Obviously △APM∽△CQM, while AP length is 8t and CQ length is 2t.
The ratio of ∴△APM to △CQM is (8t)/2t.
∴△APM is 6×(8-t)/(8+t), and△△△ CQM is 6×2t/(8+t).
The area of ∴△APM is1/2× [6× (8-t)/(8+t) ]× (8-t).
The area of △CQM is 1/2×[6×2t/(8+t)]×2t.
∑△APM is 6 times larger than△△△ CQM.
∴ 1/2×[6×(8-t)/(8+t)]×(8-t)- 1/2×[6×2t/(8+t)]×2t=6
∴t=2 or t=-8 (truncation)
When the area of triangle APM is 6 larger than that of triangle CQM, the value of t is 2.
(2) The total length of O-A-B-C is 8+8+6=24.
The topic is equivalent to the relative movement of point P at a speed of 4 units/second and point Q at a speed of 2 units/second on a straight line with a length of 24.
Meaning from the topic: |24-(4+2)t |≤6
Solution: 3≤t≤5
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