According to the meaning of the question, it only takes 1 to finish the BA section, so it takes (t- 1)s to finish the AP section, so AP=8(t- 1)=8t-8.
When the point P moves along D-A (that is, P has reached the point D and started to turn back), the distance taken by P at this time is BA+AD+DP.
According to the meaning of the question, it takes 1s and 6.25s to complete the BA segment and the AD segment respectively, so it takes (t-7.25)s to complete the DP segment.
Then DP=8(t-7.25)=8t-58, then AP=AD-DP= 108-8t.
2. When P is on BA, draw a vertical line from P and A to BC respectively, marked as h 1, h2, then h2= 12.
Then h1/H2 = BP/ba =13t13, and the solution is h 1= 12t, then s △ bpq =1/2 * bq * h/.
And S△ABQ= 1/2*BQ*h2=30t, so s = 30t-30t 2. Since p and a do not coincide, then 0 < t < 1.
When P is on AD, S =1/2 * AP *12 = 48t-48 (AP = 8t-8 has been found in the first question).
Since the time of BA+AD is 7.25=29/4, 1 < t ≤ 29/4.
3. when p is on BA (0 < t ≤ 1), let the intersection of PQ and BR be g, and S△BPG=S△BQG according to the meaning of the question.
According to the second question, s △ bpq = 30t 2, then s △ bpg = s △ bqg = 15t 2.
Because S△RQG∽S△BPG, then s △ rqg/s △ bpg (rq/BP) 2 =1/t 2, then S△RQG= 15t.
While s △ bqr =1/2 * bq *12 = 30t = s △ rqg+s △ bqg =15t2+15t, the solution is t= 1.
When P is on AD (1 < t ≤ 29/4), remember that the intersection of PQ and BR is G, and according to the meaning of the question, S quadrilateral BAPG=S△BQG.
S trapezoid abqp =1/2 (AP+bq) *12 = 78t-48 (AP = 8t-8), then s quadrilateral BAPG=S△BQG=39t-24.
Because S△BQG∽S△RPG and RP=AR-AP=BQ-AP=8-3t, then BQ/RP=5t/(8-3t).
Let the heights of BQ and RP in S△BQG and S△RPG be h 1 and h2 respectively, then h 1/h2=BQ/RP=5t/(8-3t).
And h 1+h2= 12, the solution is h 1=30t/(t+4), then s △ bqg =1/2 * bq * h1= 75t2/(t
After simplification: 3t 2- 1 1t+8 = 0, the solution is t= 1 or 8/3. To sum up: t= 1 or 8/3.
4. A simple diagram shows that when P is on BA (including the endpoint), it is impossible to satisfy C'D'//BC.
In the case of Figure 6, when P is in the AD section, according to the meaning of the question, QCOC' is a diamond, OC=QC=50-5t, OD=OC-CD=37-5t=PD.
And AP=AD-PD=8t-8, t=7.
When P is in the DA segment, PD=OD=37-5t, AP=AD-PD= 108-8t, and the solution is t=95/ 13.
In the case of fig. 7, when P is in the DA segment, PDOD' is a diamond, because AP = 108-8t, then PD=8t-58=OD.
CO=OD-CD=8t-7 1，QC=BC-BQ=50-5t，QC=OC。
Then 50-5t=8t-7 1, and the solution is t =12113.
When P is in the AD segment, because AP = 8t-8, PD=58-8t=OD, CO=OD-CD=45-8t.
Similarly, 50-5t=45-8t, and t < 0 (impossible) is obtained.
To sum up, t=7 or 95/ 13 or12113.