Now consider a line segment, translating (i.e. moving but not rotating) around a closed path, so that it returns exactly to its original position. Then if we consider a signed flow of area past the segment, the net flow for the closed path will be zero. (Easy to prove, if necessary, but fairly intuitive.) That's the second idea.

Next, let ABC be a right triangle with sides a, b, and hypotenuse c. Translate it a distance a perpendicular to side a, then a distance b perpendicular to side b, and then a distance c perpendicular to side c, so that the triangle returns to its original position. In the first motion, an area a^2 sweeps into side a and out of side c; in the second motion, an area b^2 sweeps into b and out of side c, and in the third motion, an area c^2 sweeps into side c and out of sides a and b together. This is according to the first idea. But by the second idea, the area swept in and out of side c are equal. Therefore a^2 + b^2 = c^2.

--Stuart Anderson

https://www.cut-the-knot.org/pythagoras/MechanicalProofs.shtml

The area sweeping comes up in a little different setting:

https://www.cut-the-knot.org/Curriculum/Geometry/PythFromRing.shtml

I think I'll add your musing as a comment to that page.

https://www.cut-the-knot.org/pythagoras/torque.shtml

Thank you.