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If altitude CD is \(8 \sqrt{3}\) centimeters, what is  BD?

 

 Feb 15, 2022
 #1
avatar+2668 
0

This problem is actually simple because it is a 30-60-90 triangle.

 

Because we know that \(CD = 8\sqrt3\)\(AD\) must equal \(\sqrt2\times 8\sqrt3 = 8\sqrt6\)

 

We also know that \(BD\) must equal to \({8\sqrt3}\over\sqrt2\), which can be rewritten as \(4\sqrt6\)

 

This means that \(\color{brown}\boxed{AB=12\sqrt6}\)

 Feb 15, 2022
 #2
avatar+129852 
+1

Note that triangles  ACD ,  CBD  are  similar  and each are 30-60-90 right triangles

And since  AD is opposite a 60° angle it =  sqrt 3 * CD =  sqrt 3 * 8 sqrt 3 =  24

 

And, by similar triangles, we have the following relationship

 

DB / DC  = DC / AD

 

DC^2 = DB * AD

 

(8sqrt 3)^2  = DB * 24

 

256 / 24  = DB  = 32 / 3  (cm)

 

 

cool cool cool

 Feb 15, 2022
 #3
avatar+2668 
+1

I misread the question... DB should be \(4 \sqrt6\), because \(\triangle{CDB}\) is a 30-60-90.

 Feb 15, 2022

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