Monday, September 3, 2012

Problem 802: Triangle, Median, Perpendicular, Angle, Measurement

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 802.

Online Geometry Problem 802: Scalene Triangle, Median, Perpendicular, Angle, Measurement.

15 comments:

  1. Prolongar el segmento ED, desde D, 2 unidades hasta el punto que llamamos F, de modo que el segmento EF mide 4 y por tanto el triángulo rectángulo CEF es isósceles, y así el ángulo CFE=45º. Como los triángulos CDF y ADE son congruentes, por ser CD=AD, DF=DE y ángulos opuestos por el vértice en D, se tiene que ángulo AED =àngulo CFD, esto es x=45º.

    Migue.

    ReplyDelete
  2. tan x
    = (AE sin x) / (AE cos x)
    = (AD sin ADE) / (DE - AD cos ADE)
    = (DC sin CDE) / (DE + DC cos CDE)
    = (4) / (2 + 2)
    = 1

    x = 45 degrees

    ReplyDelete
  3. By Pyth. Thm and median, we have AD = DC = sqrt(20) = 2*sqrt(5).
    Simply let angle EDC be Y.
    Hence, using sine rule,
    (2*sqrt(5))/(sinX) = 2/(sinEAD)
    = 2/(sin(Y-X))
    = 2/(sinYcosX - sinXcosY)
    Note that sinY = 2/sqrt5 and cosY = 1/sqrt5
    => 2cosX - sinX = sinX
    => cosX = sinX
    X = 45.

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  4. Extend the median to a point P such that AP is perpendicular to EC. Since AD = DC, triangle PAD is congruent to triangle ECD.
    tanX = PA/PE = 4/(2+2) = 1, X = 45

    ReplyDelete
    Replies
    1. It should be AP perpendicular to BD, parallel to EC.

      Delete
    2. Yes , thanks Jacob

      Delete
    3. thanks for correction, jacob

      Delete
  5. We draw AP⊥BD
    1.ΔAPD= ΔDEC =>ED=DP=2
    AP=EC=4
    2.Δ APE, AP=4=EP=>angle AED=45º

    Another solution
    We draw AF⊥CE=>AF‖DF
    AD=DC=>EF=FC=4
    AF=2ED=4
    ΔAEF, angle AEF=45º=>angle AED=45º

    ERINA.H NJ

    ReplyDelete
  6. Note that AD/AC=ED/EC=1/2
    So AE is the external angle bisector of angle DEC
    and x=1/2.90=45
    Peter Tran

    ReplyDelete
  7. Draw DF parallel to CE (F on AE)
    So FDE = DEC is a right angle
    DF = half CE = 2 = DE
    cot x = DE/DF = 1
    x = 45 deg

    ReplyDelete
  8. http://www.alnasiry.com/upload//uploads/images/domain-d4dc4fd216.png
    draw ED and MD=2 then AECM is parall.
    AM=ME and M<EMA=90 then x=45

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  9. Let line perpendicular to AC that passes through A meet BD at P. PA/AD=4/2=PA/(2*sqrt(5)), so PA=4*sqrt(5) while AC=2*(2*sqrt(5))=PA, because sqrt(4^2+2^2)=2*sqrt(5). Then <ACP=45=<AEP.

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  10. Extend ED to F such that DF=2. Then AECF is a parellogram, EFC is an isoceles right Tr and so x=45

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  11. Problem 802
    Suppose that M is the medium of EC, then EM=2=ED.So <EDM=45.But AD=DC then AE//DM .Therefore x=<AED=<EDM=45.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete
  12. Problem 802 (sοlution 2)
    Let AN//BD ,or AN perpendicular CE (N belongs to the extension of CE).Then NE=NC=4
    And AN=2.ED=2.2=4=NE.So <NEA=45=<AED=x.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

    ReplyDelete