See complete Problem 171 at:

www.gogeometry.com/problem/p171_trapezoid_triangle_area_midpoint.htm

Trapezoid, Midpoints, Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Friday, September 5, 2008

### Elearn Geometry Problem 171

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by Proposed Problem 170,

ReplyDelete[CED]=[ABCD]/2=[FGDC]

See the drawing : Drawing

ReplyDelete- Define S0=S(ABCD), L midpoint of CD and h the height of CBE and DAE

- S(ECL)=EL*h/2 and S(ELD)=EL*h/2 => S(ECD)=EL*h

- S(BCE)=BC*h/2 and S(AED)=AD*h/2

- EL=(BC+AD)/2

- S=EL*h + (BC+AD)*h/2 => (BC+AD)*h => S(ECD)=S0/2

- S(AIG)=S(GID) and S(BIF)=S(FIC) => S(ABFG)=S(DCFG) => S(DCFG)=S0/2

- S(DCFG)= S(ECD) =>S1+S(CHMD)+S2= S(CHMD)+S

- Therefore S=S1+S2

See the

ReplyDeletedrawingDefine S0=[ABCD], L midpoint of CD, I intersection of AB and CD and h the height of CBE and DAE

[ECL]=EL*h/2 and [ELD]=EL*h/2 => [ECD]=EL*h

[BCE]=BC*h/2 and [AED]=AD*h/2

EL=(BC+AD)/2

S0=[ECD]+[BCE]+[AED]=EL*h + (BC+AD)*h/2= EL*h + EL*h

=> [ECD]=[BCE]+[AED]=S0/2

[AIG]=[GID] and [BIF]=[FIC] => [ABFG]=[DCFG]

S0=[ABFG]+[DCFG] => [DCFG]=S0/2

S0/2=[DCFG]= [ECD] =>S1+[CHMD]+S2= [CHMD]+S

Therefore

S=S1+S2