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