Thursday, April 13, 2017

Geometry Problem 1332: Two Squares Side by Side, 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 1332: Two Squares Side by Side, Triangle, Area.

Posted by Antonio Gutierrez at 11:16 AM
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6 comments:

  1. c.t.e.oApril 13, 2017 at 12:53 PM

    Extend BC and GF at P. S=S(ABPG)-1/2 S(ABPG)-1/2 S(DEFG)-S(CPFE)-S(BCE)
    S= 36

    ReplyDelete
    Replies
    1. KARAM HASAN (KARHAS)April 22, 2017 at 5:32 PM

      How did you calculate S(BCE)?
      you have nothing to calculate.

      Delete
    2. c.t.e.oApril 23, 2017 at 10:47 AM

      name AD = a
      S = [(a+6)∙a] - [1/2(a+6)∙a] - [1/2(6∙6)] - [(a-6)∙6] - [1/2(a-6)∙a]

      Delete
  2. Peter TranApril 13, 2017 at 8:28 PM

    https://goo.gl/photos/ZgtosScujk4knoyr9

    Draw rectangle BDHK as shown on the sketch
    Area(BGE)= ½. EG*BK= ½. DH*EG= area(EDG)
    = ½. 6*6= 18

    ReplyDelete
  3. APOSTOLIS MANOLOUDISApril 13, 2017 at 11:07 PM

    Problem 1332
    Is BD//EG then (EBG)=(EDG)=DE.DG/2=6.6/2=18.

    ReplyDelete
  4. Sumith PeirisOctober 23, 2025 at 3:56 AM

    Complete Rectangle CEFH
    Let AB = a

    S(BEG) = S(BHG) - S(CEFH) - S(BEC) - S(EFG)
    S(BEG) = a(a + 6)/2 - 6(a -6) - a(a-6)/2 - 6 X 6 / 2
    S(BEG) = 18 upon simplification

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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