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Thread: very hard equilateral question

  1. #1
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    very hard equilateral question

    ABC is an equilateral triangle.
    Angle(DAB)=Angle(EBD)
    Area(ABC)=7squareroot(3)
    AE=4
    |BE|=X=?
    Attached Thumbnails Attached Thumbnails very hard equilateral question-ads-z.png  
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    Re: very hard equilateral question

    As DenisB mentioned in the other post, you have shown no work. We do not do your work for you. How about you give it a shot and tell us what you think the answer is.
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    Re: very hard equilateral question

    You don't have to do it. But if I could solve it I shouldn't have asked it. I solve 100 questions a day but there are some questions that I couldn't do.
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    Re: very hard equilateral question

    my time is limited. I will get a brain operation 40 days later I must be quick.
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    Re: very hard equilateral question

    Quote Originally Posted by kastamonu View Post
    You don't have to do it. But if I could solve it I shouldn't have asked it. I solve 100 questions a day but there are some questions that I couldn't do.
    No one asked you to solve it. You were asked to show what work you have done and to explain where you get stuck. You have not even attempted these problems. If you have 99 other problems to do, then after you finish those, attempt this one, and get back to us. When you explain where you are having difficulty, we can help you.
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    Re: very hard equilateral question

    Quote Originally Posted by kastamonu View Post
    ABC is an equilateral triangle.
    Angle(DAB)=Angle(EBD)
    Area(ABC)=7squareroot(3)
    AE=4
    |BE|=X=?
    You need to study this.
    The area of an equilateral triangle, as you can see, is $\dfrac{\sqrt{3}s^2}{4}$ where $s$ is the length of a side.
    Using the given: $\|\overline{BC}\|=s=2\sqrt{7}$.
    SHOW SOME WORK!
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    Re: very hard equilateral question

    For example I drew EC. AEB and BEC are congruent. By using congruency I tried to find something BUT I couldn't.
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    Re: very hard equilateral question

    I am sure we can solve that by congruency. This is an olympics question.
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    Re: very hard equilateral question

    Angle AEB=120 degrees.Let BK be the through B. EK=KA.If we can show that
    BE=EK,it is ok.
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    Re: very hard equilateral question

    I determined the value of $x$ using the law of cosines ...
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    Re: very hard equilateral question

    Used cos 120.
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    Re: very hard equilateral question

    (2SQR7)^2=16+x^2-2xcos120?
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    Re: very hard equilateral question

    Quote Originally Posted by kastamonu View Post
    Angle AEB=120 degrees.Let BK be the through B. EK=KA.If we can show that
    BE=EK,it is ok.
    Where did you get $m(\angle AEB)=120^o~?$
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    Re: very hard equilateral question

    By giving values to angles.CAD=y
    DAB=60-y
    ABE=y
    EBD=60-y
    EDB is exterior angle so 60+y
    60+y+60-y=120(AEB is exterior angle and it is 120 degrees.)
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