According to the experience of understanding and doing problems, right-angled triangles AB and BC are right-angled sides and AC is hypotenuse. The answer is as follows.

Solution: Let the moving time be t,? As shown in the figure

Then AP=t, BQ=2t.

BP=6-t，？ CQ= 12-2t

The area of △ABC is s1=1/2× ab× BC =1/2× 6×12 = 36.

△ mpqs2 = the area of s (△ ABC)-s (△ APM)-s (△ BPQ)-s (△ MQC).

∴? For the right triangle BPQ,

There are s (△ bpq) =1/2× BP× bq =1/2× (6-t )× (2t) = 6t-t&; sup2

For triangle APM, H 1 is the height of △APM, and m is the midpoint, so h 1= 1/2BC6.

∴S(△APM)=？ 1/2×AP×h 1 = 1/2×AP×( 1/2BQ)= 3t

Similarly, for △MQC, its area is s (△ mqc) =1/2× QC× H2 =1/2× (12-2t) × 3 =18-3t.

That is, the area of △ mpq S2 = s (△ ABC)-s (△ APM)-s (△ bpq)-s (△ MQC).

= 36-3t-(6t-t & amp； sup2)-( 18-3t)

= t & ampsup2-6t+ 18

The area of △MPQ is a quarter of that of △ABC.

So S2= 1/4(S 1).

t & ampsup2-6t+ 18= 1/4×36

Get a solution? t=3

That is, the p movement time is 3 seconds. Yes, the conditions are met.

Finished answering? Please adopt it in time if you are satisfied.

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