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Math Help - similar triangles problem

  1. #1
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    similar triangles problem

    Can anyone explain how this is solved. I know that the sides are proportional to each other but can't get much further.

    A shape is formed by drawing a triangle ABC inside the triangle ADE.
    BC is parallel to DE.
    AB = 4cm, BC = xcm, DE = x + 3cm, DB = x - 4cm
    Calculate the length of DE.
    No angles are given.
    The correct answer is 9cm but how is this arrived at?

    Thanks in advance for your help.
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  2. #2
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    Each set of corresponding sides are in the same ratio...
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  3. #3
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    similar triangles problem still a problem

    Thanks for the quick reply 'prove it' but I did know that. As I mentioned in the original message the sides are proportional to each other. What I really need is someone to walk me through the steps.

    Thanks again.
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  4. #4
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    Hello, Lewis1!

    A shape is formed by drawing a triangle ABC inside the triangle ADE.
    BC is parallel to DE.
    AB = 4cm, BC = x cm, DE = x + 3 cm, DB = x - 4 cm
    Calculate the length of DE.

    The correct answer is 9 cm.
    Code:
                  A
                  o
               4 *  *
                *     *
               *        *
            B o  *  *  *  o C
             *      x       *
      x - 4 *                 *
           *                    *
        D o   *   *   *   *   *   o E
                    x + 3

    We have: .∆ABC ~ ∆ADE


    - - - . . . BC . . .DE . . . . . .x . . .x + 3
    Hence: . --- .= .--- . . . . -- .= .------
    - - - . . . BA . . .DA . . . . . .4 . . . . x


    . . x^2 .= .4x + 12 . . . . x^2 - 4x - 12 .= .0


    . . (x - 6)(x + 2) .= .0 . . . . x ,= ,6, -2


    Hence: .x = 6


    Therefore: .DE .= .x + 3 .= .9

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  5. #5
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    Hi Soroban,
    Thanks very much. When I looked at the solution I realised just how much I have forgotten from my school maths. Thanks again for investing the time.

    Lewis
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  6. #6
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    Quote Originally Posted by Lewis1 View Post
    Thanks for the quick reply 'prove it' but I did know that. As I mentioned in the original message the sides are proportional to each other. What I really need is someone to walk me through the steps.

    Thanks again.
    It is generally against the policy of MHF to walk someone through the steps, as it is expected that students will be doing their own work. In future, post what you do know, post what you have tried and where exactly you are stuck, and if you are given a hint, use it. If you have shown some effort you will generally get more assistance.
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  7. #7
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    Sorry Prove It,
    As a newbie I wasn't aware of this policy. I didn't want to seem rude but I assumed that anyone who went to the effort of posting a thread had already made every effort to solve the problem beforehand. I accept the criticism that I could of included all the steps I had tried but in the rules it does state that posts should be concise. Thanks for the tip anyway.

    Lewis
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  8. #8
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    http://www.mathhelpforum.com/math-he...hp?do=vsarules

    What I posted is in line with rules 6 and 11.
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