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    Hi can anyone help me and find out how i have done the area of the segment wrong.

    Question:The sector has a radius of 5 cm and the angle between it is 34 degrees.
    Work out the area of the sector and segment.

    Sector:
    \dfrac{\pi \theta r^2}{360}
    \dfrac{\pi * 34 * 5^2}{360} = \dfrac{85\pi}{36}
    That is what i got for the area of the sector.

    Segment:
    \dfrac{1}{2}a b sin C

    \dfrac{1}{2}*5*5* sin 34 = 6.9899

    Area of segment = area of sector - area of triangle

    \dfrac{85\pi}{36} - 6.9899 = 0.427738 cm^2

    Although apparently the answer is 2.92cm ^2
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    (Original post by ch0wm4n)
    Hi can anyone help me and find out how i have done the area of the segment wrong.

    Question:The sector has a radius of 5 cm and the angle between it is 34 degrees.
    Work out the area of the sector and segment.

    Sector:
    \dfrac{\pi \theta r^2}{360}
    \dfrac{\pi * 34 * 5^2}{360} = \dfrac{85\pi}{36}
    That is what i got for the area of the sector.

    Segment:
    \dfrac{1}{2}a b sin C

    \dfrac{1}{2}*5*5* sin 34 = 6.9899

    Area of segment = area of sector - area of triangle

    \dfrac{85\pi}{36} - 6.9899 = 0.427738 cm^2

    Although apparently the answer is 2.92cm ^2
    Your calculation is OK.
    Are you sure that the answer relates to this question?
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    (Original post by ch0wm4n)
    Hi can anyone help me and find out how i have done the area of the segment wrong.

    Question:The sector has a radius of 5 cm and the angle between it is 34 degrees.
    Work out the area of the sector and segment.

    Sector:
    \dfrac{\pi \theta r^2}{360}
    \dfrac{\pi * 34 * 5^2}{360} = \dfrac{85\pi}{36}
    That is what i got for the area of the sector.

    Segment:
    \dfrac{1}{2}a b sin C

    \dfrac{1}{2}*5*5* sin 34 = 6.9899

    Area of segment = area of sector - area of triangle

    \dfrac{85\pi}{36} - 6.9899 = 0.427738 cm^2

    Although apparently the answer is 2.92cm ^2
    \dfrac{\pi \theta r^2}{180} is what you want for sector area where theta is in degrees.
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    (Original post by ztibor)
    Your calculation is OK.
    Are you sure that the answer relates to this question?
    ok thanks
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    (Original post by ShahzaibMuneeb)
    \dfrac{\pi \theta r^2}{180} is what you want for sector area where theta is in degrees.
    i think that is length of the arc of a sector.
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    (Original post by ch0wm4n)
    i think that is length of the arc of a sector.
    Nope that's:

    \dfrac{\pi \theta r}{180} where theta is in degrees.
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    (Original post by ShahzaibMuneeb)
    \dfrac{\pi \theta r^2}{180} is what you want for sector area where theta is in degrees.
    Nope, it is over 360.

    As a check, consider what your formula gives for a sector with angle at the centre of 180, i.e. half a circle.
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    (Original post by ghostwalker)
    Nope, it is over 360.

    As a check, consider what your formula gives for a sector with angle at the centre of 180, i.e. half a circle.
    Oh, sorry, my bad, I just realised what I was doing wrong. I was assuming that formula for measuring the area of a sector was r^2 \theta where it's actually \frac{1}{2} r^2 \theta.. which makes sense why I was getting half less in the denominator... sorry
 
 
 
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