Big Hexagon Small Hexagon

As there are $3 \times 6 = 18$ small triangles within the purple hexagon, there are $6$ more triangles outside. Therefore, the area ratio is $\dfrac{18}{18+6} = \boxed{\dfrac{3}{4}}$ .

Let the radius of the big hexagon be $1$ . Then the radius of the small hexagon $r_s = \cos 30^\circ = \dfrac {\sqrt 3}2$ and the ratio of their areas $\dfrac {A_s}{A_B} = \dfrac {r_s^2}{1^2}= \left(\dfrac {\sqrt 3}2 \right)^2 = \boxed {\dfrac 34}$ .

sin 60 = root3/2 is the ratio of the smaller to bigger side, square it and is the ratio of the area

The area of a regular hexagon is proportional to the square of the length of a side (with a constant of proportionality of 3sqrt(3)/2, which is not relevant). Therefore, the ratio of the areas is the square of the ratio of the lengths of their sides. Without loss of generality, let the length of the side of the big hexagon be 2. Since each vertex of the small hexagon lies on the midpoint of a side of the big hexagon, and each internal angle of a hexagon is 120 degrees, if L is the side length of the small hexagon the Law of Cosines yields:

L^2 = 1^2 +2^2 - 2 (1) (1)*cos(120 degrees), or

L = sqrt(3).

Therefore, the ratio of the areas is:

(sqrt(3)/2)^2 = 3/4

Since, all regular polygons are cyclic and each side subtends an angle of $60^{\circ}$ at the center, we have $6$ equilateral triangles as shown.

$\therefore \text{area} = 6 \cdot \frac {\sqrt {3}}{4}s^2$

Also, since $\frac {S\sqrt {3}}{2} = s,$

$\text {ratio of areas} = \frac {s^2}{S^2} = \boxed {\frac {3}{4}}$

The expression for area can be neglected.