Saturday, September 4, 2010

Problem 518: Parallelogram, Five Squares, Centers, 45 Degrees, Measure

Geometry Problem
Click the figure below to see the complete problem 518 about Parallelogram, Five Squares, Centers, 45 Degrees, Measure.

Problem 518: Parallelogram, Five Squares, Centers, 45 Degrees, Measure


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Complete Problem 518

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 12:20 PM
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2 comments:

  1. c .t . e. oSeptember 4, 2010 at 1:28 PM

    1)▲EBF = ▲FCG (SAS) => EF = FG, EFG = 90°
    2)FMDH paralelogram => MD = FH => EF = (DM√2)/2
    ( from ▲EFO, EG meet FH at O)

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  2. Peter TranSeptember 4, 2010 at 4:30 PM

    Note that BF=CF=DH=HA and BE=EA=DG=GC
    And angle EBF=angle GCF=angle GDH=angle EAH ( angles equal 90+ angle BCD)
    Triangles EFB, GCF, GDH and EAH are congruent ( Case SAS)
    We gave angle BFC=angle EFG= 90
    So EFGH is a square.
    Consider quadrilateral HFMD
    FM=1/2 BM=HD ( ½ of diagonal of congruent squares)
    And FM//HD ( translation of square center H to square center F)
    HFMD is a parallelogram
    EF=HF *SQRT(2)/2 = DM * SQRT(2)/2

    Peter Tran

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