1, take CG=EF=2 on BC as the symmetrical point D' of point D about OA,

2. Connect D'G to OA in E, take EF=2 on OA, and connect CF, DE and CD.

At this time, the perimeter of the quadrilateral CDEF is the smallest.

This idea comes from the problem of the shortest route to build a bridge in textbooks!

Using Pythagorean Theorem, it is easy to get: D g = root number12+6 2 = root number 37.

Using the property of parallelogram: CF=GE

D'G=D'E+GE=DE+CF= root number 37

The minimum value of the perimeter of quadrilateral CDEF =DE+CF+EF+DC= root number 37+2+ root number 13.