Then (3t+2t) × 6x1/2 = 33.

The solution is: t= 1 1/5s, and because 3t < 16cm, when p and q start from 1 1/5s, the area of quadrilateral PBCQ is 33cm2.

(2) Point Q is QT⊥AB, so QT=BC=6cm.

TQ= 10cm, so PT=8cm.

Because AB= 16cm, CQ+AP=8cm.

Let CQ=2t AP=3t.

So: 5t=8, t=8/5s.