tbrooks3
find the area of segment with a 60 degree angle and 12 radius.



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bond
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\[r ^{2}(\pi \Theta \div360\sin \Theta \div2)\]

maheshmeghwal9
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Area of Segment = ½ × ( (θ × π/180)  sin θ) × r2 (when θ is in degrees)

bond
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Why 1/2 @maheshmeghwal9


King
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24pi36sqrt3?

maheshmeghwal9
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but it is the area of sector I think @ParthKohli

maheshmeghwal9
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we have to find segment

maheshmeghwal9
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's area