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+0  
 
0
779
6
avatar+2496 

excuse me for writings

 Jan 13, 2016

Best Answer 

 #5
avatar+104793 
+15

Note that we have similar triangles, such that :

 

BDA ≈ BAC   ....so....

 

BA/DA  = BC/AC

 

And BC  = sqrt(10^2 + 20^2)  = sqrt(500)  = 10sqrt(5)    .......so....

 

10/DA  = 10sqrt(5) /20

 

10/DA  = sqrt(5) / 2    cross-multiply

 

20 = sqrt(5) * DA

 

DA = 20/sqrt(5)  = 20sqrt(5)  / 5   = 4sqrt(5)  →  (C)

 

 

 

 

cool cool cool

 Jan 13, 2016
 #1
avatar+8613 
0

I wanna help you, But I cannot see it :(

 Jan 13, 2016
 #2
avatar+2496 
0

http://s43.radikal.ru/i101/1601/c3/84168694f4a9.jpg

 Jan 13, 2016
 #3
avatar+8613 
0

Ughh. Blockeed. Can you messege me the picture? :)

 Jan 13, 2016
 #4
avatar+2496 
0

enter from the link below may be it will show

 Jan 13, 2016
 #5
avatar+104793 
+15
Best Answer

Note that we have similar triangles, such that :

 

BDA ≈ BAC   ....so....

 

BA/DA  = BC/AC

 

And BC  = sqrt(10^2 + 20^2)  = sqrt(500)  = 10sqrt(5)    .......so....

 

10/DA  = 10sqrt(5) /20

 

10/DA  = sqrt(5) / 2    cross-multiply

 

20 = sqrt(5) * DA

 

DA = 20/sqrt(5)  = 20sqrt(5)  / 5   = 4sqrt(5)  →  (C)

 

 

 

 

cool cool cool

CPhill Jan 13, 2016
 #6
avatar+2496 
+5

Thanks CPhill !

 Jan 13, 2016

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