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C2 - Sector and Segment help

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:
πθr2360\dfrac{\pi \theta r^2}{360}
π∗34∗52360=85π36\dfrac{\pi * 34 * 5^2}{360} = \dfrac{85\pi}{36}
That is what i got for the area of the sector.

Segment:
12absinC\dfrac{1}{2}a b sin C

12∗5∗5∗sin34=6.9899\dfrac{1}{2}*5*5* sin 34 = 6.9899

Area of segment = area of sector - area of triangle

85π36−6.9899=0.427738cm2\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:
πθr2360\dfrac{\pi \theta r^2}{360}
π∗34∗52360=85π36\dfrac{\pi * 34 * 5^2}{360} = \dfrac{85\pi}{36}
That is what i got for the area of the sector.

Segment:
12absinC\dfrac{1}{2}a b sin C

12∗5∗5∗sin34=6.9899\dfrac{1}{2}*5*5* sin 34 = 6.9899

Area of segment = area of sector - area of triangle

85π36−6.9899=0.427738cm2\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?
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:
πθr2360\dfrac{\pi \theta r^2}{360}
π∗34∗52360=85π36\dfrac{\pi * 34 * 5^2}{360} = \dfrac{85\pi}{36}
That is what i got for the area of the sector.

Segment:
12absinC\dfrac{1}{2}a b sin C

12∗5∗5∗sin34=6.9899\dfrac{1}{2}*5*5* sin 34 = 6.9899

Area of segment = area of sector - area of triangle

85π36−6.9899=0.427738cm2\dfrac{85\pi}{36} - 6.9899 = 0.427738 cm^2

Although apparently the answer is 2.92cm ^2


πθr2180\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
πθr2180\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.
Original post by ch0wm4n
i think that is length of the arc of a sector.

Nope that's:

πθr180\dfrac{\pi \theta r}{180} where theta is in degrees.
Original post by ShahzaibMuneeb
πθr2180\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.
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 r2θr^2 \theta where it's actually 12r2θ\frac{1}{2} r^2 \theta.. which makes sense why I was getting half less in the denominator... sorry :frown:

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