Given that in #\Delta ABC#, #A=\pi/12#, #B=\pi/6#
#C=\pi-A-B#
#=\pi-\pi/12-\pi/6#
#={3\pi}/4#
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 (\pi/6)}=\frac{c}{\sin ({3\pi}/4)}=k\ \text{let}#
#a=k\sin(\pi/12)=0.259k#
# b=k\sin(\pi/6)=0.5k#
#c=k\sin({3\pi}/4)=0.707k#
#s=\frac{a+b+c}{2}#
#=\frac{0.259k+0.5k+0.707k}{2}=0.733k#
Area of #\Delta ABC# from Hero's formula
#\Delta=\sqrt{s(s-a)(s-b)(s-c)}#
#4=\sqrt{0.733k(0.733k-0.259k)(0.733k-0.5k)(0.733k-0.707k)}#
#4=0.0459k^2#
#k^2=87.187#
Now, the in-radius (#r#) of #\Delta ABC#
#r=\frac{\Delta}{s}#
#r=\frac{4}{0.733k}#
Hence, the area of inscribed circle of #\Delta ABC#
#=\pi r^2#
#=\pi (4/{0.733k})^2#
#=\frac{16\pi}{0.537289k^2}#
#=\frac{93.5539}{87.187}\quad (\because k^2=87.187)#
#=1.073\ \text{unit}^2#