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Hedong Ermo Mathematics 24
Exceeding c as CM⊥GF in m,

∵ quadrilateral ABCD is a square,

∴AD=AB=BC=CD,∠D=∠B=∠BCD=90,

Fold △ADE in half along AE to △AFE,

∴AD=AF,DE=EF,∠D=∠AFE=90,

∴AB=AF,∠B=∠AFG=90,

In Rt△ABG and Rt△AFG,

∫AB = AFAG = AG,

∴△ABG≌△AFG(HL),

∴BG=FG,

CD = 3DE,CD=6,

∴DE=EF=2,CE=4,

Let BG=FG=x, then CG=6-x, GE=x+2,

In Rt△GCE, ∫GE2 = CG2+CE2,

∴(x+2)2=(6-x)2+42,

The solution is x=3,

∴BG=3,

BG = GF = 3,

∴CG=3,EC=6-2=4,

∴GE=32+42=5,

∵ 12CM? GE= 12GC? European community,

∴CM×5× 12=3×4× 12,

∴CM=2.4,

∴S△FGC= 12GF×CM= 185.

So the answer is 185.