Answer:
≈ 0.8 m²
Step-by-step explanation:
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FORMULA FOR FINDING AREA OF SEGMENT :-
[tex]A = r^{2} [\frac{\alpha }{360} \times \pi - (\sin \frac{\alpha }{2} \times \cos \frac{\alpha }{2} )][/tex]
Where :-
- A = Area of the segment
- r = Radius of the circle
- α (alpha) = Angle subtended by arc at the center
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Angle subtended by the minor arc at the center = 360° - 300° = 60°
Radius of the circle = 3 m
Area of the segment =
[tex]3^{2} [ \frac{60}{360} \times \pi - (\sin \frac{60}{2} \times \cos \frac{60}{2})][/tex]
[tex]= 9[\frac{1}{6} \times \pi - (\sin 30 \times \cos 30)][/tex]
[tex]= 9[\frac{\pi }{6} - (\frac{1}{2} \times \frac{\sqrt{3} }{2} )][/tex]
[tex]= 9[\frac{\pi }{6} - \frac{\sqrt{3} }{4} ][/tex]
[tex]= 9[\frac{2\pi - 3\sqrt{3} }{12} ][/tex]
[tex]= \frac{9}{12} \times (2\pi - 3\sqrt{3})[/tex]
( Assuming √3 = 1.73 & π = 3.14 )
[tex]= \frac{3}{4} \times (6.28 - 5.19)[/tex]
[tex]= 0.75 \times 1.09[/tex]
[tex]= 0.8175[/tex]
∴ Area of the segment ≈ 0.8 m² (rounding to the nearest tenth)
