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


See also:
Complete Problem 519

Level: High School, SAT Prep, College geometry

6 comments:

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

    ReplyDelete
  2. 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

    ReplyDelete
  3. 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

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

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

    ReplyDelete
  6. 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

    ReplyDelete