∴△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.