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Math Help - Geometry regarding the intersection of wires criss crossing vertical poles.

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    Geometry regarding the intersection of wires criss crossing vertical poles.

    Two vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?

    help step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.
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  2. #2
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    Do you find it odd that the distance between the poles is not given?

    In any case, you're going to need some similar triangles. If you have a method that has similar triangles, I suspect it is NOT totally worthless. Post it.
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    Quote Originally Posted by swanz View Post
    Two vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?

    help step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.
    If you know the poles are vertical, you know they are parallel. You can treat each wire as a transversal of these - check out which angles are alternate interior angles of these two transversals to see some similar triangles. Draw yourself a decent picture to see them! Then use these similar triangles to answer the question.
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  4. #4
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    Quote Originally Posted by swanz View Post
    Two vertical poles are of lengths 10 m and 6 m. They are connected by wires going from the top of each pole to the base of the other. At what height do the two connecting wires intersect?

    help step by step would be appreciated. thanks. i have tried but failed maybe something very simple i am missing. btw i am new, so if i was to post my method (which was utterly useless), i apologize in advance for not doing so.
    Assume the wires are straight (with no sag)
    Since the distance between the two is irrelevant, we can assign it what ever needed, say 1 m.

    Assume the shorter pole is coincident with the y axis and the taller pole is parallel but at x = 1

    An equation for each wire:

    y_1 = -6x_1 + 6

    y_2 = 10x_2 + 0


    Isolate x:

     x_1 = \dfrac{ y_1 - 6 }{-6}

     x_2 = \dfrac{y_2}{10}

    equate:

     \dfrac{ y - 6}{-6} = \dfrac{y}{10}

    solve for y
     y = \dfrac{ 60 }{16}

    .
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    Resolved

    thank you aidan....never even considered plotting it on the graph .....all the time i was playing around with similar triangles, transversals and loads of variables.
    nice method, can come in handy in various situations, thnx a bunch.
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    Quote Originally Posted by aidan View Post
    Since the distance between the two is irrelevant
    Well, one probably should prove that, rather than just state it.
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  7. #7
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    Quote Originally Posted by TKHunny View Post
    Well, one probably should prove that, rather than just state it.
    My Bad. I just made an assumption that it was common knowledge.
    Sorry about that.
    One proof (if needed) is here:
    PlanetMath: harmonic mean in trapezoid

    It indicates that the diagonals of a trapezoid will intersect at some point. A line through that point parallel to the parallel sides will intersect the non parallel sides. The length of this segment is a ratio of the parallel sides only:

    If the lengths of the parallel sides are "a" and "b" then the length of the interior line that is parallel to the parallel lines and passes through the intersection point is

     \dfrac{2 a b }{(a+b)}


    The intersection point is at the midpoint of the interior line.

    In our case here:

    HALF the length of the interior line is

     \dfrac{( 6 \cdot 10) }{( 6 + 10 )} = 3.75

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    I think it is second tier common knowledge. It can be assumed if you work with that sort of thing, but not generally.

    My views. I welcome others.
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  9. #9
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    That problem is "ye olde crossing ladders" problem in disguise
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