Given that in #\Delta ABC#, #A=\pi/2#, #B={\pi}/3#
#C=\pi-A-B#
#=\pi-\pi/2-{\pi}/3#
#={\pi}/6#
from sine in #\Delta ABC#, we have
#\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}#
#\frac{a}{\sin(\pi/2)}=\frac{b}{\sin ({\pi}/3)}=\frac{c}{\sin ({\pi}/6)}=k\ \text{let}#
#a=k\sin(\pi/2)=k#
# b=k\sin({\pi}/3)=0.866k#
#c=k\sin({\pi}/6)=0.5k#
#s=\frac{a+b+c}{2}#
#=\frac{k+0.866k+0.5k}{2}=1.183k#
Area of #\Delta ABC# from Hero's formula
#\Delta=\sqrt{s(s-a)(s-b)(s-c)}#
#24=\sqrt{1.183k(1.183k-k)(1.183k-0.866k)(1.183k-0.5k)}#
#24=0.2165k^2#
#k^2=110.8545#
Now, the in-radius (#r#) of #\Delta ABC#
#r=\frac{\Delta}{s}#
#r=\frac{24}{1.183k}#
Hence, the area of inscribed circle of #\Delta ABC#
#=\pi r^2#
#=\pi (24/{1.183k})^2#
#=\frac{576\pi}{1.399489k^2}#
#=\frac{1293.0129}{110.8545}\quad (\because k^2=110.8545)#
#=11.664\ \text{unit}^2#