Given that $[ABC]$ : Area of small circle = $\frac{3\sqrt3}{4}$ : $\pi$. How many parts of area of small circle is inscribed in large circle? [closed]

In the common region of two circle, $$\triangle ABC$$ has been drawn with its maximum area such that the proportion of the maximum area of $$\triangle ABC$$ and the area of small circle is equal to $$\frac{3\sqrt3}{4}:\pi$$. How many parts of area of small circle is inscribed in the large circle?

It is very unclear a matter for me that how I should construct the equilateral $$\triangle ABC$$ in the common region. Moreover, I couldn't figure out even the radius of the large circle. Will it affect the equilateral triangle?

• @jens You can check it out in GeoGebra. I used the app and found that moving outside the point $Q$, $\angle C$ results in both increasing and decreasing. Main fact is that there is a huge mistake in my diagram setup. – Anirban Niloy Feb 16 at 15:49
• As I see it, if $Q$ was placed at $O$, the ratio of the common area would be $1$. As $Q$ is moved to the right along the line $OC$ and with the $Q$ circle still intersecting the same points $A$ and $B$, the ratio will decrease until the arc of the $Q$ circle between $A$ and $B$ approaches a straight line. – Jens Feb 16 at 16:40
• I edited my diagram and brought a little change in my text and stated the point $D$ remaining still at same position. Keeping the point $D$ constant, I wanted to said about the increasing and decreasing of the radius of the large circle that obviously results in the same case for $OD$. And I told the ratio of $\triangle ABC$ and the area of small circle. And there is a little difference of the position of $A$ and $B$ by moving point $Q$. – Anirban Niloy Feb 16 at 17:09