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Math Help - similar triangle

  1. #1
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    similar triangle

    find x from the picture in attachment
    Attached Thumbnails Attached Thumbnails similar triangle-triangle.bmp  
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  2. #2
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    Talking

    Your subject line refers to similar triangles, but I see nothing saying that the given picture displays such...?

    Please reply with clarification. Thank you!
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  3. #3
    No one in Particular VonNemo19's Avatar
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    4/(x+5)=(12+4)/(3x+5)

    4(3x+5)=16(x+5)

    3x+5=4x+5

    3x=4x

    Since this is obviously impossible you can see that the triangles can not be similar. Maybe you copied the problem wrong?
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  4. #4
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    You can find x by use of the cosine law, but I cannot visualize where you would use a similar triangle function to arrive at the value of x.
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  5. #5
    No one in Particular VonNemo19's Avatar
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    I'm interested in how you would solve this with the law of cosines aidan. If you can show me, you've got a thanks coming your way.
    Last edited by VonNemo19; May 2nd 2009 at 08:53 AM. Reason: 1
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  6. #6
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    I do not have access to graphics to create a an image, but
    Label the points around the triangle, counter clockwise from the left most point. You should have A,B,C,D,E.
    A being on the extreme left and E on the extreme right, with C being at the top. Point B is between A & C. Point D is between C & E.
    Further clarifications:
    Distances:
    A-to-B is 16
    B-to-C is 8
    C-to-D is 4
    D-to-E is 12
    B-to-D is x+5
    A-to-E is 3x+5

    Cos(C) will refer to the angle \angle ACE & \angle BCD
    Cosine Law:
     c^2 = a^2 + b^2 - 2ab \cos\left(C\right)

    Equation1:
    \left( x + 5 \right)^2 = 8^2 + 4^2 - 2\times 8 \times4\times cos\left( C \right)
    Equation2:
    \left( 3 x +5 \right)^2 = \left( 16+8 \right)^2 + \left( 4+12 \right) ^2 - 2 \left( 16+8 \right) \left( 4+12 \right) cos \left( C \right)

    Two equations, unknown x and unknown angle C.

    You can expand the LHS of equation1  \left( x + 5 \right)^2 \; = \; x^2 + 10x + 25
    and the same for equation2.

    or re-write:
    \left( \; \left( x + 5 \right)^2 - 8^2 - 4^2 \right) \; \div \left( 2\times 8 \times4 \; \right) = cos\left( C \right)

    Either way will work.
    My preference is to re-write the equations, and isolate the trig function (the cosine). This effectively gives you one equation with one unknown, which reduces to a simple quadratic equation.

    Hope that helps.
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  7. #7
    No one in Particular VonNemo19's Avatar
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    that's thinking outside the box!
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  8. #8
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    Hello, beq!x!

    That picture is way out-of-scale . . . and mine is no better.

    There must be typo . . . The problem has no solution.

    Code:
                                                    A
                                                    *
                                                *    * 4
                                            *         *
                                     8  *              o E
                                    *           o       *
                                *       o                *
                            *   o        x + 5            *
                      D o                                  * 12
                    *                                       *
           16   *                                            *
            *                                                 *
        *                                                      *
    *   *   *   *   *   *   *   *   *   *   *   *   *   *   *   *
    B                           3x + 5                          C

    I agree with Aidan . . . the Law of Cosines: . \cos A \:=\:\frac{b^2+c^2-a^2}{2bc}


    In \Delta ABC\!:\;\;\cos A \;=\;\frac{24^2 + 16^2 - (3x+5)^2}{2(24)(16)} \;\;{\color{blue}[1]}

    In \Delta ADE\!:\;\;\cos A \;=\;\frac{8^2+4^2-(x+5)^2}{2(8)(4)} \;\;{\color{blue}[2]}


    Equate [1] and [2]: . \frac{832 - (3x+5)^2}{768} \;=\;\frac{80 - (x+5)^2}{64}

    This simplifies to: . x^2 + 30x + 49 \:=\:0

    Quadratic Formula: . x \;=\;\frac{-30 \pm\sqrt{704}}{2} \;=\;\frac{-30\pm8\sqrt{11}}{2} \;=\;-15 \pm 4\sqrt{11}


    We have two roots: . x \;\approx\;-1.7335,\;-28.2665

    But both roots yield a negative value for side BC \,=\,3x+5

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  9. #9
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    hmm maybe it is a typo...
    i will try to edit it on Monday or Tuesday
    thanks for help at all you!
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  10. #10
    No one in Particular VonNemo19's Avatar
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    Yes, one of the criteria that must be meet when triangles are similar is that they're corresponding sides are proportional.
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  11. #11
    No one in Particular VonNemo19's Avatar
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    Math A
    Similar Triangles


    The cat on the right is an enlargement of the cat on the left. They are exactly the same shape, but they are NOT the same size.
    These cats are similar figures.

    Objects, such as these two cats, that have the same shape, but do not have the same size, are said to be "similar".


    The mathematical symbol used to denote
    similar is .
    Do you remember this symbol as "part" of the symbol for congruent??

    Similar
    Symbol


    Definition: In mathematics, polygons are similar if their corresponding (matching) angles are equal and the ratio of their corresponding sides are in proportion.
    (This definition allows for congruent figures to also be "similar",
    where the ratio of the corresponding sides is 1:1.)




    Facts about similar triangles:






    is called the ratio of similitude

    AA
    To show triangles are similar, it is sufficient to show that two angles of one triangle are congruent (equal) to two angles of the other triangle.





    Roberts


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