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Math Help - Chord problem

  1. #1
    Member smmmc's Avatar
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    Chord problem

    Hey
    1.a chord of length 6 cm is drawn in a circle of radiue 7cm. Find:
    a)the area of the smaller region inside the circle cut off by the chord.

    My working out:
    Formula i used is r^2 (theta-sintheta)
    So r=7, theta=?
    Find theta.
    Sine theta/2 =3/7
    =50.75 degrees conversion to radians is 0.89.
    So plug out degree and radius into the equation
    r^2(theta-sintheta)
    r^2(0.89-sin0.89)
    =21.42 ??, the actual answer is 2.73 cm^2


    my diagram is below
    Attached Thumbnails Attached Thumbnails Chord problem-untitled.jpg  
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  2. #2
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    Using the chord formula as you done, we should get the angle theta as being

    14sin(\frac{\theta}{2})=6

    {\theta}=2sin^{-1}(\frac{3}{7})\approx .885822 \;\ \text{radians} or about 50.75 degrees.

    Then, using the formula for the area of a circular segment gives:

    \frac{1}{2}(7)^{2}(2sin^{-1}(\frac{3}{7})-sin(2sin^{-1}(\frac{3}{7})))\approx 2.73
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  3. #3
    Member smmmc's Avatar
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    Quote Originally Posted by galactus View Post
    Using the chord formula as you done, we should get the angle theta as being

    14sin(\frac{\theta}{2})=6

    {\theta}=2sin^{-1}(\frac{3}{7})\approx .885822 \;\ \text{radians} or about 50.75 degrees.

    Then, using the formula for the area of a circular segment gives:

    \frac{1}{2}(7)^{2}(2sin^{-1}(\frac{3}{7})-sin(2sin^{-1}(\frac{3}{7})))\approx 2.73
    hey thats the exact thing i did, but in radians, i plugged your formula into the calc and it says 1224.4 ???
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  4. #4
    Eater of Worlds
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    Make sure your calculator is not set to degrees even though you're working in radians. Alas, something is wrong there somewhere.

    That is your problem. I just checked. Your calculator is set to degrees.
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  5. #5
    Member smmmc's Avatar
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    YES! my calc is in degree mode .. thanks
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  6. #6
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    Quote Originally Posted by smmmc View Post
    Hey
    1.a chord of length 6 cm is drawn in a circle of radiue 7cm. Find:
    a)the area of the smaller region inside the circle cut off by the chord.

    My working out:
    Formula i used is r^2 (theta-sintheta)
    So r=7, theta=?
    Find theta.
    Sine theta/2 =3/7
    =50.75 degrees conversion to radians is 0.89.
    So plug out degree and radius into the equation
    r^2(theta-sintheta)
    r^2(0.89-sin0.89)
    =21.42 ??, the actual answer is 2.73 cm^2


    my diagram is below
    Using the Cosine Rule, we find

    b^2 = a^2 + c^2 - 2ac \cos{\theta}

    6^2 = 7^2 + 7^2 - 2\cdot 7\cdot 7 \cdot \cos{\theta}

    36 = 49 + 49 - 98\cos{\theta}

    36 = 98 - 98\cos{\theta}

    98\cos{\theta} = 62

    \cos{\theta} = \frac{31}{49}

    \theta = \arccos{\frac{31}{49}}.


    As long as the angle is in degrees, the Area of the sector is given by

    A_{\textrm{sector}} = \frac{\theta}{360} \pi r^2

     = \frac{1}{360}\cdot \arccos{\frac{31}{49}} \cdot \pi \cdot 7^2

     = \frac{49\pi}{360} \cdot \arccos{\frac{31}{49}}.


    The Area of the triangle is found using Heron's Formula.

    s = \frac{a + b + c}{2}

     = \frac{7 + 7 + 6}{2}

     = \frac{20}{2}

     = 10.

    A_{\textrm{triangle}} = \sqrt{s(s - a)(s - b)(s - c)}

     = \sqrt{10(10 - 7)(10 - 7)(10 - 6)}

     = \sqrt{10\cdot 3 \cdot 3 \cdot 4}

     = \sqrt{360}

     = 6\sqrt{10}.


    So the Shaded Area is given by

    A = \frac{49\pi}{360} \cdot \arccos{\frac{31}{49}} - 6\sqrt{10} \,\textrm{units}^2
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