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Math Help - Find the possible values of X using "Similar right triangles"?

  1. #1
    Newbie Lizarrdo's Avatar
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    Find the possible values of X using "Similar right triangles"?

    Find the possible values of x.

    Find the possible values of X using "Similar right triangles"?-question-13.png
    I got 6, but the answer is 12. Why is that? My teachers said there was an easy way to do it without quadratics but I don't remember.
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  2. #2
    Senior Member BAdhi's Avatar
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    there are three right angled triangles that you can use to apply phythagoras theorem. apply that theorem to two of them find the lenght of unknown sides with the variable x.

    then apply the pythagoras theorem to the other triangle to find x
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  3. #3
    Senior Member Sambit's Avatar
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    If you want to avoid quadratics, try to use the fact that the sides of the opposite of the same angles of two similar triangles are proportional in length.
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  4. #4
    Newbie Lizarrdo's Avatar
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    Quote Originally Posted by BAdhi View Post
    there are three right angled triangles that you can use to apply phythagoras theorem. apply that theorem to two of them find the lenght of unknown sides with the variable x.

    then apply the pythagoras theorem to the other triangle to find x
    So

    (x-3)^2 + x^2 = x^2

    x^2 - 9 + x^2 = x^2

    2x^2 - 9 = x^2

    -9 = -x^2

    x=3?

    I did something wrong for sure. How would I do teh square root of (x-3)?
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  5. #5
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    Hello, Lizarrdo!


    \text}Find the possible values of }x.

    Code:
                C
                *
               *| *
              * |   *   x
             *  |h    *
            *   |       *
           *    |         *
        A *  *  *  *  *  *  * B
           x-5  D    x-3

    Note that x > 5.


    We have right triangle ABC,\:\angle C = 90^o.

    The altitude to the hypotenuse is: . h = CD.
    It divides the hypotenuse into two segments: . AD = x-5,\;DB = x-3.


    Fact: The altitude to the hypotenuse is the geometric mean
    . . . . . . . . . of the segments of the hypotenuse.

    Hence, we have: . h^2 \:=\:(x-3)(x-5) .[1]


    In right triangle CDB\!:\;h^2 + (x-3)^2 \:=\:x^2 \quad\Rightarrow\quad h^2 \:=\:x^2-(x-3)^2 .[2]


    Equate [1] and [2]: . (x-5)(x-3) \:=\:x^2 - (x-3)^2

    We have: . x^2 -14x + 24 \:=\:0 \quad\Rightarrow\quad (x-2)(x-12) \:=\:0

    . . Therefore: . x \,=\,12

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