Sunday, January 4, 2009

Elearn Geometry Problem 219: Rhombus, perpendiculars, arithmetic mean

Geometry Problem 219. Rhombus, perpendiculars, arithmetic mean.

See complete Problem 219 at:
gogeometry.com/problem/p219_rhombus_perpendicular_arithmetic_mean.htm

Level: High School, SAT Prep, College geometry
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2 comments:

  1. AnonymousJanuary 8, 2009 at 7:56 AM

    clearly HM=HO=HG=x and OF=NF/2, so
    a+x=(b+c+2x)/2
    a=(b+c)/2

    ReplyDelete
  2. GregSeptember 6, 2020 at 6:28 AM

    ABCD rhombus => GHO isoscele in H => OH=HG=HM since GOM rectangle in O => OH=GM/2
    Also, ABCD rhombus => FN distance between AB and CD is the same as that between AD and BC which is 2.OE
    So a=OE-OH=(FN-GM)/2=(FG+MN)/2=(b+c)/2 QED

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