1. First, find out the range of exercise time t.. It takes 6/2=3 seconds to complete the line segment CD and 12/4=3 seconds to complete the line segment BC, so the time range is

0<t< three

2, and then find the area of trapezoidal ABCD, requiring the length and height of AD, with vertical lines from A and D to BC, with vertical scales of M and N respectively. Obviously, there is.

MN=AD, trapezoidal height AM=AB*SIN (angle ABC)=6*sin60 degrees =3 times the root number 3, BM=AB*COS (angle ABC)=6*COS60 degrees =3.

Because BM=CN and AD=MN=BC-BM-CN= 12-3-3=6, the area of trapezoidal ABCD is (AD+BC)*AM/2=27 times the root number 3.

3. The PCQ area of triangle is required, and only the length of PC and CQ is required. PC=BC-BP= 12-4t, and CQ=2t, so the PCQ area of the triangle is = 1/2.

*PC*CQ*SIN (angle C)=(6-2t)* root number 3

4. The area s of pentagonal ABPQD = the area of trapezoidal ABCD.

-

PCQ area of triangle =(27 times root number 3)—

((6-2t)* root number 3)=(2 1+2t)* root number 3.

, where 0

5. Hypothesis exists, according to the meaning of the question.

5 times the area of triangle PCQ.

=

Pentagonal ABPQD area

That is:

5*(6-2t)* Root number 3=(2 1+2t)* Root number 3

, solution

t=3/4=0.75