Smaller hexagon in a Hexagon

Similar problem can be found here: link

For each blue region, there are $2$ corresponding white regions of same area. Hence $\dfrac{1}{3}$ .

Each side of the smaller hexagon is of length $b=\dfrac {a}{\sqrt 3}$ , where $a$ is the length of each side of the bigger hexagon. Hence the required ratio is $(\frac{b}{a})^2=\boxed {\dfrac{1}{3}}$ .

$ar(ABC)$ = $ar(ABD)$

smaller hexagon= $6ar(ABC)$

Larger hexagon = $18ar(ABC)$

So,ratio = $\boxed{\frac{1}{3}}$

It is clear to see that the hexagon can be divided to six triangles, and that the sum of the triangles are equal to the sum of the angles in contact with the hexagon.

We can also notice that the triangles on contact with the large hexagon, when split to two, can rearrange and form a triangle similar to the one found above.

Therefore, as we can create three figures with the same area, and only the smaller hexagon is shaded blue, only $\frac {1} {3}$ of the large hexagon is shaded blue.