Solution: the intersection g is GN⊥CD in n, and the intersection f is FM⊥AB in m,

Make an equilateral △BCE in a square ABCD with a side length of 2,

∴AB=BC=CD=AD=BE=EC=2,∠ECB=60，∠ODC=45，

∴S△BEC= 12×2×3=3, s square =AB2=4,

Let GN=x,

∠∠NDG =∠NGD = 45，∠NCG=30，

∴DN=NG=x,CN=3NG=3x，

∴x+3x=2，

Solution: x=3- 1,

∴S△CGD= 12CD？ GN = 12×2×(3- 1)= 3- 1，

Similarly: S△ABF=3- 1

∴S Shadow =S squared ABCD-s△ ABF-s△ BCE-s△ CDG = 4-(3-1)-3-(3-1) = 6-33.

So the answer is 6-33.