solution: extend the QR of BA intersection to point m and connect AR, AP.
∵ AC = GC, BC=FC, ∠ACB=∠GCF,
∴△ABC≌△GFC,
∴.
∴∠ RHA+∠ DAH = 181,
∴∠ RHA = ∠ BAC = 31,
∴∠ QHg = 61,
Cos 31 = 4× 32 = 23.
then qh = ha = Hg = AC = 23.
in right angle △HMA, HM=AH? sin61°=23×32=3.AM=HA? Cos61 = 3.
In right-angle △AMR, Mr = AD = AB = 4.
∴ QR = 23+3+4 = 7+23.
∴ Qp = 2QR = 14+43.
PR = QR. 3 = 73+6.
The circumference of ∴△ PQR is equal to RP+QP+QR = 27+133.
So the answer is: 27+133.