Nested Hexagons

$WLOG~let~S_o=1~be~the~side~of~ the~first~hexagon..\\ A~hexagon~is ~formed~by~six~equilateral~\Delta s, ~with ~side = S.\\ \implies~ratio~of~corresponding~equilateral~\Delta s,~will~be~the~same~as~the~one~required.\\ As~seen~from~the~sketch~the~side~ratios~are~\dfrac {\sqrt3} 2.\\ \implies~area~ratios, A_1/A_o=\left ( \dfrac {\sqrt3} 2 \right )^2=\dfrac 3 4 .\\ \therefore ~we~have~area~sum~as~:~~~A = \dfrac{\sqrt3} 4* \left \{1^2+\frac 3 4+(\frac 3 4)^2+(\frac 3 4)^3 . . .\right \} \\ =\displaystyle~~\frac{\sqrt3} 4* \sum_{n=0}^{\infty} \left (\frac 3 4 \right )^n. \\ This~is~a~G.P.~so~sum~\dfrac a {1-r}=\dfrac 1{1-\frac 3 4}=4.\\ So~\dfrac A{ A_o} =\dfrac {\dfrac{\sqrt3} 4 *4}{\dfrac{\sqrt3} 4 *1}=\huge \color{#D61F06}{4}.\\$

Let $x_{0}$ be a side of the initial hexagon $\implies$ the height $h_{0}$ of the initial hexagon is $h_{0} = \dfrac{\sqrt{3}}{2}x_{0} = x_{1}$

$\implies h_{1} = \dfrac{\sqrt{3}}{2}h_{0} = (\dfrac{\sqrt{3}}{2})^2 x_{0} = x_{2} \implies h_{2} = \dfrac{\sqrt{3}}{2}h_{1} = (\dfrac{\sqrt{3}}{2})^3 x_{0} = x_{3}$ and $h_{j - 1} = (\dfrac{\sqrt{3}}{2})^{j} x_{0} = x_{j} \implies h_{j} = (\dfrac{\sqrt{3}}{2})^{j + 1} x_{0}$ .

$\begin{vmatrix}{x_{0}} && {\dfrac{\sqrt{3}}{2}x_{0}} \\ {\dfrac{\sqrt{3}}{2}x_{0}} && {(\dfrac{\sqrt{3}}{2})^2 x_{0}} \\ {(\dfrac{\sqrt{3}}{2})^2 x_{0}} && {(\dfrac{\sqrt{3}}{2})^3 x_{0}} \\ {...} && {...} \\ {(\dfrac{\sqrt{3}}{2})^{j} x_{0}} && {(\dfrac{\sqrt{3}}{2})^{j + 1} x_{0}} \end{vmatrix}$

$\implies A_{j} = 6(\dfrac{1}{2})(\dfrac{\sqrt{3}}{2})^{j}(\dfrac{\sqrt{3}}{2})^{j + 1} {x_{0}}^2 = 3(\dfrac{\sqrt{3}}{2})^{2j + 1} x_{0}^2 = \dfrac{3\sqrt{3}}{2} {x_{0}}^2 (\dfrac{3}{4})^{j} = (\dfrac{3}{4})^{j} A_{0} \implies$

$A = \sum_{j = 0}^{\infty} A_{j} = A_{0}\sum_{j = 0}^{\infty} (\dfrac{3}{4})^j = 4A_{0} \implies \dfrac{A}{A_{0}} = \boxed{4}$ .