Thursday, May 23, 2013

Problem 880: Triangle, Midpoints, Sides, Perpendiculars, Hexagon, Area

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

Click the figure below to see the complete problem 880.

Online Geometry Problem 880: Triangle, Midpoints, Sides, Perpendiculars, Hexagon, Area


  1. Let S(PQR) be the area of ΔPQR.

    Form ΔDEF.

    Since ΔBFD, ΔFAE and ΔDEC are congruent,
    and H, G, M are their orthocenters.

    S(HFD)+S(FGE)+S(DEM) = S(DEF)

    Area of hexagon = 2×S(DEF) = 12

  2. Connect EF, FD and DE
    Note that area(BFD)=Area(AFE)=Area(CED)=Area(EFD)=6
    Triangles BFD, DEC and FAE are congruence (case SSS)
    Triangle EDM congruence to triangle FBH ( case ASA)
    Triangle FGE congruence to triangle BHD ( case ASA)
    So Area(EGFHDM)= Area(EFD)+Area(FBD)= 12

  3. join F to D, D to E; A to G. Tr FHD = Tr AGE, Tr EMD = tr AGF
    => Sx = S (AFDE) = 1/2 S (ABC) = 12

  4. A beautiful solution I have :consider the triangle FDE WE have FD=1/2AC. and DE=1/2AB and FE=1/2BC
    then the area of triangle FDE is FD×DE×sin^D/2(1) and WE know that ^d=^A(FDEA is parallellogram)
    Consider the triangle ABC WE have AB×AC×sinA=2×24=48 so sinA=48/AB×AC(2)
    then WE get two equations have same sin
    So the area of triangle FDE 1/2AC×1/2AB×48/AB×AC=6
    the other triangles is similar (∆FDE~DEC)and(∆AFE~FDB)
    They are same area 6
    Let S'area of traingle FHD and S" the are of traingle EMD and S'''thé Area of triangle FGE
    triangle BHD is similar to DMC
    and BHF is similar to EMD
    and FHD is similar to EMC
    SO S'+S"+S'''=6
    And the are of hexagon is the total area of four triangle FHD.FGE.EMD and FED
    SO S=12