Question

To the nearest tenth, what is the area of the shaded segment when BN = 8 ft?

Answer 1

Answer:

The area of the shaded segment is approximately $39.3 ft^{2}$

Solution:

Note: Refer the image attached below.

As given, BN = 8 ft

So radius r = 8

The angle c = $120^{\circ}$

We know that the area of the shaded part is $=\left(\frac{r^{2}}{2}\right) \times\left(\left(\frac{\pi}{180}\right) \times c-\sin (c)\right)$

$=\left(\frac{8^{2}}{2}\right) \times\left(\left(\frac{\pi}{180}\right) \times 120-\sin (120)\right)$ (// putting the value of r and c

)

$=\left(\frac{64}{2}\right) \times\left(\left(\frac{\pi}{180} \times 120\right)-\left(\frac{\sqrt{3}}{2}\right)\right)$ (// putting value of   sin⁡(120))

$=32 \times\left(\frac{2 \pi}{3}-\left(\frac{\sqrt{3}}{2}\right)\right)$

$=32 \times\left(\frac{4 \pi-3 \sqrt{3}}{6}\right)$

$=\left(\frac{16}{3}\right) \times(4 \pi-3 \sqrt{3})$

$=5.33 \times((4 \times 3.14)-(3 \times 1.732))$  (//putting value of π and √3 )

$=5.33 \times(12.56-5.196)$

$=5.33 \times 7.36$

= 39.2288 which is approximately, 39.3

So, the area of the shaded part is $39.3 ft^{2}$