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Math Help - proof of relation of radius, chord, chord hight of a circle

  1. #1
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    proof of relation of radius, chord, chord hight of a circle

    Hello, I am ok with this 'till the last step, re-arranging for a;

    r^2 = (OS)^2 + a^2 (by Principle of Pathagoras, OS being that part of the radius below a chord and a being half the chord length, h the chord hight)
    = (r-h)^2 + a^2
    = r^2 - 2hr + h^2 + +a^2


    therefor a^2 = 2hr - h^2

    can someone explain the method involved in the last step?

    Thanks,

    Clotsworth
    Last edited by Clotsworth; November 25th 2012 at 03:11 AM.
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  2. #2
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    Re: proof of relation of radius, chord, chord hight of a circle

    Quote Originally Posted by Clotsworth View Post
    Hello, I am ok with this 'till the last step, re-arranging for a;

    r^2 = (OS)^2 + a^2 (by Principle of Pathagoras, OS being that part of the radius below a chord and a being half the chord length, h the chord hight)
    = (r-h)^2 + a^2
    = r^2 - 2hr + h^2 + +a^2


    therefor a^2 = 2hr - h^2

    can someone explain the method involved in the last step?

    Thanks,

    Clotsworth
    \displaystyle \begin{align*} r^2 &= r^2 - 2hr + h^2 + a^2 \\ 0 &= -2hr + h^2 + a^2 \\ 2hr - h^2 &= a^2 \end{align*}
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  3. #3
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    Re: proof of relation of radius, chord, chord hight of a circle

    Marvellous,
    That is most helpful, and much appreciated, thank you.

    Regards,

    Clotsworth
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