Wednesday, May 3, 2023

Geometry Problem 1537 Challenge: Can You Solve for the Missing Area in a Parallelogram using Midpoints and Intersection Points

Geometry Problem 1537. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

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Geometry Problem 1537 Challenge: Can You Solve for the Missing Area in a Parallelogram using Midpoints and Intersection Points.

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Posted by Antonio Gutierrez at 3:33 PM
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2 comments:

  1. Sumith PeirisMay 3, 2023 at 9:17 PM

    Let S(CHD) = X, S(AEG) = Y. Let the height between AD & BC be h.
    Let AF = a and EH = p

    So S(AEF) = 10 + Y = ah/2 ......(1)
    and S(AEH) = 3 +Y = ap/2 ........(2)

    (1) - (2); X = (a-p).h/2 = 10 +Y - (3 + Y) = 7

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete
  2. Peter TranMay 4, 2023 at 12:08 PM

    https://photos.app.goo.gl/hsSdBqPvrrEzUCPTA


    Let a= AF=EC
    Let h= height of parallelogram ABCD
    Let h1 and h2 are heights shown in the stretch
    Let [XYZ]= area of triangle XYZ

    Let u= EH/AF= h1/h2=u
    We have h1= ụh/(1+u)
    And h2= h/(1+u)
    Calculate [AGF]=½.a .h2= ½. (ah)/(1+u)
    Calculate [GEH]= ½. h1.EH= 1/2/ (au^2)/(1+u)
    Calculate [HCD]=½(a-EH)(h)=1/2a.h.(1-u)

    Verify that with any the value of u and h we always have [AGF]=[GEH]+[HCD]
    So [HCD]= [AGF]-[GEH]= 10- 3= 7

    ReplyDelete

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