Friday, October 28, 2016

Geometry Problem 1279 Parallelogram, Midpoint, Triangle, Area, 3/10

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

Click the figure below to view more details of problem 1279.

Geometry Problem 1279 Parallelogram, Midpoint, Triangle, Area, 3/10, tutoring.

9 comments:

  1. https://goo.gl/photos/L9wsdrtPF7pDebWf7
    Let DE extended meet BC at P
    Since EB= ½. CD => PB=BC=AD
    Triangle GAD similar to GFP => GA/GF=AD/PF= 2/3=> GA/AF=2/5 => S(DGF)/S(ADF)=3/5
    S(DGF)=3/5.S(ADF)= 3/5 x ½ S(ABCD)= 3/10. S(ABCD)

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  2. Let M be the mid point of AF.

    EM = AD/4 so AD = 4/5AM

    Now S(AEM) = 1/4 S(ABF) = 1/4(S/4) = S/16
    Hence S(AED) = 4/5 X S/16 = S/20

    S1 = S -S(ABF) - S(ADE) - S(CDF) + S(ADE)

    Hence S1 = S -S/4 - S/4 - S/4 + S/20 = 3S/10

    Sumith Peiris
    Moratuwa
    Sri Lanka

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    Replies
    1. In the 2nd line it should read as

      AG = 4/5AM not AD = 4/5AM

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  3. Replies
    1. S(AEM) = 1/4 S(ABF) = 1/4(S/4) = S/16
      Hence S(AED) = 4/5 X S/16 = S/20

      Delete
    2. Sumith
      Per observation, S(AED) is not equal S/20.
      Peter Tran

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    3. Area of triangle AED is 1/4 the area of ABCD, because triangle has same base as ABCD and height 1/2 of height of parallelogram ABCD

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    4. Extremely sorry Omid and Peter. My mistake.

      I should have said S(AEG) = S/20 which is easily seen since S(AEM) = S/16 and AG = 4/5AM

      Delete
  4. Extend AF to meet DC at P
    Since FC is half of AD, from mid-point theorem we have
    CP = CD & FP = AF = l (say) and AG = x (say)
    Also the triangles DGP and EGA are similar
    => DP/EA = PG/AG
    => 4 = 2l-x/x
    => x = AG = (2/5)l and GF = (3/5)l

    Since the area of triangle AFD = 1/2(S) => Area of triangle GFD (S1) = 3/5*1/2*S (S being the are of //gm ABCD)

    Therefore S1 = 3/10*S

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