To find the perimeter of triangle \( \triangle CDE \), we can use the lengths of the segments \( CF \), \( BD \), and \( EG \) that are tangent to the circle.
Given the following lengths:
- \( CF = 15 \) cm
- \( BD = 12 \) cm
- \( EG = 4 \) cm
We can denote the points where the triangle tangents meet as follows:
- Let \( a = CF \)
- Let \( b = BD \)
- Let \( c = EG \)
Now, denote the segments tangent to the circle from the vertices of the triangle as:
- From point \( C \), let \( CE = x \)
- From point \( D \), let \( DE = y \)
- From point \( E \), let \( CD = z \)
Because the lengths from each vertex to the point of tangency are equal:
- The length from \( C \) to the point where \( BD \) and \( CF \) meet will be \( CE = CF = 15 \) cm.
- The length from \( D \) to the point where \( CF \) and \( EG \) meet will be \( DE = EG = 4 \) cm.
- The length from \( E \) to the point where \( BD \) and \( EG \) meet will be \( CD = BD = 12 \) cm.
Now we can substitute our known values:
- \( CE = 15 \)
- \( DE = 4 \)
- \( CD = 12 \)
Now sum these lengths to find the perimeter \( P \) of triangle \( \triangle CDE \): \[ P = CE + DE + CD = 15 + 4 + 12 \]
Calculating this gives: \[ P = 31 \text{ cm} \]
Therefore, the perimeter of triangle \( \triangle CDE \) is \( \boxed{31 \text{ cm}} \).
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