See complete Problem 150

Quadrilateral, Area, Trisection of Sides. Level: High School, SAT Prep, College geometry

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## Sunday, July 27, 2008

### Elearn Geometry Problem 150

Labels:
area,
Problem 150,
quadrilateral,
trisection

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[BAH]=[BHG]=[BGD] & [DBE]=[DEF]=[DFC] so [BGDE]=S/3,

ReplyDeleteS1=[EHGF]=[EHG]+[GFE]=[EGD]+[GEB]=[BGDE]=S/3

We can use vector production - DG=a, BE=b, AB=c and S[BGDE]=(a+c+b)*(-a)/2+(2a+c+2b)*(-b)/2 = S[ABCD]/3=(c*(-3a)/2+(-3b)*(3a+3b+c)/2)/3, here a*a=0 a*b=-b*a

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