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

2 comments:

  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.

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

    So
    Area of hexagon = 2×S(DEF) = 12

    ReplyDelete
  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

    ReplyDelete