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Math Help - Geometrical meaning of √2

  1. #1
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    Geometrical meaning of √2

    See the drawing at the attached file, the sides of the triangle are 1,1,2

    Ιwant a geometric relation for 2, i.e. something like that:
    a/1=b/a=>b=a^2
    and thus
    2=1+a+b+c+...=1+a+a^2+a^4+a^8+a^16+...
    But a/1=b/ais probably wrong. Can you proove it if itís correct?
    Any other similar relation I could get?
    Attached Thumbnails Attached Thumbnails Geometrical meaning of √2-square-root-2.jpg  
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  2. #2
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    A right triangle is one in which ( leg1)^2 + (leg 2)^2 = (hypothenuse)^2
    1^2 + 1^2 = rad2^2
    hypothenuse =rad2
    Any equal legged right rt triangle has a hypothenuse= to leg* rad2


    bjh
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  3. #3
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    a=?
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  4. #4
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    If x is the length of the leg of an isosceles right triangle, then (x)(√2) is the length of its hypotenuse ... we can prove it using pythagorean theroem .
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  5. #5
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    a=?
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  6. #6
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    Quote Originally Posted by ThodorisK View Post
    a=?
    a = 1 - \dfrac{1}{\sqrt{2}}
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  7. #7
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    geometric meaning of rad2

    Hello Thodorisk,
    I read your largest triangle as a right triangle with sides 1 and 1 and hyoothenuse =rad 2

    1^2 + 1^2 =rad2^2 proving that it is a right triangle If the two legs of a right triangle are equal then the hypothenuse = leg times rad2 Call a leg a then the hypot is atimes rad2

    bjh
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  8. #8
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    No rad, sin, cos etc please.

    a is not √2-1.
    √2-1=a+b+c+...

    a=?
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  9. #9
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    It's not clear exactly what you are trying to do. I can see that you have a right triangle with legs of length 1 and so hypotenuse of length \sqrt{2}. Then, I think, you have marked of a length of 1 on the hypotenuse and drawn a line from the right angle to that point. But how is the second line, separting the segments marked "a" and "b" determined. Is it parallel to the first line from the right angle?
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  10. #10
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    That huge drawing shows what's parallel and what's not. As I am not hoping to get an answer but only warnings from the moderators, it's a waste of time for me too. Bann me and I am over with your site.
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