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: , , ,

3 comments:

  1. Peter TranMay 22, 2011 at 10:52 PM

    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

    ReplyDelete
  2. PravinMay 22, 2011 at 11:08 PM

    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

    ReplyDelete
  3. c.t.e.oFebruary 11, 2019 at 8:03 AM

    Draw altitudes h1 from D to AH, h2 from B to AE
    S(ABCD) = (AH)(h1), S(AEFG) = (AE)(h2)
    From similarity (h2)/(h1) = (AH)/(AE) => h2 = ((AH)(h1))/(AE)
    => S(AEFG) = (AH)(h1) => S(ABCD) = S(AEFG) => x = 10

    ReplyDelete

Subscribe to: Post Comments (Atom)