Monday, December 27, 2010

Problem 558: Quadrilateral, Trisection, Sides, Wittenbauer Parallelogram, Area

Geometry Problem
Click the figure below to see the complete problem 558.

Problem 558: Quadrilateral, Trisection, Sides, Wittenbauer Parallelogram, Area.
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Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 4:55 PM
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2 comments:

  1. PravinDecember 27, 2010 at 8:01 PM

    Denote area common to ABCD and KLMN by S’
    We denote area XYZ.. by XYZ.. itself.
    Using KE’ // AC // NH, it is easy to see
    AEH’ = KEE’ + HNH’. Similarly,
    BFE’ = LFF’ + KEE’,
    CGF’ = MGG’ + LFF’,
    DHG’ = NHH’ + MGG’.
    Add: S – S’ = 2(KLMN – S’)
    KLMN = (S + S’)/2 ….…. (i)
    Next,
    AEH’ = ABD/9 by similar triangles and AE = AB/3
    CGF’ = CBD/9,
    BFE’ = ABC/9,
    DHG’ = ADC/9.
    Add: S – S’ = 2S/9
    So S’ = 7S/9 ……… (ii)
    Hence by (i) and (ii):
    KLMN = [S+(7S/9)]/2 = 16S/18 = 8S/9

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  2. c .t . e. oDecember 28, 2010 at 2:12 PM

    1) E'F//EF'//AC & HG'//H'G//AC => E'F//HG' (Thales T )
    => KLMN parallelogram
    2) EF' = 2/3 AC, hp = 2/3 hq (altitudes of par, quad)
    => 1/2 Sp = EF' ∙hp = 2/3∙2/3 = 4/9 S
    => Sp = 4/9S + 4/9S = 8/9 S

    ReplyDelete

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