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 #2
avatar+775 
+2

The same question was asked here on the same website.

 #4
avatar
+2

You plagiarized both of your answers.

https://artofproblemsolving.com/wiki/index.php/2004_AMC_12B_Problems/Problem_19

Nothing new here, you’ve plagiarized many times before.

 Is this the only way you can correctly answer the questions?

 

You’ve implied others have plagiarized, but you blatantly do it yourself, so you are not only a plagiarizer you are a hypocrite!

 

Why are you here?

Dec 24, 2018
 #2
avatar+775 
+2

Consider a trapezoidal (label it ABCD as follows) cross-section of the truncate cone along a diameter of the bases:

 

image here.

 

Above, E, F, and G are points of tangency. By the Two Tangent Theorem, BF = BE = 18 and CF = CG = 2, so BC = 20.

 

We draw H such that it is the foot of the altitude  Segment HD to Segment AB:

 

image here.

 

By the Pythagorean Theorem, \(r = \dfrac{DH}{2} = \dfrac{\sqrt{(20)^2 - (16)^2}}{2} = \boxed{6}\).

 

Hope this helps,

 

- PM

 

coolcoolcool

Dec 23, 2018
 #2
avatar+6251 
+1
Dec 23, 2018

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