Big Hexagon Small Hexagon - 2

Since the ratio values are integers, we can formulate the triangular lattices as shown above. From there, we see that there are $6 \times 2 = 12$ triangles unshaded, whereas there are $4 \times \dfrac{6}{2} + 6 + 12 \times 2 = 42$ triangles shaded. Thus, the area ratio is $\dfrac{42}{42 + 12} = \boxed{\dfrac{7}{9}}$ .

Let $AB = 1$ , then $GB = \frac{2}{3},\quad BH = \frac{1}{3}$

Then by the cosine rule, $GH^2 = GB^2 + BH^2 - 2\cdot{GB}\cdot{BH}\cdot\cos{120^\circ}$

$GH^2 = (\frac{2}{3})^2 + (\frac{1}{3})^2 - 2\cdot\frac{2}{3}\cdot\frac{1}{3}\cdot(-\frac{1}{2}) = \frac{7}{9}$

Since area is proportional to the square if the side, the solution is $\boxed{\frac{7}{9}}$ .

Note that the two hexagons share a center. Let the radius or side length of the big hexagon be $1$ . Then the apothem of the big hexagon $a_B = \dfrac {\sqrt 3}2$ and the radius of the small hexagon is given by $r_s^2 = \left(\dfrac {\sqrt 3}2 \right)^2 + \left(\dfrac 12-\dfrac 13 \right)^2 = \left(\dfrac {\sqrt 3}2 \right)^2 + \left(\dfrac 16 \right)^2 = \dfrac 79$ . And the ratio of areas $\dfrac {A_s}{A_B} = \dfrac {r_s^2}{1^2} = \boxed{\frac 79}$ .

$a_{purple} = \sqrt {(\frac {\sqrt 3}{2}a_{green})^2 + (\frac {1}{6}a_{green})^2} \\ = \frac {\sqrt 7}{3} a_{green}$

$S_{purple} : S_{green} = a_{purple}^2 : a_{green}^2 = 7:9$