$\dfrac{x_1}{2} = 6 \cos 30^\circ$
$x_1 = 6\sqrt{3} ~ \text{cm}$
$R_2 = 6 \sin 30^\circ$
$R_2 = 3 ~ \text{cm}$
Note: All equilateral triangles are similar; and for similar figures
$$\dfrac{A_2}{A_1} = \left( \dfrac{x_2}{x_1} \right)^2 = r ~ \text{of GP}$$
where x1 and x2 = corresponding linear dimensions
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
