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

## Link List

Online Geometry theorems, problems, solutions, and related topics.

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