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Math Finding area of sector!

 
 
Termoi
 
Reply Wed 9 May, 2012 07:53 am
Hello, I have this question to which it's solution I'm not quite sure of. The question is based on a diagram, and I can't really present it but I'll explain it to my very best, thank you very much for trying to understand my bad presentation of the question.

The diagram is a circle, with center O and points A and B lying on the circle. Radius is 6cm therefore OA=OB=6cm, such that angle AOB = 79 degrees.
Find the area of the shaded segment of the circle contained between the arc AB and the chord [AB].

The solution was a simple sentence, [ (79degrees/360 degrees) x pi x 6^2 ] - [(1/2)(6)(6)sin79]
I understand that [(1/2)(6)(6)sin79] was derived from the rule 1/2absinc and we're finding the area of the triangle.
Therefore, [ (79degrees/360 degrees) x pi x 6^2 ] must be the area of the sector. However, when I looked up the formula for the area of sector, it's (1/2)(delta)(r)^2, where delta is the angle measured in radians.
Primarily, how did [ (79degrees/360 degrees) x pi x 6^2 ] come about then?
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raprap
 
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Reply Wed 9 May, 2012 12:03 pm
@Termoi,
2 pi radians is 360 degrees. If you divide by 2, then pi radians is 180 degrees or 1 radian is 180/pi degrees.

Rap
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markr
 
  1  
Reply Thu 10 May, 2012 12:39 am
@Termoi,
Another way to look at it is that the area of the sector is a fraction (79/360) of the area of the circle. The area of the circle is pi*6^2
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