Dividing a square into $n$ equal size parts with minimal fence

Question

Let's divide a square into $n$ parts with equal area. How to do this using minimum amount of fence? (I assume the dividing should be done with line segments(?) so the meaning of fence length is sum of their lengths).

Here is an example with $n=5$. I don't know if it is minimal though:

Answer 1

A minimal configuration will satisfy the two-dimensional version of Plateau’s laws: each fence will be straight line or circular arc, and on either end, it will meet two others at 120° angles, or the boundary at a 90° angle.

Your configuration (left) has fences that don’t meet at 120°, so it can’t be minimal. Its length is $2 + \frac{4}{\sqrt{5}} - \frac{4}{\sqrt{10}} \approx 2.52394$. By rounding out the inside square enough to make the fences meet at 120° (right), we decrease the length to $2 + 2 \sqrt{\frac{\pi + 3 - 3\sqrt{3}}{15}} \approx 2.50211$.