Solution:

It is known that if P is PE⊥RQ and RQ is E, RE=QE=8/2=4,

PE =√(pq ^2-qe^2)=√(25- 16)=3

tg∠PQE=PE/QE=3/4

Let PQ of isosceles △PQR intersect with CD in F, then

It is known that ABCD is a square, and points B, C, Q and R are on the same straight line L.

∴CD⊥CB ,CF⊥CQ

CF=CQ*tg∠PQE=3CQ/4

According to the known conditions, it can be concluded that

CQ=t seconds * 1 cm/sec =tcm.

CF=t*3/4=3t/4

∴S=CF*CQ/2=(t*3t/4)/2=3t^2/8