Look at the following figure carefully,
As the triangle is equilateral ($AC$ is the angle bisector). So, $\angle ACD = 30^{\circ}$
$$\tan 30^{\circ} = \frac{AD}{DC} = 2AD\ (\because DC = 1/2) $$
$$\therefore AD = \frac{1}{2\sqrt{3}}$$
This is the radius of the bigger circle, let its area be $A_1$
$$\therefore A_1 = \frac{\pi}{12}$$
To calculate the radius of the next smaller circle (let it be $x$), please note that
$$AC = \frac{1}{\sqrt{3}}$$
$$AB =\frac{1}{2\sqrt{3}} +x$$
$$\therefore BC = AC - AB =\frac{1}{2\sqrt{3}} -x $$
Note that triangles $ADC$ and $BCE$ are similar.
$$\therefore \frac{AD}{AC} = \frac{BE}{BC}$$
$$\frac{1}{2\sqrt{3}} \times \sqrt{3} = x \times \left( \frac{2\sqrt{3}}{1-2\sqrt{3}x} \right)$$
$$\therefore x = \frac{1}{6\sqrt{3}}$$
Similarly we can find the radii of the next circles. They would be $\frac{1}{18\sqrt{3}}$,
$\frac{1}{54\sqrt{3}}, ...$
Now, the main answer,
The sequence $\frac{1}{6\sqrt{3}},\frac{1}{18\sqrt{3}}, \frac{1}{54\sqrt{3}}, ... $
can be generally written as $\frac{1}{6\sqrt{3}(3)^{n-1}}$
Total area of these circles,
$$T = \frac{\pi}{12} + 3\sum_{n=1}^{\infty} \pi {\left(\frac{1}{6\sqrt{3}(3)^{n-1}} \right)}^2 $$
Notice that,
$$\sum_{n=1}^{\infty} \pi {\left(\frac{1}{6\sqrt{3}(3)^{n-1}}\right)}^2 = \sum_{n=1}^{\infty} \pi {\left( \frac{1}{108}\right)}{\left(3^{-(n-1)}\right)}^2$$
=$$\sum_{n=1}^{\infty} \pi {\left( \frac{1}{108}\right)}{\left(3^{-2(n-1)}\right)}$$
=$$\sum_{n=1}^{\infty} \pi {\left( \frac{1}{108}\right)}{\left(\frac{1}{9}\right)}^{n-1}$$
This is a GP with $a = \frac{\pi}{108}$ and $r = \frac{1}{9}$
For infinite terms, the sum of this GP = $\frac{a}{1-r} = \frac{\pi}{96}$
Now, finally,
$$T = \frac{\pi}{12} + 3 \times \frac{\pi}{96} = \frac{11\pi}{96}$$
Best Answer
If the radius of circle is $R$, then I get the length of the triangle as $(6+2\sqrt{3})R$.
Drop a perpendicular from the center of the rightmost bottom circle to the bottom side.
The distance from the bottom right vertex of the triangle to the base of the perpendicular is $D = \sqrt{3}\ R$, as we get a triangle of angle $30^{\circ}$ and we get $\tan(30^{\circ}) = \frac{R}{D}$. Since $\tan(30^{\circ}) = \frac{1}{\sqrt{3}}$, we get $D = \sqrt{3}\ R$.
Now drop perpendiculars from the centers of the bottom circles to the bottom side and add up the lengths of the segments formed on the bottom side.
Area of an equilateral triangle of side $a$ is $\sqrt{3}\ a^2/4$.