Friday, April 3, 2020

Geometry Problem 1467: Square, Rectangle, Triangle, Area

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

Details: Click on the figure below.

Geometry Problem 1467: Square, Rectangle, Triangle, Area, Auxiliary Lines, iPad apps, Tutoring.
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4 comments:

  1. c.t.e.oApril 4, 2020 at 6:25 AM

    From sim triangles BC/GJ = EB/AG & EB/EA = AG/AB
    => BC/GJ = EA/AB

    ReplyDelete
  2. Peter TranApril 4, 2020 at 7:16 PM

    NOte that Triangles ABE and EHC are similar ( case AA)
    so AB/AE=FH/BC => AB.BC=AE.FH
    and area (ABCD)= area(FHJG)

    ReplyDelete
  3. georg.wengler@gmx.atApril 5, 2020 at 6:43 AM

    AB=s, EB=a, EF=a', FH= s'
    EA=d=sqrt(s²+a²)=FG
    a':a= s:d => a'=as/d
    s':a'=s:a => s'=s/a*as/d=s²/d
    => FH*FG=s'*d = s²

    ReplyDelete
  4. Arindam BanerjeeNovember 8, 2020 at 9:14 PM

    Triangle ACE=1/2 of rectangle AJHE, triangle ABE=1/2 of rectangle AGFE. So, triangle ACB=triangle ACE - triangle ABE=1/2(rectangle AJHE - rectangle AGFE) = 1/2 of rectangle FHJG. Again, So, triangle ACB=1/2 of square ABCD. So, rectangle FHJG=square ABCD

    ReplyDelete

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