Sunday, May 22, 2011

Geometry Problem 604: Parallelograms, Triangle, Trapezoid, Area

Geometry Problem
Click the figure below to see the complete problem 604.

Geometry Problem 604: Parallelograms, Triangle, Trapezoid, Area.

Posted by Antonio Gutierrez at 8:55 PM
Labels: , , ,

2 comments:

  1. http://img37.imageshack.us/img37/1471/problem604.png
    Let K is the projection of B over AE and L is the projection of E over AB ( see picture)
    We have area(GAEF)=Area(ABH)+10=AE .BK= AB. EL= 2.* Area(ABE)=Area(ABCD)
    but Area(ABCD)=x+ Area(ABH)
    So x=10
    Peter Tran

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  2. Extend GF and DE to meet at K
    ∆AGB≡∆EFK
    Using notation (…) for area
    (EFK) = (AGB) = 3
    (BHEK) = 7 + 3 = 10
    Denote (HCE) by y
    (BCK) = 10 + y
    Next ∆ADE ≡ ∆BCK
    (ADE )= (BCK)
    x+ y = 10 + y
    x = 10

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