If MH intersects BC at point R, AM=AD-MD=a-40sinθ, AG=AB-BG=a-40cosθ,

So, S=AG? AM = a2-40a(sinθ+cosθ)+ 1600 sinθ？ cosθ

Let t = sin θ+cos θ = 2 sin (θ+45), then sinθcosθ=t2? 12,

So S=a2-40at+ 1600t2? 12=800(t-a40)2+a22-800。

∵00≤θ≤900

∴ 1≤t≤2

∴ When t= 1, that is, θ = 45, S has a maximum value of a2-40a.

At this time, point H is at the midpoint of EF, and the maximum rectangular area is a2-40a. ..