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Thread: circle's annular sector

  1. #1

    circle's annular sector

    Hello,

    I was wondering if anyone could help me find the perimeter and area of a 54 degree annular sector with inner radius of 20 and outer radius of 30. This is what I got so far:

    I supposed area = area of outer circle-area of inner circle so I used A=pi times 30 to the power of 2 and then pi times 20 to the power of 2 which gave me 500pi.

    Then, I got 56/360= x/500pi which gave me 100pi square units.

    So, my question is, am I on the right track or not? If not, what should I do? Also, how do I find the perimeter?

    Thanks for your time.

  2. #2
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    Re: circle's annular sector

    I believe what you're talking about when you say "annular sector" is a section of a double ring geometrical figure, whose inner radius is 20 and outer 30; Correct?

    I had to do some digging to find out exactly what one of those was; Thank you wikipedia.

    Related to a torus! Nifty!

    To area:

    [tex]A = \pi(R^2 - r^2)\,[/tex]

    Which is essentially like saying, the area of the larger circle minus the area of the "hole" taken out of the center.

    So in this case:

    [tex]A = \pi(30^2 - 20^2)\ = 500\pi,[/tex]

    Which is absolutely correct.

    I'll tackle more of this tomorrow, I'm up too late as it is!

    --Pontifex
    Ever wonder where we get all these nifty mathematical symbols? Behold the glory of [tex:3g09vodw]TeX[/tex:3g09vodw].

  3. #3

    Re: circle's annular sector

    That's exactly right Pontifex. Sorry for not clarifying that lol. So, this means I'm on the right track, but I wonder if the rest is right. By the way, sorry because I don't know how to write pi and other stuff on the computer lol... thanks a lot!!

  4. #4
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    Re: circle's annular sector

    The area of the sector of a circle of [tex]\theta[/tex] radians is:

    [tex]A_{\theta} = \frac{r^2}{2}\theta[/tex]

    So the modified formula for the area of an annular sector would be:

    [tex]A_{\theta}= \frac{R^2-r^2}{2}\theta[/tex]

    Since you are working in degrees, replace [tex]\theta[/tex] with [tex]\frac{\pi \theta^o}{180}[/tex]

  5. #5
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    Re: circle's annular sector

    For the perimeter we will have four borders to add.

    Since you are working in degrees, I will suppose [tex]\theta[/tex] is in degrees.

    First, add the two partial circumferences (inner + outer): [tex]\frac{\pi\theta r}{180}+\frac{\pi\theta R}{180}=\frac{\pi\theta}{180}(R+r)[/tex]

    The other two borders are what adjoin the circumferences together. They are just the difference in radii, and there are two: [tex]2(R-r)[/tex].

    So your perimiter will be the sum of all:

    [tex]P_{\theta} = \frac{\pi\theta}{180}(R+r) + 2(R-r)[/tex]

  6. #6

    Re: circle's annular sector

    let me see if I understand: so area would be R2-r2/2 times pi times 0degrees/180? Forgive me b/c I don't know how to type...so in my case the 0 would be 54? Am I right?

  7. #7
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    Re: circle's annular sector

    Let me (hopefully) shed some light on what's going on here with the help of a picture:
    [attachment=0:1xun2e6u]480x320_Annular Sector.jpg[/attachment:1xun2e6u]

    As you can see we have your Annular Sector there, a thin slice of a "doughnut".

    Area:

    So your area would be the space inside the slice there, which daon helpfully provided:

    [tex]A = {R^2 - r^2 \over 2}\theta[/tex]

    Which - in English - is the difference of the larger circle's radius and the inner circle forming the "doughnut", then the [tex]\theta[/tex](pronounced "theta" and written using TeX, like this: '\theta' [minus the quotes]) there is the fraction of the complete "doughnut" you're asking about (54 degrees in this case). But! Since we're talking about degrees instead of radians, we have to convert between the two in order to get the right answer. Which is where all this talk about [tex]\theta={\pi\theta^{\circ}\over180}[/tex] comes from (Where [tex]\theta[/tex] is radians and [tex]\theta^{\circ}[/tex] is the number of degrees you have).

