9. As shown in the figure, there is an equilateral triangle with a side length of 1. The first time, the equilateral triangle formed by connecting the midpoints of the three sides is removed; the second time, for the three remaining equilateral triangles, the triangles formed by connecting their midpoints are removed; ... After the fourth time, a total of $\qquad$ triangles have been removed. The sum of the side lengths of all the removed triangles is $\qquad$.


9. Solution: In the first step, 1 triangle is removed, resulting in 3 small triangles, with the side length of the removed triangle being $3 \times \frac{1}{2}$;

In the second step, 3 triangles are removed, resulting in 9 small triangles, with the side length of the removed triangles being $3 \times 3 \times \frac{1}{4}$;

In the third step, 9 triangles are removed, resulting in 27 small triangles, with the side length of the removed triangles being $9 \times 3 \times \frac{1}{8}$;

In the fourth step, 27 triangles are removed, with the side length of the removed triangles being $27 \times 3 \times \frac{1}{16}$; Therefore, a total of $1+3+9+27=40$ (triangles) are removed after four steps, and the sum of the side lengths of all removed triangles is: $3 \times \frac{1}{2} + 9 \times \frac{1}{4} + 27 \times \frac{1}{8} + 81 \times \frac{1}{16} = 12 \frac{3}{16}$