At a crossroads

The picture above is of the region you can visit. I'm not going to create a detailed solution right now (I might edit later) but my method involved assuming a length of path along one of the lines, and then constructing a circle where the path ended of the longest possible radius. One of these circles, near a vertex, is shown in the picture.

By constructing these circles, it is possible to obtain equations for the lines that make up the perimeter since they are tangent to these circles. With a little algebra, you can obtain that the unlabelled points in the figure have the coordinates $\displaystyle \left( \pm \frac { \sqrt { 3 } }{ \sqrt { 2 } +1 } ,\quad \pm \frac { \sqrt { 3 } }{ \sqrt { 2 } +1 } \quad \right)$ The square constructed by connecting the labelled points has area 6. The height of the shallow triangles not included by $R$ but by the square can now be found as $\displaystyle \sqrt { 2 } \left( \frac { \sqrt { 3 } }{ 2 } -\frac { \sqrt { 3 } }{ \sqrt { 2 } +1 } \right)$ Thus, the area of all of them combined is $\displaystyle 2\sqrt { 2 } \left( \frac { \sqrt { 3 } }{ 2 } -\frac { \sqrt { 3 } }{ \sqrt { 2 } +1 } \right) \sqrt { 6 }$ Subtracting from the larger square gives $\displaystyle 6-2\sqrt { 2 } \left( \frac { \sqrt { 3 } }{ 2 } -\frac { \sqrt { 3 } }{ \sqrt { 2 } +1 } \right) \sqrt { 6 } = 12\sqrt{2}-12$ So $a+b+c=26$ .