Friday, September 19, 2014

Geometry Problem 1044: Triangle, Parallelogram, Midpoint, Area, Polya's Mind Map

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view the complete problem 1044.

Online Math: Geometry Problem 1044: Triangle, Parallelogram, Midpoint, Area, Polya's Mind Map.

4 comments:

  1. Let Q is the midpoint of OB
    Triangles BAH is congruent to COG and DEF ( translation transformation)
    So BH=CQ=DF and BH//CQ//DF
    And MQ= ½ BH, NP= 1/DF => MPNQ and BHFD are parallelograms
    Note that triangle BCD is congruent to HGF and BAH is congruent to DEF
    So area(MNP)= ½ Area(MPNQ)= 1/8 Area(BHFD)
    = 1/8(the sum of the areas of parallelograms OABC, OCDE, OEFG, and OGHA)
    =120/8= 15

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  2. S(BAH) = S(DEF)
    S(HGF) = S(BCD)

    Thus
    S(BDFH) = sum of area of //gram = 120

    S(MNP) = 1/4 S(HDF) = 1/8 S(BDFH) = 15

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  3. We denote S(XYZ...) by the area of XYZ... Connect C,E,G,A to obtain quadrilateral CEGA. Then S(CEGA)=1/2( S(OABC)+S(OCDE)+S(OEFG)+S(OGHA))=60. Let Q be the midpoint of AC. Then MPNQ is parallelogram. Since $S(CMQ)=1/4S(CEA) and S(GPN)=1/4S(GEA) then S(CMQ)+S(GPN)=1/4S(CEGA). By the same agurment S(MEN)+S(PQA)=1/2S(CEGA). Thus S(CMQ)+S(GPN)+S(MEN)+S(PQA)=1/2S(CEGA). This implies S(MPNQ)=1/2 S(CEGA). Hence S(MNP)=1/4S(CEGA)=1/4 x 60=15.

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  4. S(BAH) = S(DEF)
    S(HGF) = S(BCD)

    120
    = sum of 4 parallelograms
    = S(BHFD)
    = 2*S(HDF)
    = 2*4*S(MNP)

    S(MNP) = 15

    ReplyDelete