Geometry Problem

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

Click the figure below to see the complete problem 839.

## Friday, December 28, 2012

### Problem 839: Parallelogram, Perpendicular, Diagonal, Metric Relations

Labels:
diagonal,
metric relations,
parallelogram,
perpendicular,
triangle

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∠EBA = 180-∠ABC = 180-∠ADC = ∠ADF

ReplyDeleteTriangle EDA ~ Triangle ADF (AA)

Hence EB/AB = DF/AD

DF = 15/4

x=3.75

ReplyDeleteLet G be a point such that EAGC is rectangle.

ReplyDeleteThen CD=4, DG=3.

∵ A,F,G,C concyclic

∴ DF×DC=DA×DG

⇒ 4x=15, x=15/4

Aria paralelogramului ABCD=BC.AE=AF.DC=>AF=BC.AE/DC=5√7/4

ReplyDeleteAE^2 = 4^2 −3^2 = 7,AD^2 = AF^2 +FD^2 ,x^2 = 5^2 −(5√7/4)^2 = 5^2(1-7/16)=25.9/16=>

x=15/4