Since triangles \( \triangle CDE \) and \( \triangle PQR \) are similar (\( \triangle CDE \sim \triangle PQR \)), the lengths of corresponding sides are proportional.
From the information given:
- \( CD = 9 \) m
- \( EC = 15 \) m
- \( PQ = 15 \) m
Since \( CD \) and \( PQ \) are corresponding sides, we can set up the ratio of the lengths:
\[ \frac{CD}{PQ} = \frac{9}{15} \]
To find the length of \( RP \), we need to determine the corresponding side \( DE \) in \( \triangle PQR \).
We know that the ratio of \( CD \) to \( PQ \) can be simplified:
\[ \frac{CD}{PQ} = \frac{9}{15} = \frac{3}{5} \]
Let \( RP = x \). Now let's assume that the corresponding side \( DE \) (in \( \triangle CDE \)) corresponds to a side in \( \triangle PQR \). The corresponding side will maintain the same ratio of \( \frac{3}{5} \).
We can find \( RP \) corresponding to \( EC \) since \( \frac{EC}{QR} = \frac{15}{RP} \):
Since \( EC \) gives a side, let's set them proportionate:
Given that \( EC \) is 15 and corresponds to some side in \( \triangle PQR \):
\[ \frac{CD}{PQ} = \frac{RP}{EC} \]
So, we can express it more accurately while calculating what side in \( PQR \) should correspond to 15 in similarity: \( RP = x = \frac{3}{5} \times 15\):
Calculating,
\[ x = \frac{3}{5} \times 15 \] \[ x = 9 \]
Thus, we can conclude that the length of \( RP \) is 9 m.