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Math Help - concentric circles with radii, outside and inside areas

  1. #1
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    concentric circles with radii, outside and inside areas

    i need help with this.

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  2. #2
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    Quote Originally Posted by aaaaaa1234 View Post
    i need help with this

    Sounds like the area between every 2nd consecutive circle is shaded.

    Therefore my suggestion is

    Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)

    Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]

    Should be fairly easy from there!

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    Quote Originally Posted by pickslides View Post
    Sounds like the area between every 2nd consecutive circle is shaded.

    Therefore my suggestion is

    Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)

    Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]

    Should be fairly easy from there!

    im not sure i totally get it, so is the answer 1156? if not, then how do u do it? thanks. if you could solve that step by step, i'd really appreciate it. thanks you
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    Hello, aaaaaa1234!

    48 concentric circles are drawn having radii 1, 2, 3, ... 48.
    The area outside the circles of odd radii but inside the circles of even radii are shaded.
    What is the area of the shaded region?
    A circle has area: . \pi r^2

    The shaded area is: . A \:=\:(\pi\!\cdot\!2^2 - \pi\!\cdot\!1^2) + (\pi\!\cdot\!4^2 - \pi\!\cdot\!3^2) + (\pi\!\cdot\!6^2-\pi\!\cdot\!5^2) + \hdots + (\pi\!\cdot\!48^2 - \pi\!\cdot\!47^2)

    . . = \;3\pi + 7\pi + 11\pi + \hdots + 95\pi \;=\; \underbrace{(3 + 7 + 11 + \hdots + 95)}_{\text{arithmetic series}}\pi

    The series has: first term a = 3, common difference d = 4, and n = 24 terms.

    . . Its sum is: . \tfrac{24}{2}[2(3) + 23(4)] \:=\:1176


    Therefore: . A \;=\;1176\pi
    Last edited by mr fantastic; April 22nd 2009 at 12:39 AM. Reason: No edit - just flagging the post as having been moved from a thread where the same question was posted.
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    Quote Originally Posted by pickslides View Post
    Sounds like the area between every 2nd consecutive circle is shaded.

    Therefore my suggestion is

    Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)

    Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]

    Should be fairly easy from there!

    When visualising the problem you should see similar to an achery target.

    Every second circle (or hoop section) coloured in. My equation calculates the area of each hoop (as the difference of the area of the outer part and inner part of each coloured circle) and summing them together to find total area. The area of the first hoop (the biggest hoop) is simply

    Area = (\pi.48^2-\pi.47^2)

    then the next biggest

    Area = (\pi.46^2-\pi.45^2)

    then the next biggest

    Area = (\pi.44^2-\pi.43^2)

    and so on until you get to the end

    Area = (\pi.2^2-\pi.1^2)

    Add all these guys together and you get

    Total Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)

    In the next line of my calculation I simply removed \pi as a common factor.

    Total Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]

    I arrived at the solution Total Area = \pi.1176 \approx 3692.64 units^2
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  6. #6
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    Quote Originally Posted by pickslides View Post
    When visualising the problem you should see similar to an achery target.

    Every second circle (or hoop section) coloured in. My equation calculates the area of each hoop (as the difference of the area of the outer part and inner part of each coloured circle) and summing them together to find total area. The area of the first hoop (the biggest hoop) is simply

    Area = (\pi.48^2-\pi.47^2)

    then the next biggest

    Area = (\pi.46^2-\pi.45^2)

    then the next biggest

    Area = (\pi.44^2-\pi.43^2)

    and so on until you get to the end

    Area = (\pi.2^2-\pi.1^2)

    Add all these guys together and you get

    Total Area = (\pi.48^2-\pi.47^2)+(\pi.46^2-\pi.45^2)+\cdots+(\pi.2^2-\pi.1^2)

    In the next line of my calculation I simply removed \pi as a common factor.

    Total Area = \pi[(48^2-47^2)+(46^2-45^2)+\cdots+(2^2-1^2)]

    I arrived at the solution Total Area = \pi.1176 \approx 3692.64 units^2
    i think i get it, so for like pi[(48^2-47^2) you would do 48-47 and you end up with 1^2?
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    Quote Originally Posted by aaaaaa1234 View Post
    i think i get it, so for like pi[(48^2-47^2) you would do 48-47 and you end up with 1^2?
    No that arithmetic isn't very robust.

    48^2-47^2 \neq (48-47)^2

    The arithmetic you should use is.

    48^2 = 2304

    47^2 = 2209

    So

    48^2-47^2 = 2304-2209 = 95

    You would have to repeat that calculation for every hoop before summing together.

    I would let ms excel do the work for me
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    would the answer be 1176pi or 1176pi unit squares?
    Last edited by aaaaaa1234; April 21st 2009 at 06:38 PM.
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  9. #9
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    1176pi units squared is the answer
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