Friday, September 10, 2010

Problem 519: Triangle, Squares, Altitude, Perpendicular, Rectangle, Area

Geometry Problem
Click the figure below to see the complete problem 519 about Triangle, Squares, Altitude, Perpendicular, Rectangle, Area.

Problem 519: Triangle, Squares, Altitude, Perpendicular, Rectangle, Area


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Complete Problem 519

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 11:20 AM
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6 comments:

  1. c .t . e. oSeptember 10, 2010 at 1:44 PM

    ▲BPH ~ ▲ABH => PB/BH = BH/AB
    ▲BMH ~ ▲BHC => BM/BH = BH/BC
    => PB∙AB = BM∙BC

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  2. Peter TranSeptember 10, 2010 at 3:04 PM

    In right triangle ABH with altitude HP we have BH^2=BP.BA=S1 ( relations in right triangles)
    Similarly in right triangle BHC we have BH^2=BM.BC= S2
    So S1=S2

    Peter Tran

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  3. Emil EkkerSeptember 10, 2010 at 4:22 PM

    Triangle ABH and BHP are similar --->
    AB:BH=BH:BP ---> S1=ABxBP=BH^2

    Triangle BCH and BHM are similar --->
    BC:BH=BH:BM ---> S2=BCxBM=BH^2 ---> S1=S2

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  4. AnonymousSeptember 11, 2010 at 6:41 AM

    a trigonometric solution
    BH=csinA=asinC
    BM=BHsinC
    BP=BHsinA
    S1=c.BP=acsinAsinC
    S2=a.BM=acsinAsinC
    .-.

    ReplyDelete
  5. PravinDecember 8, 2010 at 5:25 PM

    Angles BPM = BHM = C;
    Triangles BPM, BCA are ///.
    BP:BC = BM:BA,
    BP.BA = BM.BC,
    BP.BD = BM.BG
    S1 = S2

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  6. APOSTOLIS MANOLOUDISJune 21, 2016 at 10:29 AM

    Problem 519
    Is BH^2=BP.BA=BP.BD=S_1, BH^2=BM.BC=BM.BG=S_2,Therefore S_1=S_2.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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