?f=r)
. In other words, if
![h](https://tex.z-dn.net/?f=h)
increases, then
![s](https://tex.z-dn.net/?f=s)
decreases, so we can never eventually have
![s\to r](https://tex.z-dn.net/?f=s%5Cto%20r)
(
![r](https://tex.z-dn.net/?f=r)
is fixed).
That leaves the first option. Indeed, as
![n](https://tex.z-dn.net/?f=n)
increases, we get a polygon that looks increasingly rounder and more like a perfect circle. At the same time, that means
![h](https://tex.z-dn.net/?f=h)
gets larger, but would be bounded above by the circle's perimeter. So as
![h](https://tex.z-dn.net/?f=h)
increases indefinitely, it will eventually "be equal" (in the limit sense) to
![r](https://tex.z-dn.net/?f=r)
, so that
![rh\to r^2](https://tex.z-dn.net/?f=rh%5Cto%20r%5E2)
.