Friday, December 28, 2012

Problem 839: Parallelogram, Perpendicular, Diagonal, Metric Relations

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

Click the figure below to see the complete problem 839.

Online Geometry Problem 839: Parallelogram, Perpendicular, Diagonal, Metric Relations

Posted by Antonio Gutierrez at 8:26 AM
Labels: , , , ,

4 comments:

  1. W FungDecember 28, 2012 at 8:58 AM

    ∠EBA = 180-∠ABC = 180-∠ADC = ∠ADF
    Triangle EDA ~ Triangle ADF (AA)
    Hence EB/AB = DF/AD
    DF = 15/4

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  2. UnknownDecember 28, 2012 at 9:02 AM

    x=3.75

    ReplyDelete
  3. Jacob HA (EMK2000)December 28, 2012 at 5:29 PM

    Let G be a point such that EAGC is rectangle.
    Then CD=4, DG=3.

    ∵ A,F,G,C concyclic
    ∴ DF×DC=DA×DG
    ⇒ 4x=15, x=15/4

    ReplyDelete
  4. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 29, 2012 at 4:04 AM

    Aria paralelogramului ABCD=BC.AE=AF.DC=>AF=BC.AE/DC=5√7/4
    AE^2 = 4^2 −3^2 = 7,AD^2 = AF^2 +FD^2 ,x^2 = 5^2 −(5√7/4)^2 = 5^2(1-7/16)=25.9/16=>
    x=15/4

    ReplyDelete

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