Friday, December 28, 2012

Problem 838: Parallelogram, Perpendicular, Diagonal, Metric Relations

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 838.

Online Geometry Problem 838: Parallelogram, Perpendicular, Diagonal, Metric Relations

Posted by Antonio Gutierrez at 8:20 AM
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5 comments:

  1. W FungDecember 28, 2012 at 8:55 AM

    cosBAD = cosEBA = 3/4
    By cosine law,
    x^2 = 4^2 + 5^2 - 2(4)(5)cosBAD
    x = sqrt11

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  2. UnknownDecember 28, 2012 at 8:57 AM

    sqrt(11)

    ReplyDelete
  3. Jacob HA (EMK2000)December 28, 2012 at 4:50 PM

    Using Pythagoras Theorem,
    AE^2 = 4^2 − 3^2 = 7
    AC^2 = 8^2 + 7 = 71

    By Parallelogram Law,
    2×(4^2 + 5^2) = x^2 + 71
    x^2 = 11
    x = √11

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  4. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 30, 2012 at 4:43 AM

    Notam cu P punctul in care paralela la BD prin E intalneste pe AD.Vom obtine
    EP=BD=X;EB=DP=5-3=2.Aplicam teorema lui Pitagora
    AE^2 = 4^2 − 3^2 = 7
    X=EP^2=AE^2+AP^2=7+2^2=11 =>x = √11

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  5. APOSTOLIS MANOLOUDISAugust 22, 2016 at 11:42 PM

    Problem 838
    Draw BF perpendicular in AD,then AF=3 and FD=5-3=2.Is BD^2=BF^2+FD^2=(4^2-3^2)+2^2=16-9+4=11.So x=√11.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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