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Circle problem - perimater of a minor sector

  1. Feb 3, 2005 #1
    Code (Text):
                     ..      
                A .'    '.B  
                 . \    / .  
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    The radius of the circle is 5. The perimeter of the minor sector AOB is [tex]P{\pi} + Q[/tex]. Find P and Q
     
    Last edited: Feb 4, 2005
  2. jcsd
  3. Feb 3, 2005 #2

    HallsofIvy

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    The circumference of the circle is, of course, [itex]10\pi[/itex]. I don't see any way of determining P and Q without knowing what part of the entire circle AOB is. Are you given the central angle? Surely there must be some conditions on P and Q? If P and Q could be any real numbers, then even if we know exactly what the perimeter is, we could choose one of P and Q to be anything we want and then solve for the other.

    I take it by "perimeter" you mean the whole perimeter, both the curved part and the two radii. A plausible answer would be that Q= 10 (the length of the two radii) and P would be [itex]\frac{5\theta}{\pi}[/itex].
     
  4. Feb 3, 2005 #3

    saltydog

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    How about this:

    [tex] P \pi +Q=5\theta [/tex]

    with theta being the radian measure of the angle (inside the slice of pie):

    Thus for "any" Q:

    [tex] P=\frac {5\theta-Q}{\pi} [/tex]

    Is is really for any Q?
     
    Last edited: Feb 3, 2005
  5. Feb 3, 2005 #4

    HallsofIvy

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    Why "[itex]5\theta[/itex]"?
     
  6. Feb 3, 2005 #5

    saltydog

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    Ohhh, they mean the whole perimeter around the slice of pie and not just the arc length. In that case may I suggest:


    [tex] P \pi +Q=(5\theta+10) [/tex]
    (the arc length +2*radius)

    with theta being the radian measure of the angle (inside the slice of pie):

    Thus for "any" Q:

    [tex] P=\frac {5\theta+10-Q}{\pi} [/tex]
     
    Last edited: Feb 3, 2005
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