Let in #\Delta ABC#, #\angle A={3\pi}/8#, #\angle B=\pi/2# hence
#\angle C=\pi-\angle A-\angle B#
#=\pi-{3\pi}/8-\pi/2#
#={\pi}/8#
For maximum perimeter of triangle , we must consider the given side of length #4# is smallest i.e. side #c=4# is opposite to the smallest angle #\angle C=\pi/8#
Now, using Sine rule in #\Delta ABC# as follows
#\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}#
#\frac{a}{\sin ({3\pi}/8)}=\frac{b}{\sin (\pi/2)}=\frac{4}{\sin ({\pi}/8)}#
#a=\frac{4\sin ({3\pi}/8)}{\sin (\pi/8)}#
#a=4(\sqrt2+1)# &
#b=\frac{4\sin ({\pi}/2)}{\sin (\pi/8)}#
#b=4\sqrt{4+2\sqrt2}#
hence, the maximum possible perimeter of the #\triangle ABC # is given as
#a+b+c#
#=4(\sqrt2+1)+4\sqrt{4+2\sqrt2}+4#
#=8+4\sqrt2+4\sqrt{4+2\sqrt2}#