∴△OTF≌△DTF，

∴∠TOF=∠TDF=∠ADE，

AD = OG，∠A=∠TGO=90，

∴Rt△AED≌Rt△GTO，

∴ED=OT，

OA = DG，AE=TG，

∴DT=EO，

∴ED=DT=OT=OE，

The quadrangle OEDT is a diamond.

Solution: Using Figure 2'Rt△DBC, we can get: (10-x)2= 102-62.

The solution is x=2 or x= 18 (irrelevant, omitted).

Using the results of Figure 2 and (1), x=6,

So 2≤x≤6，

According to the meaning of the question: x2+(6-L 4 )2=(L 4 )2,

So L= 12+x2 3 (2≤X≤6),

Since the function value L increases with the increase of X on the right side of the coordinate axis (L axis), when x=6, L takes the maximum value.

L max = 12+62 3 =24,

A: When X is 6, the circumference L of the diamond OEDT is the largest, and the maximum circumference L is 24.