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Trapezoid ABCD has bases AB and CD.  Find x.

 

 Apr 24, 2022
 #1
avatar+124594 
-1

Draw altitude AE

 

Then  DE  =  (3  -1)  / 2  = 1

And we have right triangle  ADE with legs of sqrt 2 , 1  and  the trapezoid height as the other leg

 

So

 

h =  sqrt [( sqrt 2)^2  - 1 ]  =  1

 

And   we also have right triangle   AEC  where  EC =   2

 

So

 

x =  sqrt [ 2^2 + 1^2 ]   =   sqrt  5

 

cool cool cool

 Apr 24, 2022
 #2
avatar+2437 
+1

Draw line \(AM\) so it is perpendicular to \(DC\)

 

Trapezoid \(ABCD\) is isceles, meaning \(DM = 1\)

 

Applying the Pythagorean Theoram to \(\triangle AMD\), we find that \(AM = 1\)

 

Applying the Pythagorean Theorem to \(\triangle AMC\), we find \(\color{brown}\boxed{x=\sqrt{5}}\)

 Apr 24, 2022

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