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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.
Let Q is the midpoint of OBTriangles 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 parallelogramsNote that triangle BCD is congruent to HGF and BAH is congruent to DEFSo 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
S(BAH) = S(DEF)S(HGF) = S(BCD)ThusS(BDFH) = sum of area of //gram = 120S(MNP) = 1/4 S(HDF) = 1/8 S(BDFH) = 15
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.
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