∠∠PFG =∠E，∠PGF =∠FBE = 90°，FG=BC=EB，
∴△EBF≌△PFG.
Let FB=a, then PG=a,
The sum of the areas of △EFB and quadrilateral AFPD is y = 10 (x+a).
∫PC = 10-x，
∴a 10？ x= 1020，
∴a=5？ 12x。
∴y= 10(x+a)=50+5x.
(2) the point p moves on the CD,
∴0≤x≤ 10.