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Math Help - Arc length

  1. #1
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    radians....

    hey...

    if a sector of a circle has a perimeter of 20cm,

    how do you prove the "area = 25.(r-5)^2"

    thankyou
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  2. #2
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    Area of a sector of a circle

    Remember that S = r\theta

    S = arc length
    r = radius
    theta = angle of arc/sector

    And the area of a sector is \pi*r^2\frac{\theta}{2\pi}

    which simplifies to r^2\frac{\theta}{2}

    we know from the given information that S = 20cm so we can write theta in terms of r.

     20cm = r\theta
     \theta = \frac{20cm}{r}

    Lets substitute theta into the previous equation and we get:

    r^2\frac{\frac{20cm}{r}}{2}

    which simplifies to:

    r*10cm

    I beleive I have made a mistake somewhere . . . can anyone find it?
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  3. #3
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    MathGuru, perimeter of sector is
    P = r+r+s
    P = 2r +r*theta
    So,
    2r +r*theta = 20
    r*theta = 20 -2r
    theta = (20 -2r)/r

    Using your area A = [(r^2)theta]/2,
    A = [(r^2)/2](theta)
    A = [(r^2)/2][(20 -2r)/r]
    A = r(10-r) ----***

    ---------
    I'd do it another way.

    P = 2r +s
    2r +s = 20
    s = 20 -2r
    s = 2(10-r)

    A = (1/2)(r)(s)
    A = (1/2)(r)[2(10-r)]
    A = r(10-r) ----same as in your way.

    -----------------
    That A is not the same as the A = 25(r-5)^2, which is given in the question.

    I say then that A = 25(r-5)^2 is wrong.
    And so, the question is wrong.

    --------------
    Another reason why A = 25(r-5)^2 is wrong.

    When r = 5:

    A = 25(5 -5)^2 = 0.
    There is no area, no sector, when r = 5?
    Incredible.

    P = 20 = 2r +s
    20 = 2*5 +s
    s = 10
    So, if r = 5, and s = 10, why is there no sector?
    Last edited by ticbol; July 4th 2005 at 10:41 PM.
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  4. #4
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    soz i typed it incorrectly :S its A = 25 - (r - 5)^2

    there lol soz bout that :S
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  5. #5
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    Umm, so A = 25 -(r-5)^2 ?

    That figures.

    I found A = r(10-r).
    I will convert that into A = 25 -(r-5)^2.

    A = r(10 -r)
    A = 10r -r^2
    r^2 -10r +A = 0
    Complete the square of the r-terms,
    [r^2 -10r] + A = 0
    [r^2 -10r +(-10/2)^2 -(-10/2)^2] +A = 0
    [r^2 -10r +(-5)^2 -(-5)^2] +A = 0
    [(r -5)^2 -(25)] +A = 0
    (r-5)^2 -25 +A = 0
    A = 25 -(r-5)^2 -----***

    Proven.
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