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#