Dimensions of paper needed to roll a cone (Updated with clarifications) I'm looking for a way to calculate the dimensions of a piece of paper needed to roll up into a cone shape. Please consider the following diagram I created (nothing is drawn to scale):

In this example, I'm trying to create a cone shape, that is 100 mm tall and has a top diameter of 3 mm, and a bottom diameter of 12 mm (FIG 1). FIG 2 shows a piece of paper that is 0.1 mm thick shaped like what I would expect the final piece to look like. However, sides A and C would be much longer than illustrated.
Essentially, what I need is a way to figure out the lengths of sides A, B, and C in FIG 2, such that when the paper is rolled tightly, I could achieve the cone in FIG 1.
NOTE: I'm not trying to just create a "shell" of the cone as shown in FIG 3. I would like to roll the paper shape tightly into the cone to provide a solid structure (as shown in FIG 4). 

EDIT: I'm adding some clarification. I did a "real world" example to try to illustrate what I'm trying to accomplish.


*

*I cut out a shape similar to the one is FIG 2 above.
image 1

*I roll it as shown here.
image 2

*This is what the cone looks like when done rolling.
image 3

*The top of the cone shows a diameter of 2.1 mm
image 4

*And the bottom of the cone shows a diameter of 3.4 mm
image 5
I hope this makes things more clear.
EDIT 2: Maybe the numbers in the initial example are just not realistic. So how about something like 2 mm for the top diameter and 5 mm for the bottom one.
 A: If we slightly simplify the situation, we can say that at the top (and at the bottom) the paper is arranged in concentric circles. If $t=0.1$mm is the thickness of the paper, we get a radius $r_n = n \cdot t$ for the $n$th circle. The circumference of the $n$th circle is $2 \pi r_n$. The total length of the paper (for $N$ circles) is the sum of the circumferences
$$
l = \sum_{n=1}^N 2 \pi r_n = 2 \pi \cdot t \cdot \sum_{n=1}^N n = 2 \pi \cdot t \cdot \frac{N(N+1)}{2} 
$$
and the diameter is $d = 2 N t$.
So if you want a diameter $d$, then you need $N=\frac{d}{2t}$ many layers resulting in a length of
$$l = 2 \pi \cdot t \cdot \frac{\frac{d}{2t}\left(\frac{d}{2t}+1\right)}{2} = \pi \cdot \frac{d}{2} \left(\frac{d}{2t}+1\right).$$
According to the pictures in your clarification, $A$ and $C$ are independent of each other, thus the above calculation can be done seperatly for the top and the bottom.
At the top, you want the diameter to be $d=3$mm, resulting in a length of $A\cong 65$mm.
At the bottom, you want the diameter to be $d=12$mm, thus $C \cong 1100$mm would do the trick.
However, rolling up a meter of paper perfectly tight is probably not physically feasible, thus the calculation might not agree with experimental results.
