Given that in \Delta ABC, A=\pi/8, B=\pi/4
C=\pi-A-B
=\pi-\pi/8-\pi/4
={5\pi}/8
from sine in \Delta ABC, we have
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
\frac{a}{\sin(\pi/8)}=\frac{b}{\sin (\pi/4)}=\frac{c}{\sin ({5\pi}/8)}=k\ \text{let}
a=k\sin(\pi/8)=0.383k
b=k\sin(\pi/4)=0.707k
c=k\sin({5\pi}/8)=0.924k
s=\frac{a+b+c}{2}
=\frac{0.383k+0.707k+0.924k}{2}=1.007k
Area of \Delta ABC from Hero's formula
\Delta=\sqrt{s(s-a)(s-b)(s-c)}
48=\sqrt{1.007k(1.007k-0.383k)(1.007k-0.707k)(1.007k-0.924k)}
48=0.125k^2
k^2=384
Noe, the in-radius (r) of \Delta ABC
r=\frac{\Delta}{s}
r=\frac{48}{1.007k}
Hence, the area of inscribed circle of \Delta ABC
=\pi r^2
=\pi (48/{1.007k})^2
=\frac{2304\pi}{1.007^2k^2}
=\frac{2304\pi}{1.007^2\cdot 384}\quad (\because k^2=384)
=18.588\ \text{unit}^2
