Answer: 353.77
Step-by-step explanation:
Let's start by considering the sector ACB.
The area of this sector is [tex]\pi \cdot 24^{2} \cdot \frac{120}{360}=192\pi[/tex]
Now, let's consider the area of triangle ACB. Its area is given by:
[tex]\frac{1}{2} \cdot 24 \cdot 24 \cdot \sin 120^{\circ}=144\sqrt{3}[/tex]
So the area of the shaded segment is [tex]192\pi -144\sqrt{3} \approx 353.77[/tex]
