Geometry Problem

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

See also:

Complete Problem 518

Level: High School, SAT Prep, College geometry

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## Saturday, September 4, 2010

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

See also:

Complete Problem 518

Level: High School, SAT Prep, College geometry

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Geometry Problem

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

See also:

Complete Problem 518

Level: High School, SAT Prep, College geometry

Labels:
45 degrees,
center,
measurement,
parallelogram,
square

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1)▲EBF = ▲FCG (SAS) => EF = FG, EFG = 90°

ReplyDelete2)FMDH paralelogram => MD = FH => EF = (DM√2)/2

( from ▲EFO, EG meet FH at O)

Note that BF=CF=DH=HA and BE=EA=DG=GC

ReplyDeleteAnd 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