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Math Help - pythag problem

  1. #1
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    pythag problem

    Hi all, I have an answer which is puzzeling me

    from the answer page it reads;

    if L is the length of the shorter edge, by pythag` theorem, 2L^2 =a^2, so L^2=1/2a^2

    Hence the shorter edge L, has length a/sqrt 2

    if it helps the picture is a right angled triangle with a hyp` of length given as a, and the two shorter lenghts given as L

    what I can`t get is why the shorter edge L, has length a/sqrt 2

    Dave
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  2. #2
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    Re: pythag problem

    Quote Originally Posted by davellew69 View Post
    what I can`t get is why the shorter edge L, has length a/sqrt 2
    Because 2L^2 =a^2.
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  3. #3
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    Re: pythag problem

    Thanks for the reply , but stilll stuck
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  4. #4
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    Re: pythag problem

    Please indicate the last claim of the proof that you understand and the first one that you don't.
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  5. #5
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    Re: pythag problem

    im struggling to see how edge L, has length a/sqrt 2
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  6. #6
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    Re: pythag problem

    Quote Originally Posted by davellew69 View Post
    im struggling to see how edge L, has length a/sqrt 2
    Sorry, you did not specify the last claim of the proof that you understand. This must mean that you don't understand why L^2 + L^2 = a^2. This is an instance of the Pythagoras theorem because we have a right-angled triangle with two sides of length L and the hypotenuse of length a.

    So, we have 2L^2 = a^2. Divide both sides by 2. Now, take the square root of both sides.

    I am just repeating the proof you gave. I am still curious to know which exactly part of it you don't understand.
    Last edited by emakarov; July 21st 2012 at 10:44 AM.
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  7. #7
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    Re: pythag problem

    i get the pythag , just not the ending
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  8. #8
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    Re: pythag problem

    why do we divide both sides by 2?
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  9. #9
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    Re: pythag problem

    Look, is it too hard for you to say: "This equality I understand, but how this one follows from it I don't understand."
    We have the following theorems about square root:

    (1) If x\ge0 , then \sqrt{x^2}=x .
    (2) If x\ge0 and y>0 , then \sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}.

    So, we have the following derivation.

    1. From the Pythagoras theorem, 2L^2=a^2. Divide both sides by 2.
    2. L^2=\frac{a^2}{2} . Take square root of both sides.
    3. \sqrt{L^2}=\sqrt{\frac{a^2}{2}} . Apply (2)
    4. \sqrt{L^2}=\frac{\sqrt{a^2}}{\sqrt{2}} . Apply (1)
    5. L=\frac{a}{\sqrt{2}} .

    Could you say which of the facts (1), (2) and steps 1-5 were not clear to you?
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  10. #10
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    Re: pythag problem

    Quote Originally Posted by davellew69 View Post
    why do we divide both sides by 2?
    Because in the final claim L=\frac{a}{\sqrt{2}} , L is alone in the left-hand side, but in the Pythagoras theorem 2L^2=a^2 ,  L^2 is multiplied by 2. We are trying to solve the latter equation for L, so we are trying to isolate L.
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  11. #11
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    Re: pythag problem

    thanks , all makes sense :0 )
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