Tuesday, July 22, 2008

Elearn Geometry Problem 139

Area of triangle

See complete Problem 139
Triangle Area, Orthic Triangle, Semiperimeter, Circumradius. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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3 comments:

  1. AnonymousDecember 10, 2009 at 5:15 AM

    Trough trigonometry
    AE=c cosA; AF=b cosA
    in triangle AFE with the law of cosine:
    FE²=AF²+AE²-2AF.AEcosA
    =cos²A.(b²+c²*2bc cosA)=a²cos²A
    then
    FE=a cosA= 2RsinAcosA=Rsin(2A)
    the same way ED=R sin(2C); FD= Rsin(2B)
    p= R(sin2A+sin2B+sin2C).(0.5)
    pR= 0.5R²(sin2A+sin2B+sin2C)
    but 0.5R²sin2A is the area of isoscele triangle
    OBC; OB=OC=R and ang(BOC)=2ang(BAC)=2A
    S=(ABC)=(AOB)+(BOC)+(COA)=pR
    .-.

    ReplyDelete
  2. Peter TranDecember 1, 2010 at 4:24 PM

    http://img607.imageshack.us/img607/2342/problem139.png
    Connect OA, OB, OD, OF and OE
    Area(ABC)=Area(ODBF)+Area(OEFA)+Area(OACE)
    Each quadrilateral have diagonals perpendicular to each other ( see Problem 138)
    And area of each quadrilateral = ½ * diagonal1 *diagonal2
    So Area(ABC)=1/2*OB.FD+1/2*OA.FE+1/2*OC.ED
    =1/2*R*(FD+FE+ED)= p.R

    Peter Tran

    ReplyDelete
  3. PravinDecember 7, 2010 at 5:12 PM

    Nice, but for a Typo:
    Area(ABC)=Area(ODBF)+Area(OEFA)+Area(ODCE)

    ReplyDelete

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