A soldier needs to check if there are any mines in the interior or on the sides of an equilateral triangle $ABC.$ His detector can detect a mine at a maximum distance equal to half the height of the triangle. The soldier leaves from one of the vertices of the triangle. Which is the minimum distance that he needs to traverse so that at the end of it he is sure that he completed successfully his mission?
1. **Define the problem and initial setup:**
   - Consider an equilateral triangle \(ABC\) with side length \(a\).
   - The height \(h\) of the triangle is given by \(h = \frac{\sqrt{3}}{2}a\).
   - The soldier's detector can detect mines within a radius of \(\frac{h}{2} = \frac{\sqrt{3}}{4}a\).

2. **Identify key points and segments:**
   - Let \(D\), \(E\), and \(F\) be the midpoints of sides \(AB\), \(BC\), and \(CA\) respectively.
   - The soldier starts at vertex \(A\).

3. **Construct the inner triangle and midpoints:**
   - Construct the inner triangle \(DEF\).
   - Let \(K\) and \(L\) be the midpoints of segments \(DE\) and \(EF\) respectively.

4. **Calculate distances:**
   - The vertical distance from \(A\) to \(K\) is \(2h - \frac{h}{2} = \frac{3h}{2}\).
   - The horizontal distance from \(A\) to \(K\) is \(\frac{a}{8}\).

5. **Use the Pythagorean theorem to find \(AK\):**
   \[
   AK^2 = \left(\frac{3h}{2}\right)^2 + \left(\frac{a}{8}\right)^2
   \]
   Substitute \(h = \frac{\sqrt{3}}{2}a\):
   \[
   AK^2 = \left(\frac{3 \cdot \frac{\sqrt{3}}{2}a}{2}\right)^2 + \left(\frac{a}{8}\right)^2
   \]
   \[
   AK^2 = \left(\frac{3\sqrt{3}a}{4}\right)^2 + \left(\frac{a}{8}\right)^2
   \]
   \[
   AK^2 = \left(\frac{3\sqrt{3}a}{4}\right)^2 + \left(\frac{a}{8}\right)^2 = \frac{27a^2}{16} + \frac{a^2}{64}
   \]
   \[
   AK^2 = \frac{27a^2}{16} + \frac{a^2}{64} = \frac{108a^2}{64} + \frac{a^2}{64} = \frac{109a^2}{64}
   \]
   \[
   AK = \sqrt{\frac{109a^2}{64}} = a\frac{\sqrt{109}}{8}
   \]

6. **Calculate \(KL\):**
   - Since \(KL\) is the distance between midpoints of \(DE\) and \(EF\), it is \(\frac{a}{4}\).

7. **Sum the distances \(AK\) and \(KL\):**
   \[
   AK + KL = a\frac{\sqrt{109}}{8} + \frac{a}{4} = a\left(\frac{\sqrt{109}}{8} + \frac{2}{8}\right) = a\frac{\sqrt{109} + 2}{8}
   \]

Conclusion:
The minimum distance the soldier needs to traverse to ensure he has checked the entire area for mines is \(a\frac{\sqrt{109} + 2}{8}\).

The final answer is \(\boxed{a\frac{\sqrt{109} + 2}{8}}\)