Sunday, January 4, 2009

Elearn Geometry Problem 218: Right triangle, Altitude and Distances

Geometry Problem 217: Right triangle, Altitude and Projections.

See complete Problem 218 at:
gogeometry.com/problem/p218_right_triangle_altitude_perpendicular.htm

Level: High School, SAT Prep, College geometry

Posted by Antonio Gutierrez at 7:22 AM
Labels: , , ,

5 comments:

  1. UnknownJanuary 6, 2009 at 6:06 PM

    DFBE is a rectangle so its diagonals meet at the midpoint O of BD. Then
    h=2·OD.
    But EHGF is a trapezium (trapezoid) hence OD=1/2(a+b).
    Finally h=2.OD=a+b

    SG

    ReplyDelete
  2. APOSTOLIS MANOLOUDISAugust 9, 2016 at 6:13 AM

    Problem 218
    Let FP perpendicular in BD (D,P,B are collinear).Then triangleBFP=triangleHDE(FB=DE,<BFP=<BAC=<EDC ).So BP=EH and FG=PD.
    Therefore BD=BP+PD=EH+FG.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  3. AnonymousJanuary 9, 2018 at 2:40 AM

    We see that the triangles AFD ABC and DEC are similar , therefore we get :

    a/AD=h/AC
    aAC=hAD

    and

    b /DC =h/AC
    bAC=hDC

    adding these to equations and using the fact that AD+DC =AC we get

    (a+b)AC=(AD+DC)h
    (a+b)AC=ACh
    a+b=h

    ReplyDelete
  4. ArifJanuary 23, 2018 at 2:22 AM

    In problem 217 we see that GD=HD :

    GD=HD =d

    Then from similar triangles DGF and EHD we get :

    DH/EH=FG/GD
    d/b=a/d
    d^2=ab

    Now let :

    FD=BE= x
    FB=DE= y

    and apply pythagoras to triangle DGF

    x^2=a^2+d^2
    x^2=a^2+ab

    and apply pythagoras to triangle EHD

    y^2=a^2+d^2
    y^2=a^2+ab

    Then from pythagoras we get the relation :

    h^2=x^2+y^2
    h^2=a^2+2ab+b^2
    h^2=(a+b)^2
    h=a+b

    ReplyDelete
  5. ArifJanuary 28, 2018 at 7:51 AM

    I made a typo when applying pythagoras to triangle EHD.
    It should be

    y^2=b^2+ab

    not

    y^2=a^2+ab

    ReplyDelete

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