Solution: (1) according to the meaning of the question, make AQ⊥BC and BC intersect at Q,

Easy to get: BQ=3, from Pythagorean theorem, easy to get AQ = 4；;

Then s △ ABC =1/2× 6× 4 =12;

(2) Let AQ and PD intersect at point m and EF intersect at point n; PD∨BC，

∴△APD∽△ABC，

∴AP∶PD=AB∶BC，

And AP=x, AB=5, BC=6,

Available: PD=6/5x, PM = 3/5x；;

It is easy to get AM=4/5x, then AN=AM+MN=AM+HF=x,

Y = s trapezoidal PBCD-S? PFED-S trapezoidal BFEC

= 1/2(6/5x+6)(4-4/5x)-6/5x * 6/5x- 1/2(6/5x+6)(4-x)=-3/25x？ +3/5x=-3/25(x-5/2)？ +3/4;

So when x=5/2, y gets the maximum value, which is 3/4.