What fraction is shaded? Conrguent $\frac{5}{6}$ circles in a circle. What fraction is shaded?
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Solution:

Let $r$ be the radius of the small circles and $R$ the radius of the big one.
The colored section is three times five sixths of the area of one of the small circles.
Colored Section area= $3\times\dfrac{5}{6}\times\pi r^2=\dfrac{5\pi r^2}{2}$
The radius $R$ of the big circle is equal to $r$ plus the radius of the circumscribed circle of equailateral triangle ABC, whose side is $2r$.
The radius of the circumscribed circle of an equilateral triangle is the length of the sides divided by $\sqrt{3}$. Since the side here measures $2r$, the radius of the circumscribed circle is $\dfrac{2r}{\sqrt{3}}$.
So we have  $R = r+\dfrac{2r}{\sqrt{3}}$
The area of the big circle is $\pi \times R^2$, which here is equal to $(r+\dfrac{2r}{\sqrt{3}})^2$
which, when expanded, gives
Big Circle area = $\dfrac{\pi r^2(7+4\sqrt{3})}{3}$
To obtain the shaded fraction, we need to divide the area of the colored region by the area of the big circle:
Shaded fraction = $\dfrac{\dfrac{5\pi r^2}{2}}{\dfrac{\pi r^2(7+4\sqrt{3})}{3}}$
Shaded fraction = $\dfrac{5\pi r^2}{2} \times \dfrac{3}{\pi r^2(7+4\sqrt{3})}$
Shaded fraction = $\dfrac{15}{2(7+4\sqrt{3})} \simeq 53.847 \% $
I think it's wrong. In the drawing the smaller circles are not tangent to the largest

Take the upper right point $A$ of the red circle, the center $B$ of the yellow circle and the center $C$ of the big circle as vertices of a triangle.  Call the radius of the smaller circles $r$.  Then $AB=3r$, $BC=r\sqrt3$ and $\angle CBA=30^\circ$.  Law of cosine gives $R=AC$, the radius of the big circle.
My results: $R^2=13r^2/3$ and the ratio is $15/26$.
