To solve for $x$ in this problem, let’s use the fact that $DE \parallel BC$ implies that $\triangle ADE$ is similar to $\triangle ABC$ by the Basic Proportionality Theorem (or Thales' theorem).

Step-by-Step Solution

Since $DE \parallel BC$, we can write the proportionality condition as:

$\frac{AD}{AB} = \frac{DE}{BC}$

Using the given values in the triangle:

1. $AD = x + 4$
2. $AB = x + 3$
3. $DE = 2x - 1$
4. $BC = x + 1$

Now, substitute these values into the proportion:

$\frac{x + 4}{x + 3} = \frac{2x - 1}{x + 1}$

Solving the Proportion

Cross-multiply to eliminate the fraction:

$(x + 4)(x + 1) = (x + 3)(2x - 1)$

Expanding both sides:

1. Left side: $(x + 4)(x + 1) = x^2 + x + 4x + 4 = x^2 + 5x + 4$

2. Right side: $(x + 3)(2x - 1) = 2x^2 - x + 6x - 3 = 2x^2 + 5x - 3$

Now we have:

$x^2 + 5x + 4 = 2x^2 + 5x - 3$

Subtract $x^2 + 5x$ from both sides:

$4 = x^2 - 3$

Add 3 to both sides:

$x^2 = 7$

Take the square root of both sides:

$x = \sqrt{7}$

Conclusion

The correct answer is:

D. $\sqrt{7}$

Would you like further details or explanations on any part of the solution?

Here are five related questions you might find useful:

1. How can the Basic Proportionality Theorem be applied in other geometric configurations?
2. What are other properties of similar triangles that might be useful in solving proportions?
3. How does the similarity of triangles help in finding unknown values in parallel line problems?
4. Can the Pythagorean theorem be used in triangles that aren't right triangles?
5. What are some real-life applications of the Basic Proportionality Theorem?

Tip: In geometry, recognizing parallel lines and similar triangles often simplifies complex problems through proportionality.