See complete Problem 164 at:
www.gogeometry.com/problem/p164_parallelogram_triangle_area.htm
Parallelogram, Trapezoid, Diagonal, Triangles, Areas. Level: High School, SAT Prep, College geometry
Post your solutions or ideas in the comments.
Tuesday, August 26, 2008
Elearn Geometry Problem 164
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Tr. ABD = Tr. BDC since AD=BC and the height is common. So S + Tr. ABF = S1 + S2 + Tr. DEF or S = S1 + S2 since Tr. ABF = Tr. DEF which have the same base and height.
ReplyDeleteAjit: ajitathle@gmail.com
s+tr EFD=1/2 ABCD
ReplyDeletes1+s2+tr EFD=1/2 ABCD
So s+tr EFD=s1+s2+tr EFD wich means s=s1+s2
If point E can be any point on BC, then let say that point E is on point B, meaning segment AE = segment AB, point F = point B, the area of S1 = 0 and the area of S = tr. ABD. This means that S = S2 because tr. ABD = tr. BDC. So, S = S1 + S2, also written as S = 0 + S2 since S1 = 0.
ReplyDeleteThe "solution" proposed by Anonymous, June 20, 2011 is not valid. He only proved that S = S1 + S2 in the case that E and B are the same point. What happens in the general situation?
ReplyDeleteTriangle BEF ~ Triangle DAF
ReplyDeleteLet 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
S1=ab/2, S2=(k^2-1)*ab/2, implies that S1+S2=abk^2/2
S=ka(kb)/2=abk^2/2=S1+S2