Given a segment of length $7\sqrt3$ .
Is it possible to use only compass  to construct a segment of length $\sqrt7$?
1. **Construct a segment of length \(7\sqrt{3}\):**
   - Place the segment \(AB\) of length \(7\sqrt{3}\) on a plane.

2. **Construct a perpendicular line:**
   - Using a compass, construct a perpendicular line passing through point \(A\).

3. **Construct an equilateral triangle:**
   - Construct an equilateral triangle \(BCD\) with base \(BC\) on the new perpendicular line such that \(BA\) is an altitude. This can be done by setting the compass width to the length of \(BA\) and drawing arcs from points \(B\) and \(C\) to intersect at point \(D\).

4. **Obtain a segment of length 7:**
   - Since \(BA = 7\sqrt{3}\) and \(BA\) is the altitude of the equilateral triangle \(BCD\), the side length of the equilateral triangle \(BCD\) is \(7\). This is because the altitude of an equilateral triangle with side length \(s\) is \(\frac{\sqrt{3}}{2}s\), and solving \(\frac{\sqrt{3}}{2}s = 7\sqrt{3}\) gives \(s = 7\).

5. **Divide the segment into sevenths:**
   - Using a compass and straightedge, divide the segment of length 7 into 7 equal parts, each of length 1.

6. **Construct a segment of length 8:**
   - Place segments of lengths 7 and 1 next to each other to make a straight line segment of length 8. Label the points \(X\), \(Y\), and \(Z\) such that \(XY = 8\) and \(XZ = 7\).

7. **Draw a circle with diameter \(XY\):**
   - Draw a circle with diameter \(XY\). The center of the circle is the midpoint of \(XY\), which is at \(4\) units from both \(X\) and \(Y\).

8. **Extend a perpendicular through \(Z\):**
   - Extend a perpendicular line through point \(Z\) to intersect the circle at point \(P\).

9. **Apply the Power of a Point (PoP) theorem:**
   - By the Power of a Point theorem, the length of the segment \(PZ\) is given by:
     \[
     PZ = \sqrt{XZ \cdot ZY} = \sqrt{7 \cdot 1} = \sqrt{7}
     \]

Thus, we have constructed a segment of length \(\sqrt{7}\) using only a compass and straightedge.

The final answer is \( \boxed{ \sqrt{7} } \)