Solution:
There are 8 diagonals coming from the first vertex, 8 more from the next, 7 from the next, 6 from the next, 5 from the next, etc., and 1 from the last, for $8+8+7+6+5+4+3+2+1=44$ total.

Third method: Each vertex has 8 diagonals touching it. There are 11 vertices. Since each diagonal touches two vertices, this counts every diagonal twice, so there are $\frac{8 \cdot 11}{2}=44$ diagonals.