etc

The first small problem: let the radius of the great circle be R and the side length of the square be A. Since the FE of this square is all on the arc, the midpoint of the bottom of the square coincides with the center of the circle, and the two images are symmetrical left and right with the same axis of symmetry. Therefore, connecting F with the center of a semicircle forms a right triangle. By using Pythagorean theorem, the square of R = the square of A..

+

(A/4)

Square of, R/A= root number 5.

Than 2

The second small problem

I really don't know.

............., hey, we haven't learned that much.

Make do with it. Give it to me if you want it.