11. (5 points) Arrange 12 rectangular cardboard pieces, each 4 cm long and 3 cm wide, into a larger rectangle (including squares). When arranging, ensure there are no overlapping parts and no gaps. The shortest perimeter of the larger rectangle is $\qquad$ cm, and the longest is $\qquad$ cm.
【Analysis】To form a rectangle with the shortest perimeter, it is required that during the assembly process, the rectangle shares as many sides as possible, and preferably the longer sides, and the difference between the length and width should be as small as possible (or form a square);

To form a rectangle with the longest perimeter, it is required that during the assembly process, the rectangle shares as few sides as possible, and preferably the shorter sides; according to this to solve.

【Solution】Solution: The shortest perimeter of the rectangle formed is as shown in the figure:
This is a square with a side length of $3 \times 4=12$ cm, and its perimeter is:
$12 \times 4=48$ (cm);
The longest perimeter of the rectangle formed is as shown in the figure:
Its length is: $12 \times 4=48$ (cm)
Width is: 3 cm;
Perimeter is: $(48+3) \times 2=102$ (cm).
Therefore, the answers are: 48, 102.
【Comment】Finding the methods for the longest and shortest perimeter assembly is the key to solving this problem.