Sunday, August 1, 2010

Problem 489: Parallelogram, Triangle, Quadrilateral, Area

Geometry Problem
Click the figure below to see the complete problem 489 about Parallelogram, Triangle, Quadrilateral, Area.

Problem 489. Parallelogram, Triangle, Quadrilateral, Area
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Complete Problem 489

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 6:14 PM
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2 comments:

  1. VihaanAugust 2, 2010 at 12:33 AM

    Tr. EFC/Tr.EFD = EC/ED = EB/EF as BC//FD. But EB/EF=DA/DF as AB//DC. So Tr. EFC/Tr.EFD =DA/DF=Tr.EAD/Tr.EFD or Tr. EFC = Tr.EAD and thus Tr. EFC - Tr. EGC = Tr.EAD - Tr.EGC or A2 = A1 as reqd.
    Vihaan, Dubai

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  2. c .t . e. oAugust 2, 2010 at 12:38 AM

    ▲AFB ~ ▲BCE => FD/AF = h/h2 (h, h2 alt FDE, FDC )
    => FD∙h2 = AF∙h => SFDC = SAFE =>
    SAGCD = SFGE

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