Given that in #\Delta ABC#, #A=\pi/12#, #B={5\pi}/12#
#C=\pi-A-B#
#=\pi-\pi/12-{5\pi}/12#
#={\pi}/2#
from sine in #\Delta ABC#, we have
#\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}#
#\frac{a}{\sin(\pi/12)}=\frac{b}{\sin ({5\pi}/12)}=\frac{c}{\sin ({\pi}/2)}=k\ \text{let}#
#a=k\sin(\pi/12)=0.259k#
# b=k\sin({5\pi}/12)=0.966k#
#c=k\sin({\pi}/2)=k#
#s=\frac{a+b+c}{2}#
#=\frac{0.259k+0.966k+k}{2}=1.1125k#
Area of #\Delta ABC# from Hero's formula
#\Delta=\sqrt{s(s-a)(s-b)(s-c)}#
#12=\sqrt{1.1125k(1.1125k-0.259k)(1.1125k-0.966k)(1.1125k-k)}#
#12=0.125k^2#
#k^2=96#
Now, the in-radius (#r#) of #\Delta ABC#
#r=\frac{\Delta}{s}#
#r=\frac{12}{1.1125k}#
Hence, the area of inscribed circle of #\Delta ABC#
#=\pi r^2#
#=\pi (12/{1.1125k})^2#
#=\frac{144\pi}{1.2376k^2}#
#=\frac{365.521}{96}\quad (\because k^2=96)#
#=3.8075\ \text{unit}^2#