Calculate the area of a triangle with four circles inside The four shown circles have the same radius and each one is tangent to one side or two sides of the triangle. Each circle is tangent to the segment which is inside the triangle ABC. Besides, the central lower circle is tangent to its neighbor circles. If AC = 12 cm, what is the value of the area of the triangle ABC?
I tried to assign angle variables to the triangle to compute the sides of the triangle, and so to use Heron's formula to calculate the area, but after all, it appears that some data is missing. I haven't found any book or article which treats these kind of problems.
Many thanks in advance

 A: Let us construct a similar configuration by choosing the radius $r$ of the involved circles and the "critical distance" $d$:

The area of the innermost triangle is $4r^2$ and the innermost triangle is similar to $ABC$. The angle bisectors of $ABC$ are also the angle bisectors of the innermost triangle, so the ratio between a side length of $ABC$ and the length of the parallel side in the innermost triangle is $\frac{i+r}{i}$, with $i$ being the inradius of the innermost triangle. It follows that the area of $ABC$ is 
$$ 4r^2\left(\frac{i+r}{i}\right)^2 $$
and $\frac{i+r}{i}=\frac{AC}{4r}$. But... wait! This gives that the area of $ABC$ is just $\color{red}{\frac{1}{4}AC^2}$, we do not need to know neither $r$, or $i$, or $d$!!!

In particular, the distance of $B$ from $AC$ is exactly $\frac{AC}{2}$.
A: By a little modification we can change the figure to a more easier form. We can move the upper circle to the right and exactly above the middle circle without changing the area of the triangle.

A: In $\triangle ABC$, base is $12$, height on the base is $h$. In the inner triangle, base is $4r$ and height is $2r$ (height above the line segment is $r$, height below the segment is also $r$). Thus the inner triangle has height half its base.
Since the two triangles are similar, $h=6$ always.
