What fraction is shaded?

Question

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

Answer 1

We need to find the radius of the large circle and the rest is straight forward:

Answer 2

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 $15/26$.

Answer 3

Using your original drawing. Let the radius of the smaller circle $r=1$. The radius of the big circle is:
$AD+AG=\frac{1}{\cos 30°}+\cos 30°=\frac{7\sqrt 3}{6}$.

Thus, the ratio of the colored area to the area of the big circle is $$\frac{5 \over 2}{49 \over 12}=\frac{30}{49}$$