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
