See complete Problem 170 at:

www.gogeometry.com/problem/p170_trapezoid_triangle_area.htm

Trapezoid, Midpoint, Triangle, Area. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Friday, September 5, 2008

### Elearn Geometry Problem 170

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M is the mid-pt of CD such that BC//EM//AD

ReplyDeletemaps A, E and B to A', E' and B' respectively by rotating through M by half pi

BCA'//EME'//ADB'

2S_1=[CEE']+[EE'D]=[BEE']+[EE'B']=[BEB'E']=[BEE'A']=S

extend ED and AD intersect at a point such as F

ReplyDeleteEC=EF

A(BEC)=A(EFA)

A(CED)=A(EFD)=A(CFD)/2=A(BACD)/2

Solution to problem 170.

ReplyDeleteLet be h the altitude of the trapezoid.

We have

S(BCE) + S(ADE) = BC.(h/2)/2 + AD.(h/2)/2 = (1/2).(BC+AD).h/2 = S/2.

Then S1 = S – S(BCE) – S(ADE) = S – S/2 = S/2.

Do you have a particular theorem that was used for the solution? We have this project that have that kind of problem and we have to find the theorem used. Thank you, I will be very grateful if you answer

DeleteDraw a // line from E to be CD at F, by mid-pt theorem, EF=(BC+AD)/2

ReplyDeleteLet the height of the trapezoid be h

S1=(EF)h/2=[(BC+AD)/2]*(h/2)=S/2