A regular octagon is a square with the 4 isosceles right triangles
chopped off the corners as you see below.



One side of the square is x+4+x or 2x+4.

So the area of the square is (2x+4)²

We look at one of the four isosceles triangles that we are to cut
off the corners of the square.



We use the Pythagoreant theorem to find side 8:

x² + x² = 4²
    2x² = 16
     x² = 8
      x = 
      x = 
      x = 


So the area of the square is
                 _            _             _              _
(2x + 4)² = (2*2V2 + 4)² = (4V2 + 4)² = [4(V2 + 1)]² = 16(V2 + 1)² sq. in.

The area of one of the triangles is:

·base·height = ·x·x = ·x² = ·8 = 4 sq. in.

So the 4 triangles that we cut from the square have total area of 
4·4 or 16 sq. in.

Therefore the area of the regular octagon is

Area of square minus area of the four triangles =
    _
16(Ö2 + 1)² - 16 = 
     _
16[(Ö2 + 1)² - 1] = 
         _
16[2 + 2Ö2 + 1 - 1] = 
         _
16[2 + 2Ö2] =
          _
16[2(1 + Ö2) =
        _
32(1 + Ö2) or about 77.25 sq. in. 

Edwin