【Solution】Solution:
Divide the hexagonal chessboard into inner and outer parts (see the division method in the following figures), and then discuss by classification:
(1) The inner and outer parts are separately densely paved: The outer ring has 2 ways of dense paving, and the inner small hexagon also has 2 ways of dense paving, so there are $2 \times 2=4$ ways at this time;
(2) 2 triangles inside are connected to 2 triangles on the outer ring, and these 2 triangles must be adjacent or opposite:

When these 2 triangles are adjacent, there are 6 ways of dense paving;
When these 2 triangles are opposite, there are 3 ways of dense paving;
There are $6+3=9$ ways in total at this time;
(3) 4 triangles inside are connected to 4 triangles on the outer ring, and since the remaining 2 triangles inside need to form a rhombus, these 2 triangles must be adjacent, so there are 6 ways of dense paving at this time:
(4) When 6 triangles inside are connected to 6 triangles on the outer ring, there is 1 way of dense paving at this time; In summary, there are $4+9+6+1=20$ ways in total for this problem.

Therefore, the answer is: 20.

Figure 1
Figure 2
10. (16 points) Figure 1 is a rhombus paper piece composed of 2 small equilateral triangles; Figure 2 is a fixed regular hexagonal board $A B C D E F$, which is made up of 24 equally sized small equilateral triangles. Now, 12 rhombus paper pieces are used to completely cover the hexagonal board, there are $\qquad$ different covering methods.