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)}
7=\sqrt{1.1125k(1.1125k-0.259k)(1.1125k-0.966k)(1.1125k-k)}
7=0.125k^2
k^2=56
Now, the in-radius (r) of \Delta ABC
r=\frac{\Delta}{s}
r=\frac{7}{1.1125k}
Hence, the area of inscribed circle of \Delta ABC
=\pi r^2
=\pi (7/{1.1125k})^2
=\frac{49\pi}{1.23766k^2}
=\frac{124.3787}{56}\quad (\because k^2=56)
=2.221\ \text{unit}^2
