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In rectangle ABCD, shown here, \(\overline{CE}\) is perpendicular to \(\overline{BD}\). If \(BC = \sqrt 3 \)and \(DC = 3\), what is \(CE\)?

 Apr 22, 2019
 #1
avatar+101838 
+1

Note that triangle BCD is right  ....with legs  BC and DC 

 

So  DB  =  √[ BC^2 + DC^2 ]  =  √ [ 3 + 9 ]  = √12  = 2√3

 

And triangle  BEC  is similar to triangle BCD

 

So....

 

CE / BC = DC/ BD

 

CE / √3  = 3 /[ 2√3]

 

CE   = 3 / 2   =  1.5

 

EDIT TO CORRECT AN ERROR!!!

 

 

cool cool cool

 Apr 22, 2019
edited by CPhill  Apr 22, 2019
 #2
avatar+10413 
+1

What is CE?

laugh

 Apr 22, 2019

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