Friday, October 28, 2016

Geometry Problem 1278 Square, 3-4-5 Right Triangle, Midpoint

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

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Geometry Problem 1278 Square, 3-4-5 Right Triangle, Midpoint.

3 comments:

  1. Proeblem 1278
    Let AB =2 then AE = EB = BF = FC = 1.Is triangleABF=triangleDAE so 〖ΕD〗^2=〖ΑF〗^2=5, or ED=sqrt5 and AF,DE are perpendicular,(<BAF=<ADE).But the triangles AGE,EAD are similar ,then EG/GA=AE/AD=1/2 so if EG=x, AG=2x and x^2+〖(2x)〗^2=1 or x=(sqrt5)/5.But GF=AF-AG= sqrt5-2(sqrt5)/5=3(sqrt5)/5,DG=ED-EG= sqrt5-(sqrt5)/5=4(sqrt5)/5, and FD= sqrt5.So the triangle FGD is a 3-4-5 triangle .
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

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  2. Let 2a be the length of the side of the square.
    Tr.s ABF & ADE are congruent SAS.
    Hence < EAG = < EDA so < FGD = 90.

    Now DG = AD^2/DE = 4a/ sqrt5
    DF = a. Sqrt5
    Hence GD/DF = 4/5
    It follows that DFG is a 3-4-5 right triangle

    Sumith Peiris
    Moratuwa
    Sri Lanka

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  3. Let DF=1, DG=x. AF is perpendicular to DE. From the right triangle EAD we get EG*GD=AG^2, so AG=sqrt((1-x)x). From the triangle FGD we get GF=sqrt(1-x^2). But GF=1-AG, so we get an equation
    sqrt(1-x^2)=1-sqrt((1-x)x). Solving for x we get x=4/5. So DG=4/5, EG=3/5 and ED=5/5.
    Piotr Chrzastowski-Wachtel, University of Warsaw, Poland.

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