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Deriving an equation

  1. May 23, 2007 #1

    danago

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    Gold Member

    Two ladders, one red, and the other green, are 2 and 3m long respectively. The base of the red ladder is resting on the side of a narrow hallway and leaning on the wall of the other side of the hallway. The green ladder is doing the same, but on the opposite side, such that the two ladders cross eachother and form an 'X' shape. The point of intersection is 1m above the floor. Show how the following equation is derived, and thus, find the width of the hallway:
    [tex]
    \frac{1}{{\sqrt {4 - x^2 } }} + \frac{1}{{\sqrt {9 - x^2 } }} = 1
    [/tex]



    Ive been working on this for quite a while, and cant seem to get anywhere. Using pythagoras' theorem, i can show that the red ladder rests [tex]
    {\sqrt {4 - x^2 } }
    [/tex] meters up the wall, and the green ladder rests [tex]
    {\sqrt {9 - x^2 } }
    [/tex] meters up the wall.

    Ive been trying to find expressions for other lengths so that i can create an equation, but i have been unsuccessfull in doing so.

    If anybody is able to shed some light on the problem, id be very thankful.

    Thanks in advance,
    Dan.
     
  2. jcsd
  3. May 23, 2007 #2

    Curious3141

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    Homework Helper

    Hint : Let the horiz. distance along the floor of the intersection point from the left wall be a, and that from the right wall be b.

    Now a + b = x, which is the width of the hallway.

    Consider similar right triangles to get expressions for a and b in terms of x. Put those in the previous equation, cancel out the x's and see what you get.
     
  4. May 23, 2007 #3

    danago

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    Gold Member

    Ahhh ofcourse. Didnt even think to take notice of the similar triangles. Got the equation now :smile:

    Thanks alot for the help.
     
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