∵ trapezoid ABCD is an isosceles trapezoid.

∴CE=(BC-AD)/2=3, then DE=√(CD? -CE？ )=4.

Therefore: s trapezoid ABCD = (ad+BC) * de/2 = (6+12) * 4/2 = 36.

(2) If the intersection D is DF∨AB and BC is in F, BF = AD = 6 and CF = BC-BF = 6.

At t seconds, PD=t, CP = CD-PD = 5-t; ? CQ=2t。

∫PQ∨AB，DF∨AB。

∴pq∑df, CP/CD = CQ/CF, (5-t)/5 = (2t)/6, t = 15/8 (seconds);

③ When pq ⊥ QC and DE⊥EC, PQ is parallel to DE.

∴ CP/CD = CQ/CE, (5-t)/5 = (2t)/3, t = 15/ 13 (seconds);

When PQ ⊥ CD: ≈ QPC = ∠ DEC = 90; ∠C=∠C。

∴⊿ QPC ∽⊿ DEC, CP/CE = CQ/CD, (5-t)/3 = (2t)/5, t = 25/ 1 1 (seconds).

To sum up, when the three points P, Q, C Q and C form a right triangle, the point P leaves the point D 15/ 13 seconds or 25/ 1 1 second. ?