What is the length of segment line fe rounded to the nearest tenth?

Well, Aml says they're congruent.  I don't think they CAN be ... There's no way
you can slide one around and make it fit exactly on top of the other one.  They
appear to be mirror-images of each other ... maybe.

But AML says they're congruent.  So let's assume that the longest sides of both
triangles have the same length.  Then the length of FE in the standing-up triangle
is the same as the distance between points 'A' and 'C' in the lying-down triangle.

Point A . . . . . (0, 1)
Point C . . . . . (3, 2) .

The distance between them is
 
     square root of (square of difference in 'x' + square of difference in 'y').

Difference in 'x' . . . . . 3
Difference in 'y' . . . . . 1

      Distance = √(3² + 1²)

                    =  √(9 + 1)

                    =  √10   =   3.1622776...      

Rounded to the nearest tenth:   3.2