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Ede4ka [16]
3 years ago
9

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

Mathematics
1 answer:
yan [13]3 years ago
8 0

<u>Answer:</u>

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

<u>Solution:</u>

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}

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