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Thread: Trapezoids

  1. #16
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    Re: Trapezoids

    How will I use this proportion?
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  2. #17
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    Re: Trapezoids

    Quote Originally Posted by Idea View Post
    CD meets BE at F

    AC = BF = a
    BE = b

    We let h and k be the distances from C and D to the side BE

    4 = area BCF - area BDF = 1/2 (h - k) a

    Use the proportion

    \frac{k}{b-a}=\frac{h}{b}
    How did you find this proportion?
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  3. #18
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    Re: Trapezoids

    Trapezoids-traps.jpg

    EDF and EAB are similar triangles therefore \frac{\text{EF}}{\text{EB}}=\frac{k}{h}
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  4. #19
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    Re: Trapezoids

    I beg your pardon. I am solving nearly 70 questions a day. My brain stopped.

    ABE=9
    CDA=4
    We don't know the area of CFE.
    Area of BECD is 5.
    To use the above ratio we need the area of FCE or BDE or DFE
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  5. #20
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    Re: Trapezoids

    area BCD = 4 = area ADC

    area BCE = 9 = area ABE
    Last edited by Idea; May 30th 2017 at 03:59 AM.
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  6. #21
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    Re: Trapezoids

    ADC and DFE are similar.
    CD/AB=2/3
    DC/DF=2
    ABC=ACE
    DFE=2
    CDE=2
    ABC=6
    Is that right?
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  7. #22
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    Re: Trapezoids

    Quote Originally Posted by kastamonu View Post
    adc and dfe are similar.
    Cd/ab=2/3
    dc/df=2
    abc=ace
    dfe=2
    cde=2
    abc=6
    is that right?
    dfe = 1
    cde = 2
    bdf = 2

    abc= 6
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  8. #23
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    Re: Trapezoids

    I made a typo DFE is 1. Many Thanks.
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