1, which is relatively simple. If the perpendicular from D to BC is E, then QC=(BC-AD)/2=3, CD=5, CE=3, so DE=4 (Pythagorean Theorem) and the area S=36.
2. The intersection point D is DF||AB, and the intersection point BC is F. It is not difficult to get CF=6 from the topic surface. At this time, let a point on the CD be P and a point on BC be Q, and get PQ||DF according to the meaning of the topic. Let point p leave point D x seconds, then CP distance is 5-x, CQ distance is 2x, CQ/CF = CP/CD- >.
2x/6 = (5-x)/5-> x =15/8 =1.875 seconds.
3. Let P run for y seconds, let PQ be perpendicular to BC, and let the vertical foot be Q, then CQ/Ce = CP/CD-> 2Y/3 = (5-y)/5-> Y =15/13 =1.15 seconds.