Tuesday, October 8, 2013

Geometry Problem 931: Square, Circle, Semicircle, Diameter, Triangle, Angle, Metric Relations

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

Click the figure below to see the complete problem 931.

Online Geometry Problem 931: Square, Circle, Semicircle, Diameter, Triangle, Angle, Metric Relations

Posted by Antonio Gutierrez at 7:21 AM
Labels: , , , , ,

7 comments:

  1. Μιχάλης ΝάννοςOctober 8, 2013 at 9:27 PM

    http://users.sch.gr/mnannos/gogeometry/931.jpg

    ReplyDelete
  2. nakaOctober 9, 2013 at 1:15 AM

    Connect BD.
    Observe that in ∠DBE = ∠CBE - ∠DBE = α
    and ∠BDE = 45 - ∠EDA = 45 - ∠EAB = 45 - α
    Since ∠ABE = 45 - α, we have ∆ABE~∆BDE
    so AE/EB = AB/BD => 3/x = 1/sqrt(2) => x = 3sqrt(2)

    ReplyDelete
  3. UnknownOctober 9, 2013 at 5:57 AM

    x= 3 root 2
    by sin law

    ReplyDelete
  4. UnknownOctober 14, 2013 at 8:06 AM

    [IMG]http://www.mathmontada.net/vb/uploaded/1488_1381763088.jpg[/IMG]

    ReplyDelete
  5. AnonymousOctober 14, 2013 at 11:18 AM

    http://bleaug.free.fr/gogeometry/931.png

    by construction BE = AE.sqrt(2) = AE.sqrt(2)

    bleaug

    ReplyDelete
  6. Sumith PeirisMay 31, 2019 at 11:45 AM

    Triangles ABE and BED are similar and x/3 = BD / AB = sqrt 2 and so x = 3sqrt2

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  7. Sumith PeirisMay 31, 2019 at 11:45 AM

    Further DE=6

    ReplyDelete

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