Originally Posted by

**Opalg** I can't read that attachment, but here's how I did the calculation. Let AB=BC=CD=2a. Then the triangle has base 2a and height √3a, and so

$\displaystyle \boxed{\text{Area of triangle }=\sqrt3a^2}$.

Let O denote the centre of the circle. Then $\displaystyle \frac a{OC} = \cos(\pi/6) = \frac{\sqrt3}2$, so that $\displaystyle OC = \frac2{\sqrt3}a$, and $\displaystyle OD = \Bigl(\frac2{\sqrt3}+2\Bigr)a$. Therefore the area of the C-shaped ring is $\displaystyle \tfrac23\pi(OD^2-OC^2) = \frac{8\pi a^2}9\bigl((1+\sqrt3)^2-1\bigr)$.

$\displaystyle \boxed{\text{Area of C-ring }=\tfrac89\pi a^2(3+2\sqrt3)}$.

That gives the ratio of the two areas as $\displaystyle \frac{9\sqrt{3}}{8\pi(3+2\sqrt{3})}$, as before.