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In rectangle ABCD, shown here, CE is perpendicular to BD . If BC = sqrt(3) and DC = 2, what is CE?

 

 May 1, 2022

Best Answer 

 #1
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By the Pythagorean Theorem, \(BD = \sqrt 7\)

 

Let \(CE = x\) and \(DE = y \)

 

We have the following system: 

 

\(x^2+y^2=4\)

\(x^2+(\sqrt7 - y)^2=3\)

 

Solve for x by solving this system. 

 May 2, 2022
 #1
avatar+2440 
0
Best Answer

By the Pythagorean Theorem, \(BD = \sqrt 7\)

 

Let \(CE = x\) and \(DE = y \)

 

We have the following system: 

 

\(x^2+y^2=4\)

\(x^2+(\sqrt7 - y)^2=3\)

 

Solve for x by solving this system. 

BuilderBoi May 2, 2022

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