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Christophicus
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Report Thread starter 14 years ago
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A chord AB subtends an angle of theta radians at the centre O of a circle of radius 6.5cm. Find the area of the segment enclosed by the chord AB and the minor arc AB when theta = 4pi/3 = 180/pi*4pi/3 degrees = 240 degrees

Easy enough.
Area = 0.5r^2(theta - sin theta)
Area = 0.5*6.5^2(4pi/3 - sin240)
Area = 106.8cm^2

The book says 25.9cm^2

Who is right?
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john !!
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look at the question closer.. it asks for the area of the segment formed by the MINOR arc, and as it stands the angle subtended is reflex.
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Christophicus
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(Original post by mik1w)
look at the question closer.. it asks for the area of the segment formed by the MINOR arc, and as it stands the angle subtended is reflex.
ah good point, thanks.
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