Jigsaw Area

Think of each connector as a circle sector ("pie", or here more like a Pacman shape) with an isosceles triangle:! The area of the circle sector is $\frac{240^\circ}{360^\circ}\cdot \pi r^2 = \tfrac23\pi r^2;$ the triangle has a width of $2\cdot r\:\sin 60^\circ = \sqrt 3 r$ and a height of $r\:\cos 60^\circ = \tfrac12 r$ , with area $\tfrac12\cdot \sqrt 3 r \cdot \tfrac12 r = \tfrac14\sqrt 3 r^2.$ Adding these areas together and multiplying by four, we find for the total area of the four connectors $4\cdot (\tfrac23\pi r^2 + \tfrac14\sqrt 3 r^2) = (\tfrac83\pi + \sqrt 3)r^2.$ It follows that $c = \tfrac83\pi + \sqrt 3\approx \boxed{10.10963}.$