The diagram above illustrates $\Delta ABD$ inscribed in the regular rhombus $ACEF$ of $60^{\circ}$ and $120^{\circ}$ angles, where $D$ is the midpoint of $\overline{EC}$ , and $\overline{BD}$ is perpendicular to $\overline{AC}$ .

What is the
**
ratio of red area to green area
**
?

$\dfrac{3}{16}$
$\dfrac{13}{16}$
$\dfrac{1}{3}$
None of the choices.
$\dfrac{3}{13}$

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In the given diagram, set $|\overline{AB}| = 1$ , so $|\overline{BD}| = \dfrac{1}{\sqrt{3}}$ . Since $\Delta ABD \sim \Delta DBC$ , $|\overline{BC}| = \dfrac{1}{3}$ , which is clearly $3$ times shorter than $|\overline{AB}|$ .

Let's now look at the next diagram, which involves section of rhombus into small triangles...

Here, the area of $\Delta ABD$ is three times the area of $\Delta BDC$ , which is shown by the sub-rectangle of six small triangles hovering $\Delta ABD$ . Since $\Delta ABD$ covers half the rectangle, this is equivalent to half the area of the rectangle.

Counting the number of small triangles for different-colored region, we see that there are $13$ of $\Delta BDC$ that are not colored red. Thus, the ratio is $\boxed{\dfrac{3}{13}}$ .

One can also view the rhombus as the triangle by moving two small triangles to the dashed regions.