Answer:
The area of the shaded segment is approximately [tex]39.3 ft^{2}[/tex]
Solution:
Note: Refer the image attached below.
As given, BN = 8 ft
So radius r = 8
The angle c = [tex]120^{\circ}[/tex]
We know that the area of the shaded part is [tex]=\left(\frac{r^{2}}{2}\right) \times\left(\left(\frac{\pi}{180}\right) \times c-\sin (c)\right)[/tex]
[tex]=\left(\frac{8^{2}}{2}\right) \times\left(\left(\frac{\pi}{180}\right) \times 120-\sin (120)\right)[/tex] (// putting the value of r and c
)
[tex]=\left(\frac{64}{2}\right) \times\left(\left(\frac{\pi}{180} \times 120\right)-\left(\frac{\sqrt{3}}{2}\right)\right)[/tex] (// putting value of sin(120))
[tex]=32 \times\left(\frac{2 \pi}{3}-\left(\frac{\sqrt{3}}{2}\right)\right)[/tex]
[tex]=32 \times\left(\frac{4 \pi-3 \sqrt{3}}{6}\right)[/tex]
[tex]=\left(\frac{16}{3}\right) \times(4 \pi-3 \sqrt{3})[/tex]
[tex]=5.33 \times((4 \times 3.14)-(3 \times 1.732))[/tex] (//putting value of π and √3 )
[tex]=5.33 \times(12.56-5.196)[/tex]
[tex]=5.33 \times 7.36[/tex]
= 39.2288 which is approximately, 39.3
So, the area of the shaded part is [tex]39.3 ft^{2}[/tex]
