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20 1 1 The detailed process of the eighth problem of one model mathematics in grade three in Haidian.
In my opinion, this question chooses A:

When t=0, PC=AC=4, and the square is equal to 16.

When P reaches the point C to the vertical foot of AC, PC is the shortest, so PC=ACxBC/AB= 12/5, and the square is equal to 144/25, and less than 9 is less than 16, so when T is greater than 0 and less than 5, the parabolic symmetry axis T =12 remains. PC=BC=3, then the square is equal to 9, and then P slides on BC, PC=AB+BC-t=8-t, then the square is equal to the square of (8-t), so it is also a parabola, and the opening is upward, so choose A. 。