Two circles, both with the same radius $r$, are placed in the plane without intersecting each other. A line in the plane intersects the first circle at the points $A,B$ and the other at points $C,D$, so that $|AB|=|BC|=|CD|=14\text{cm}$. Another line intersects the circles at $E,F$, respectively $G,H$ so that $|EF|=|FG|=|GH|=6\text{cm}$. Find the radius $r$.
1. **Identify the given information and setup the problem:**
   - Two circles with the same radius \( r \).
   - A line intersects the first circle at points \( A \) and \( B \), and the second circle at points \( C \) and \( D \) such that \( |AB| = |BC| = |CD| = 14 \text{ cm} \).
   - Another line intersects the circles at points \( E \) and \( F \), and \( G \) and \( H \) such that \( |EF| = |FG| = |GH| = 6 \text{ cm} \).

2. **Define the centers of the circles:**
   - Let \( O_1 \) and \( O_2 \) be the centers of the two circles.

3. **Midpoints and projections:**
   - Let \( M \) be the midpoint of \( O_1O_2 \).
   - Let \( N \) be the projection of \( B \) onto \( O_1O_2 \).
   - Let \( P \) be the midpoint of \( EF \).

4. **Right-angled triangle \( \triangle MNF \):**
   - Since \( |AB| = 14 \text{ cm} \), the distance from \( B \) to the line \( O_1O_2 \) is half of \( |AB| \), so \( MN = 7 \text{ cm} \).
   - Since \( |EF| = 6 \text{ cm} \), the distance from \( F \) to the line \( O_1O_2 \) is half of \( |EF| \), so \( MF = 3 \text{ cm} \).

5. **Calculate \( NF \):**
   \[
   NF = \sqrt{MN^2 + MF^2} = \sqrt{7^2 + 3^2} = \sqrt{49 + 9} = \sqrt{58}
   \]

6. **Midline of \( \triangle PO_1M \):**
   - Since \( NF \) is the midline of \( \triangle PO_1M \), the length of \( PO_1 \) is twice \( NF \):
   \[
   PO_1 = 2 \cdot \sqrt{58}
   \]

7. **Right-angled triangle \( \triangle PO_1F \):**
   - In \( \triangle PO_1F \), we have:
     \[
     PF = 3 \text{ cm}, \quad PO_1 = 2 \cdot \sqrt{58}
     \]
   - Using the Pythagorean theorem:
     \[
     O_1F^2 = PF^2 + PO_1^2 = 3^2 + (2 \cdot \sqrt{58})^2 = 9 + 4 \cdot 58 = 9 + 232 = 241
     \]
   - Therefore, the radius \( r \) is:
     \[
     r = \sqrt{241}
     \]

The final answer is \( \boxed{13} \)