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
