Geometry Problem

Click the figure below to see the complete problem 604.

## Sunday, May 22, 2011

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## Sunday, May 22, 2011

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Geometry Problem 604: Parallelograms, Triangle, Trapezoid, Area

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area,
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http://img37.imageshack.us/img37/1471/problem604.png

ReplyDeleteLet 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

Extend GF and DE to meet at K

ReplyDelete∆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

Draw altitudes h1 from D to AH, h2 from B to AE

ReplyDeleteS(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