To find the perimeter of triangle \( \triangle CDE \), we can use the lengths of the segments from the vertices of the triangle to the points of tangency with the circle.
In triangle \( \triangle CDE \):
- Let \( a = CF = 15 \) cm (the length from vertex \( C \) to the point of tangency \( F \)),
- Let \( b = BD = 12 \) cm (the length from vertex \( B \) to the point of tangency \( D \)),
- Let \( c = EG = 4 \) cm (the length from vertex \( E \) to the point of tangency \( G \)).
Since the tangents from a single point outside the circle to the points of tangency are equal, we can assign:
- Let \( x \) be the length from \( C \) to point \( F \) (which we already have as 15),
- Let \( y \) be the length from \( D \) to point \( D \) (which we already have as 12),
- Let \( z \) be the length from \( E \) to point \( G \) (which we already have as 4).
The two tangents from a vertex to the points of tangency on the triangle's sides can be combined to find the lengths of the sides of the triangle.
Thus, the segments can be summed up for each side:
- Side \( CE \) will be \( CF + EG = 15 + 4 = 19 \) cm,
- Side \( CD \) will be \( CF + BD = 15 + 12 = 27 \) cm,
- Side \( DE \) will be \( BD + EG = 12 + 4 = 16 \) cm.
Now we can calculate the perimeter of triangle \( \triangle CDE \): \[ \text{Perimeter} = CE + CD + DE = 19 + 27 + 16 \] \[ = 62 \text{ cm}. \]
Therefore, the perimeter of triangle \( \triangle CDE \) is \( \boxed{62} \) cm.