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

## Link List

Online Geometry theorems, problems, solutions, and related topics.

Labels:
area,
parallelogram,
trapezoid,
triangle

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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