设AB=3x,则BC=4x,CD=12x,DA=13x
连接AC
则由勾股定理得AC=√AB^2+BC^2=(3X)^2+(4X)^2=5x
∵AC^2+CD^2=(5X)^2+(12X)^2=169X^2
DA^2=(13X)^2=169X^2
∴AC^2+CD^2=DA^2
∴∠DCA=90°
S四边形=S△ABC+△DCA=AB×BC/2+AC×CD/2
=36x^2=25
x=5/6
则C四边形ABCD=AB+BC+CD+DA=3X+4X+12X+13X=32X
=32×5/6=80/3
不懂,请追问,祝愉快O(∩_∩)O~