(2)S trapezoid ABCD=27√3/4(cm? )

(3) At t seconds, BP=2tcm, CE = (6-2t) cm; CQ = Chinese medicine.

Then s triangle PCQ = CP * CQ * SIN ∠ C = (1/2) * (6-2t) * T * SIN60 = (3 √ 3t-√ 3t? )/2.

Therefore: S(ABPQD)=S trapezoid -S triangle PCQ=

That is, S=(√3/2)t? -(3√3/2)t+27√3/4。 (0 & ltt & lt3)

(4) When PQ divides the trapezoidal area into two parts: 1:5:

S triangle PCQ=( 1/6)S trapezoid ABCD = (1/6) * (27 √ 3/4) = 9 √ 3/8.

Namely: (3√3t-√3t? )/2=9√3/8，(2t-3)？ =0，t=3/2。

Therefore, when t=3/2 seconds, PQ divides the trapezoidal ABCD area into 1:5.