Thursday, March 2, 2023

Geometry Problem 1517: Unlocking Triangle Side Length: Solving with a Median and Two Angles. Difficulty Level: High School.

Geometry Problem 1517. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

Geometry Problem 1517: Unlocking Triangle Side Length: Solving with a Median and Two Angles. Difficulty Level: High School.

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5 comments:

  1. Replies
    1. https://photos.app.goo.gl/CkcFLDMpaKUyBK3a9

      Delete
  2. Drop perpendiculars DP and DQ from D to AB and from D to BC, P on AB and Q on BC respectively.

    Since AD = DC, S(ABD) = S(BDC)

    So AB.DP/2 = BC.DQ/2

    Hence (c/2)d/(sqrt2) = (1/2).2. (sqrt3./2)d
    Simplifying c = sqrt 6

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  3. Construct parallelogram BAEC.
    Drop perpendicular AH, from A to BE.
    AEH is a 30-60-90 triangle, and ABH is a 45-45-90 triangle.
    BC=AE=2, AH=sqrt 3, AB=sqrt 6.

    ReplyDelete
  4. AB/AD = sin D / sin 45.
    DC/BC = sin 60 / sin 45.
    Multiply side by side
    AB/BC = sqrt 3 / sqrt 2 => AB = sqrt 6.

    ReplyDelete