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20 12 mathematical answer to the 26th question in Yulin senior high school entrance examination.
When 1.t=2 and cq=2, the square of cp = the square of QP-the square of CQ =20-4.

16, so cp=4, so oc=8, so point d coordinates (8,4). Because neither P nor Q is at the endpoint, its value is 0.

2.s=32

Cq=t, DQ = 4-T. Because triangle adq is similar to triangle ecq, ce: ad = CQ: dq, so ce=8t/(4-t).

Dq=fd=4-t, so S = 8 * (8-2t)/2+2 * (4-t) * 8t/(2 (4-t)) = 32.

3。 Still thinking. . . I don't feel so limited. . .