1. **Problem statement:** In triangle $ABC$, points $D$ and $E$ lie on sides $AC$ and $BC$ respectively. Given $BE=4$, $EC=10$, $AD=x$, and $DC=11$, find $x$ such that $DE \parallel AB$. 2. **Key concept:** When a segment $DE$ is drawn parallel to one side $AB$ of triangle $ABC$, it divides the other two sides proportionally. This is called the Basic Proportionality Theorem (or Thales' theorem). 3. **Formula:** If $DE \parallel AB$, then $$\frac{AD}{DC} = \frac{BE}{EC}$$ 4. **Substitute known values:** $$\frac{x}{11} = \frac{4}{10}$$ 5. **Solve for $x$:** $$x = 11 \times \frac{4}{10} = \frac{44}{10} = 4.4$$ 6. **Answer:** The value of $x$ is $4.4$. Therefore, the correct choice is D) 4.4.