Geometry Problem

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

Click the figure below to see the complete problem 931.

## Tuesday, October 8, 2013

### Geometry Problem 931: Square, Circle, Semicircle, Diameter, Triangle, Angle, Metric Relations

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http://users.sch.gr/mnannos/gogeometry/931.jpg

ReplyDeleteConnect BD.

ReplyDeleteObserve that in ∠DBE = ∠CBE - ∠DBE = α

and ∠BDE = 45 - ∠EDA = 45 - ∠EAB = 45 - α

Since ∠ABE = 45 - α, we have ∆ABE~∆BDE

so AE/EB = AB/BD => 3/x = 1/sqrt(2) => x = 3sqrt(2)

x= 3 root 2

ReplyDeleteby sin law

[IMG]http://www.mathmontada.net/vb/uploaded/1488_1381763088.jpg[/IMG]

ReplyDeletehttp://bleaug.free.fr/gogeometry/931.png

ReplyDeleteby construction BE = AE.sqrt(2) = AE.sqrt(2)

bleaug

Triangles ABE and BED are similar and x/3 = BD / AB = sqrt 2 and so x = 3sqrt2

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Further DE=6

ReplyDelete