Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 1030.

## Saturday, July 19, 2014

### Geometry Problem 1030: Quadrilateral, Triangle, Area, Midpoint, Parallel Lines

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http://s9.postimg.org/ti7j7071r/pro_1030.png

ReplyDeleteLet S(XYZ) denote area of triangle XYZ

Extend BC to F such that BF=BC => AF //BE

Since BE//AD => F, A, D are collinear

S(ABF)= S(ABC)= 1

And S(DBF)=S(DBC)= 4 => S(ABD)= 4-1 =3

So (ABCD)= S(ABD)+S(DBC)= 7

1)Continue BE till it intersect CD at F.

ReplyDelete2)As given it is easy to see that S(BCE) = S(BAE) = 0.5

3)Beacuse F is the mid point of BCD -> S(BCF) = S(BDF ) = 2. S(CEF) = 2 - 0.5 = 1.5

4)CEF is similiar to CAD thus the proportions of the areas is 2^2 = 4 -> S(CAD) = 6

5)S(ABCD) = S(ABC) + S(CAD) = 6 +1 = 7

Notam cu F intersectia dreptelor AC cu BD si cu:x=S(BEF),a=S(BFA)=S(DFE) proprietatea trapezuluiABED,

ReplyDeletey=S(AED)=>S(ABE)=S(BEC)=a+x,S(AED)=S(DEC)=y+a si din datele problemei =>2a+2x=S(ABC)=1si

S(BDC)=2a+2x+y+a=4=>S(ABD)=y+a=4-1=3 de unde=>S(ABCD)= S(ABD)+S(DBC)= 7