Sum of Areas of Equilateral Triangles Inscribed in Circles

$x_1 = 6\sqrt{3} ~ \text{cm}$
 

$R_2 = 6 \sin 30^\circ$

$R_2 = 3 ~ \text{cm}$
 

The areas of all equilateral triangles form an IGP
$a_1 = A_1 = \frac{1}{2}{x_1}^2 \sin 60^\circ$

$r = \left( \dfrac{R_2}{R_1} \right)^2 = \left( \dfrac{3}{6} \right)^2 = 1/4$
 

$S = \dfrac{a_1}{1 - r} = \dfrac{\frac{1}{2}(6\sqrt{3})^2 \sin 60^\circ}{1 - 1/4}$

$S = 62.35 ~ \text{cm}^2$   ←   answer