Given that in \Delta ABC, A=\pi/12, B={7\pi}/8
C=\pi-A-B
=\pi-\pi/12-{7\pi}/8
={\pi}/24
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 ({7\pi}/8)}=\frac{c}{\sin ({\pi}/24)}=k\ \text{let}
a=k\sin(\pi/12)=0.259k
b=k\sin({7\pi}/8)=0.383k
c=k\sin({\pi}/24)=0.1305k
s=\frac{a+b+c}{2}
=\frac{0.259k+0.383k+0.1305k}{2}=0.38625k
Area of \Delta ABC from Hero's formula
\Delta=\sqrt{s(s-a)(s-b)(s-c)}
8=\sqrt{0.38625k(0.38625k-0.259k)(0.38625k-0.383k)(0.38625k-0.1305k)}
8=0.006392k^2
k^2=1251.634
Now, the in-radius (r) of \Delta ABC
r=\frac{\Delta}{s}
r=\frac{8}{0.38625k}
Hence, the area of inscribed circle of \Delta ABC
=\pi r^2
=\pi (8/{0.38625k})^2
=\frac{64\pi}{0.1492k^2}
=\frac{1347.699}{1251.634}\quad (\because k^2=1251.634)
=1.0767\ \text{unit}^2
