∵ quadrilateral ABCD is a square, and AC and BD intersect at F.

∴ab=bc=cd=ad,af=bf=cf=df=(√2/2)*ab

∫ The quadrangular prism takes the quadrangular ABCD as the bottom.

∴A'A⊥ Plane ABCD

∴A'A⊥AF

Similarly, C'C⊥CF

∫AA ' =(√2/2)* AB

∴AF=AA'

∴A'F=√2*A'A=√2*(√2/2)*AB=AB

Similarly, C'F=AB.

∫A ' c ' = AC =√2 * AB

⊿∴ A 'c 'f, A'F=C'F=AB, A'C'=√2*AB.

∴∠A'FC'=90

∴A'F⊥C'F

(2) This problem is very problematic. When AF rotates on AA' axis and CF rotates on CC' axis, AF and CF are in the same plane. No matter how you cut it, it is impossible to cut off a part of the volume of the quadrilateral, which is obviously wrong. It should be the rotation of f and C.

What is the volume of the cutting geometry of the original geometry when A'F rotates on AA' and C'F rotates on CC'? Then you can use the solution.

Cut two cones with geometric volume of 1/4. The formula of vertebral body volume is: (1/3)* base area * height, and the base area is: 2*( 1/4)*π*AF? =2*( 1/4)*π*(√2/2)? *AB？ =π*AB？ /4, the height is: A'A=(√2/2)*AB, so the volume of cutting geometry is: (π*AB? /4)*(√2/2)*AB=(√2/8)*π*AB？ .

The volume of the original geometry is the volume of a quadrangular prism, and the volume formula of the prism is: the bottom area * height, and the bottom area is AB? , the height is A'A=(√2/2)*AB, and the original geometric volume is AB? *(√2/2)*AB=(√2/2)*AB？ .

Therefore, the geometric volume of cutting is [(√2/8)*π*AB? ]/[(√2/2)*AB？ ]=π/4