    Perimeter:

    So the perimeter (as per the figure) is:

    [tex]P_{total} = a+b+P_1+P_2[/tex]

    [tex]a[/tex] and [tex]b[/tex] are easy, they're just difference between [tex]R[/tex] and [tex]r[/tex]; And there are two of them so we see where the:

    [tex]a+b = 2(R - r)[/tex]

    Came from.

    The curved portions [tex]P_1[/tex] and [tex]P_2[/tex] are a bit harder. But we can find them easily with the help of arc length. Which is [tex]L = \theta r[/tex], where [tex]\theta[/tex] is the portion of the circle we want to measure and [tex]r[/tex] is the radius. Since we're talking about degrees instead of radians in this case, we use the conversion already mentioned:

    [tex]\theta={\pi\theta^{\circ}\over180}[/tex]

    So our formula ends up being:

    [tex]L_\circ = {\pi\theta^{\circ} r \over 180}[/tex]

    There are two lengths there with differing radii, so that's where Daon pulls the second part of the equation from:

    [tex]P_1+P_2 = {\pi\theta^{\circ} \over 180}{(R +r)}[/tex]

    Hope that's clear.

    --Pontifex
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    Ever wonder where we get all these nifty mathematical symbols? Behold the glory of [tex:3g09vodw]TeX[/tex:3g09vodw].

  8. #8
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    Re: circle's annular sector

    Quote Originally Posted by Pontifex
    Let me (hopefully) shed some light on what's going on here with the help of a picture:
    [attachment=0:2ffyl8kh]480x320_Annular Sector.jpg[/attachment:2ffyl8kh]

    As you can see we have your Annular Sector there, a thin slice of a "doughnut".

    Area:

    So your area would be the space inside the slice there, which daon helpfully provided:

    [tex]A = {R^2 - r^2 \over 2}\theta[/tex]

    Which - in English - is the difference of the larger circle's radius and the inner circle forming the "doughnut", then the [tex]\theta[/tex](pronounced "theta" and written using TeX, like this: '\theta' [minus the quotes]) there is the fraction of the complete "doughnut" you're asking about (54 degrees in this case). But! Since we're talking about degrees instead of radians, we have to convert between the two in order to get the right answer. Which is where all this talk about [tex]\theta={\pi\theta^{\circ}\over180}[/tex] comes from (Where [tex]\theta[/tex] is radians and [tex]\theta^{\circ}[/tex] is the number of degrees you have).

    Perimeter:

    So the perimeter (as per the figure) is:

    [tex]P_{total} = a+b+P_1+P_2[/tex]

    [tex]a[/tex] and [tex]b[/tex] are easy, they're just difference between [tex]R[/tex] and [tex]r[/tex]; And there are two of them so we see where the:

    [tex]a+b = 2(R - r)[/tex]

    Came from.

    The curved portions [tex]P_1[/tex] and [tex]P_2[/tex] are a bit harder. But we can find them easily with the help of arc length. Which is [tex]L = \theta r[/tex], where [tex]\theta[/tex] is the portion of the circle we want to measure and [tex]r[/tex] is the radius. Since we're talking about degrees instead of radians in this case, we use the conversion already mentioned:

    [tex]\theta={\pi\theta^{\circ}\over180}[/tex]

    So our formula ends up being:

    [tex]L_\circ = {\pi\theta^{\circ} r \over 180}[/tex]

    There are two lengths there with differing radii, so that's where Daon pulls the second part of the equation from:

    [tex]P_1+P_2 = {\pi\theta^{\circ} \over 180}{(R +r)}[/tex]

    You need to add the straight edges (indicated by a & b - in your picture) for perimeter (daon's final answer includes that)

    Hope that's clear.

    --Pontifex
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

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