Tuesday, November 8, 2016

Geometry Problem 1285 Triangle, Quadrilateral, 60, 150, 90 Degree Angles, Congruence, Measurement

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

This entry contributed by Sumith Peiris, Moratuwa, Sri Lanka.

Click on the figure below to view more details of problem 1285.


Geometry Problem 1285 Triangle, Quadrilateral, 60, 150, 90 Degree Angles, Congruence, Measurement

13 comments:

  1. Problem 1285
    The sides AB and CD intersects at point K. Let M the medium of KB.Then <KBC=30 or KM=MB=KC.Is triangle ACD=triangle KAD (triangle KAD is equilateral) ,so CD=KB and
    AB=KC or CD=2AB=2AD/3.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete
    Replies
    1. Sory triangle ACD=KBD.(KD=BD.AC=BD,<ADC=<BKD).

      Delete
    2. Pure Geometry solution

      Extend AB to meet CD at E.
      Since AC = BD, the triangles EBD and DCA are similar (SSA)
      Therefore BE = CD => EC = AB

      Consider the 30-60-90 triangle BEC and
      let AB = EC = x
      => BE = CD = 2x and ED = AD = 3x
      Hence 2AD=6AB=3CD

      Delete
    3. Since AC = BD, the triangles EBD and DCA are similar (SSA)

      Could u explain how this is so?

      Delete
  2. Extend AB to meet CD at E and for the equilateral triangle AED
    Let AD = x
    Consider the 30-60-90 triangle BEC
    Let CE =y
    => BE = 2y and BC = √3y
    => AB = x-2y and CD = x-y

    Applying consine rule to ABC
    => AC^2 = (x-2y)^2+3y^2-2(x-2y)(√3y)cos150
    => AC^2 = x^2+y^2-xy ----------(1)

    From Right triangle BCD we have BD^2=3y^2+(x-y)^2
    => BD^2 = x^2+4y^2-2xy ----------(2)

    Since (1) = (2)
    => x^2+y^2-xy = x^2+4y^2-2xy
    => xy = 3y^2
    => x = 3y

    Therefore AB = y and CD = 2y and AD = 3y
    Hence 2AD=6AB=3CD

    ReplyDelete
  3. To Sumit :

    Since AC = BD, the triangles EBD and DCA are similar (SSA)

    Could u explain how this is so?
    ------------------------------------------------------------------


    In triangle EBD, m(BED) = 60, ED = Length of the side of equlateral triangle
    In triangle DCA, m(ADC) = 60, AD = Length of the side of equlateral triangle
    and given AC = BD

    Hence triangle EBD and DCA are congruent by SSA propertry (two sides ED=AD , AC=BD & Angle m(BED)=m(ADC) = 60)

    ReplyDelete
    Replies
    1. For congruency by SAS the angle must be the included angle.
      Here this is not the case.

      Delete
    2. Thank you good to know that SSA alone doesnt work to prove congruence.

      Delete
  4. Problem 1285
    The sides AB and CD intersects at point K then triangle KAD is equilateral (<ADC=360-60-150-90=60). Draw BE//AD( the point E lies on the CD).Then ABED is isosceles trapezoid (AB=ED).So AC=BD=AE or triamgle ACE is isosceles .Then <AED=<ACK and <DAE=<KAC.But triangles KAC=DAE and AB=ED=KC, KB=CD. But <CBK=30 then CD=KB=2KC or 3CD=6AB.
    And AD=AK=AB+BK=AB+2AB=3AB or 2AD=6AB.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

    ReplyDelete