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Math Help - Strange Problem

  1. #1
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    Strange Problem

    Hello, I'm trying to solve this problem, I really didn't know if it was Algebra or Trigonometry, so if this is the wrong section, I'm sorry.

    On the next figure, the segments m1 and m2 are parallel, what's the value of X?



    I tried solving using a system of ecuations, but I can't cause I have 3 unknowns, m1, m2 and x. Tried lots of things but couldn't solve it.

    Thanks in advance.
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  2. #2
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    Hello, norifurippu!

    In the figure, the segments m1 and m2 are parallel.
    What is the value of x ?
    Code:
          *
          |   *    8
       m1 |       *
          |           *             9
          * - - - - - - - * - - - - - - - - - *
                  x           *               |
                                  *           | m2
                                 2x   *       |
                                          *   |
                                              *
    We don't need the lengths of m1 and m2.

    Since m1 \parallel m2, the two triangles are similar.

    Hence: . \frac{x}{8} \:=\:\frac{9}{2x} \quad\Rightarrow\quad 2x^2 \:=\:72 \quad\Rightarrow\quad x^2 \:=\:36 \quad\Rightarrow\quad x \:=\:6

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  3. #3
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    Thank you Soroban.
    So we know the 2 triangles are similar only because m1 and m2 are parallel? Is there anything else we need to know 2 triangles are similar?
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  4. #4
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    Hello again, norifurippu!

    So we know the 2 triangles are similar only because m1 and m2 are parallel?
    Is there anything else we need to know 2 triangles are similar?
    Here's a more complete explanation . . .

    Code:
              B
              o
             /    *
         m1 /         *
           /              *
        A o - - - - - - - - - o - - - - - - - - - - - - - - o E
                              C   *                        /
                                      *                   /
                                          *              / m2
                                              *         /
                                                  *    / 
                                                      o
                                                      D

    Since m1 \parallel m2\!:\;\;\angle A \,=\,\angle E . (alternate-interior angles)

    Also: . \angle BCA \,=\,\angle DCE . (vertical angles)

    Hence: . \angle B \,=\,\angle D


    Therefore: . \Delta ABC \:\sim\: \Delta CDE . (a.a.a.)

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  5. #5
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    Now I get it! Thanks a lot Soroban!
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