(1): let PQ =δs, then δs = 50-AP-QB = 50-25t-25t. From the meaning of the question, Δ δs = 0 and the solution is t= 1, that is, when t= 1s, point P coincides with point Q.

(2): When D is on QF, ∫FQ∨BC∴δdpq∽δcab, ∴DP/CA=PQ/AB.

∫ Square APDE ∴ AP = DP ∴ AP/CA = PQ/AB = 25t/100 = δ s/25, and the solution is t=8/9.

(3): To solve this small problem, we must first judge whether we should discuss it. The overlapping part of the square APDE and the trapezoidal QBCF may be trapezoidal or irregular. The critical point is the time t' when D is on BC, so we must find out t' first. In (1), according to the obtained time t= 1s, AP=DE=25 and ∫DE∨AB, DE = 1/2ab can be obtained.

There are two situations: ① 8/9 < t < 65438+② 1 ≤ t < 2. I can only tell you so much, and the rest will take a lot of time, and all of them need to use similar triangles's knowledge, so there is no way to make it clear, so we need to make auxiliary lines. Ask if you want to know anything else, or add me qq.