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1. In the diagram below, if AD=30, what is BC?

 

2. We know BE=CE and AE=2DE. What is AD/BC?

 

3. Given that AF=\(4sqrt(3)\) and FC=5sqrt(3) ,what is BC?

 

Guest Oct 28, 2018
 #1
avatar+90968 
+1

First one  

 

Let the intersection of    AD  and BC   =  E

 

Triangle  AEB  is similar to Triangle   DEC

 

Let DE  =  x     and let  AE  =  30 - x

 

So

 

AE / AB   =  DE / DC

 

( 30 - x) / 20   =  x / 4       cross-multiply

 

4 (30 - x)   =  20x     simplify

 

120 - 4x  =  20x       add 4x to both sides

 

120  =  24x       divide both sides by  24

 

5  =  x   =  DE

 

So....CE  = sqrt [ DE^2  - DC^2]  =  sqrt [ 5^2  - 4^2 ] =  sqrt [ 9]  =  3

 

And BE is 5 times this  =  15

 

So....BC  =  BE + EC  =  15 +  3   =   18

 

 

cool cool cool

CPhill  Oct 31, 2018
edited by CPhill  Oct 31, 2018
 #2
avatar+90968 
+1

Second one

 

Triangle ABE  similar to triangle ECD

 

So

AE / BE  =  DE / CD

 

2DE / BE  =  DE / CD     ......so......

 

2DE / DE   =   BE / CD

 

2  = BE / CD

 

So   BE  =  2CD   ⇒   CD  =  BE / 2   =  EC / 2

 

AD  =  sqrt  ( AE^2  + DE^2]  =  sqrt [ (2DE)^2 + DE^2]  =  sqrt(5)DE

 

And

 

DE  =  sqrt  [ EC^2  + (EC/2)^2]  =   EC sqrt(5) / 2    =  BEsqrt(5)/2

 

So

 

DE  = BE sqrt(5) / 2

2DE /sqrt(5)  = BE

So

BC  = 2BE

So

2BE  =    4DE / sqrt(5)  =  BC

 

So

 

AD / BC  =  sqrt (5) DE  /  [  4DE / sqrt (5) ]

 

AD /BC  =     sqrt (5) * sqrt(5) DE  / [ 4DE ]  =  5DE / [ 4 DE ]   =   5 / 4

 

 

 

cool cool cool

CPhill  Oct 31, 2018

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