Tuesday, August 26, 2008

Elearn Geometry Problem 165



See complete Problem 165 at:
www.gogeometry.com/problem/p165_parallelogram_triangle_area.htm

Parallelogram, Diagonal, Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.
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2 comments:

  1. AjitApril 16, 2009 at 8:46 PM

    As proved in Problem 163, Tr. ABF = V(S1*S2)
    Since ABCD is a parallelogram S = 2*Tr ABD =
    2(S2 +Tr. ABF)= 2(S2 +V(S1*S2)) = 2S2 + 2V(S1*S2)
    Ajit: ajitathle@gmail.com

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  2. MarcoFebruary 4, 2023 at 12:37 AM

    Triangle BEF ~ Triangle DAF
    Let BE:AD=1:k, so height of triangle BEF : height of triangle DAF also =1:k
    Let BE=a, AD=ka, height of triangle BEF=b and height of triangle DAF =kb
    By the area formula,
    S1=ab/2, S2=abk^2/2
    S=akb(k+1)
    So, 2(S2)+2sqrt((S1)(S2))=abk^2+abk=abk(k+1)=S

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