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Mathematics in the real test paper of the junior high school entrance examination.
The simplest way to solve this problem is to add equations by Pythagorean theorem.

Solution: ∵ABCD is rectangular.

∴∠A=∠D=∠ABC=∠DCB

∴△ APB and△ PDC are right triangles.

∫△PHF is a right triangle.

∴△ APB, △ PDC and △ PHF are all right triangles.

Let AP be x, then PD =10-X.

∴BP? =X? +4? ,PC? =( 10-X)? +4?

BC? = 10? = 100

△ PBC is a right triangle.

∴X? +4? +( 10-X)? +4? = 100

∴X? - 10X+ 16=0

∴X? - 10X+25-25+ 16=0

∴(X-5)? =9

∴X-5= 3

X = 8 or 2

A: Yes, the AP length is 8 or 2.