To find the perimeter of triangle \( \triangle CDE \), we need to use the lengths of the segments tangent to the circle.
In triangle geometry involving a circle inscribed within a triangle, the tangents from a point to the circle are equal in length. Let's denote:
- \( s_1 \) as the length of the tangent from point \( C \) to the points where the circle touches sides \( CD \) and \( CE \).
- \( s_2 \) as the length of the tangent from point \( D \) to the points where the circle touches sides \( DE \) and \( DC \).
- \( s_3 \) as the length of the tangent from point \( E \) to the points where the circle touches sides \( EC \) and \( ED \).
The lengths provided are:
- \( CF = 15 \) cm
- \( BD = 12 \) cm
- \( EG = 4 \) cm
From the tangents, we can write:
- \( CD = CF = 15 \) cm
- \( DE = BD = 12 \) cm
- \( EC = EG = 4 \) cm
However, we are not directly concerned with these tangent segments but rather with the equal segments that can be established from the points of tangency.
Let \( x \) be the length of the segment from \( D \) to the point of tangency on \( CE \) (next to point \( E \)), and modify it for others. We can summarize the segments as follows:
- From \( C \), the tangents to the circle touch at \( CD \) and \( CE \): \( s_C = CF = 15 \) cm
- From \( D \), the tangents touch at \( DE \) and \( DC \): \( s_D = BD = 12 \) cm
- From \( E \), the tangents touch at \( EC \) and \( ED \): \( s_E = EG = 4 \) cm
Using this, we find:
\[ CD + DE + EC = s_C + s_D + s_E \]
Thus,
\[ \text{Perimeter of } \triangle CDE = (s_C + s_E) + (s_D + s_E) + (s_E + s_C) \]
Since we have the segments already calculated:
\[ \text{Perimeter of } \triangle CDE = CF + BD + EG + CF + BD + EG \] This would mean:
\[ 15 + 12 + 4 + 15 + 12 + 4 = 2(CF + BD + EG) \]
Calculating that value:
\[ = 2(15 + 12 + 4) = 2(31) = 62 \text{ cm} \]
Thus, the perimeter of \( \triangle CDE \) is \( 62 , \text{cm} \).
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