There are several equilateral triangles in this setup, namely `A E H , ` `D E H , ` `A F B , ` and `C F B . ` Because of this, the segments `A D ` and `A C ` has the length `2 sqrt ( 3 ) ` (twice a height of an equilateral triangle with the side of length 2).

This way, we can divide the pentagon by three nonoverlapping triangles `A E D , ` `A D C , ` and `A B C . ` The areas of `A E D ` and `A B C ` are the same as the areas of equilateral triangle `A E H , ` which is `sqrt ( 3 ) .`

The triangle `A D C ` has sides `2 , ` `2 sqrt ( 3 ) ` and `2 sqrt ( 3 ) ; ` in other words, it is isosceles. Its base is `2 , ` while the corresponding height is `sqrt ( ( 2 sqrt ( 3 ) )^2 - 1^2 ) = sqrt ( 11 ) .`

So, the area of `A D C ` is `1 / 2 * 2 * sqrt ( 11 ) = sqrt ( 11 ) ` and the area of the pentagon is `sqrt ( 3 ) + sqrt ( 3 ) + sqrt ( 11 ) = sqrt ( 12 ) + sqrt ( 11 ) .`

Now we see `m = 12 , n = 11 , ` and `m + n = 23 .`