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Math Help - [SOLVED] Pythagoras theorem

  1. #1
    adam gerrett
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    [SOLVED] Pythagoras theorem

    okay, ive got a big square, with a square inside it diagonally. leaving 4 right angled triangles. one at each cornor of the large square.

    sides of the internal squares are 'z' and the 3 sides of triangle are x y and z.. z is the hypotenuse...

    i have shown that the large square's area can be written as (x+y) squared and also as z squared + 2xy.

    now i have to use these results to prove pythagoras theorum?!?!

    can any1 help with that? like now.. lol i have to give it in 2moro at skool.
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  2. #2
    Senior Member
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    The page has 64 proofs of the Theorem. Is yours proof number 4?
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  3. #3
    MHF Contributor
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    Quote Originally Posted by adam gerrett
    okay, ive got a big square, with a square inside it diagonally. leaving 4 right angled triangles. one at each cornor of the large square.

    sides of the internal squares are 'z' and the 3 sides of triangle are x y and z.. z is the hypotenuse...

    i have shown that the large square's area can be written as (x+y) squared and also as z squared + 2xy.

    now i have to use these results to prove pythagoras theorum?!?!

    can any1 help with that? like now.. lol i have to give it in 2moro at skool.
    Ummm, you were able to show that the area, A, of the large square is:
    A = (x+y)^2 -----------(1)

    Or,
    A = z^2 +2xy -------------(2) ?

    Very good. In fact, those two equations are both correct.

    Each side of the large square is (x+y), so, A = (side)^2 = (x+y)^2.

    Each of the 4 right triangles has an area of (1/2)(x)(y) = (1/2)xy,
    so the 4 rigth triangles have a collective area of 4*(1/2)xy = 2xy.

    The area of the inner, smaller square of side z is z^2, hence the areaof the large square is
    A = 2xy +z^2
    or, A = z^2 +2xy.

    Now, since A = A, then,
    (x+y)^2 = z^2 +2xy
    Simplifying,
    x^2 +2xy +y^2 = z^2 +2xy
    The 2xy cancells out,
    x^2 +y^2 = z^2 ----------on any of the 4 right tgriangle. The Pythagorean Theorem!

    On any of the 4 right triangles,
    ---z is the hypotenuse
    ---x and y are the two legs
    z^2 = x^2 +y^2
    Phythagorean Theorem.
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  4. #4
    Grand Panjandrum
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    Don't post the same question multiple times, it wastes the time of the helper
    and if ThePerfectHacker catches you he will ban you

    RonL
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  5. #5
    Global Moderator

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    Quote Originally Posted by CaptainBlack
    and if ThePerfectHacker catches you he will ban you
    First he is going to really mad.
    -----------
    I made up my own proof to the pythagorean theorem using circles maybe I will post it. I am just warning you that I might not because it is difficult to draw geometrical shapes through the internet.
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  6. #6
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    As promised, here is my proof.
    I searched the 69 proofs and did not seem to be there.


    2)Triangle ABC with right angle at C.
    3)Using AC draw a circle with that radius at C.
    5)The circle in 2) intersect hypotenuse AB at D because it is longer than the radius.
    7)Extend AD so that it intersect this circle at E.
    11)EB is a secant by definition.
    13)BC is a tangent at C because C is a right angle.
    17)Thus, BD\cdot EB=BC^2.
    19)Thus, BD\cdot (EA+AD+BD)=BC^2
    23)Thus, BD\cdot (AC+AC+BD)=BC^2
    29)Thus, BD\cdot (2AC+BD)=BC^2
    31)Thus, 2AC\cdot BD+BD^2=BC^2
    37)Thus, AC^2+2AC\cdot BD+BD^2=AC^2+BC^2
    39)Thus, (AC+BD)^2=AC^2+BC^2
    41)Thus, (AD+BD)^2=AC^2+BC^2
    43)Thus, AB^2=AC^2+BC^2

    Notes,
    7)Postulate #2
    13)A line perpendicular to the radius on the circumfurence is the tangent.
    17)The theorem of tangent and secants.
    19)Because EB=EA+AD+DE since it is a line.
    23) EA=AD since it is radius of circle.
    29)Add them.
    31)Distributive property of the real numbers.
    37)Addition is well-defined.
    39)A complete square, again distributive.
    41)Radius of circle.
    43)Addition of line segments.

    ---
    Tell me if you need a picture I think it is good enough.
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