Inside out area

• The area of triangle will be $\dfrac{\sqrt{3}}{{4}}\cdot s^2$ .
• Subtracting the $60^{\circ}$ sector which is equal to $\dfrac{1}{6} \pi r^2$ , yields the area of the circle.
• We can then solve as follows:

\[\begin{align} \bigg(\dfrac{\sqrt{3}}{{4}}\cdot s^2 \bigg) - \bigg(\dfrac{1}{6} \pi r^2\bigg) &= \pi r^2 \\ \\ \bigg(\dfrac{\sqrt{3}}{{4}}\cdot s^2 \bigg) &= \bigg(\dfrac{7}{6} \pi r^2\bigg) \\ \\ \dfrac{s^2}{r^2} &= \bigg(\dfrac{7}{6} \pi \dfrac{4}{\sqrt{3}}\bigg) \\ \\ \dfrac{s}{r} &= \sqrt{\bigg(\dfrac{7}{6} \pi \dfrac{4}{\sqrt{3}}\bigg)} \\ \\ \dfrac{s}{r} &\approx \boxed{2.91}

\end{align}\]

The area of the triangle is (s^2) sqrt(3)/4. To get the area inside the triangle but outside the circle, we must subtract the area of a sector with central angle 60 degrees. This is given by .5 (Pi/3)*r^2. Since this difference must be the area of the circle, we have:

Sqrt(3) .25 s^2 - Pi (r^2)/6 = Pi r^2

Now solve for s/r = 2.91 (approximately)