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avatar+1442 

In triangle ABC, AB = 10 and AC = 17. Let D be the foot of the perpendicular from A to BC. If BD:CD = 2:5, then find AD.

 

Thanks you guys!

 #1
avatar+98130 
+1

Let  BC  =  x

So BD   =  (2/7)x

And CD  =  (5/7)x

 

So

 

AD  =  √ (AB^2  - BD^2]   = √   ( 10^2  - [ (2/7)x]^2 )   and

AD  =  √  (AC^2  - CD^2 ) = √ ( 17^2 - [ (5/7)x ]^2 )

 

Since   AD   =  AD......we have that

 

√   ( 10^2  - [ (2/7)x]^2 )  =  √ ( 17^2 - [ (5/7)x ]^2 )     square  both sides

 

 ( 10^2  - [ (2/7)x]^2 )  =   ( 17^2 - [ (5/7)x ]^2 )     simplify

 

100 -  (4/49)x^2  =  289 - (25/49)x^2     rearrange

 

(25/49)x^2  - (4/49)x^2  =  289 - 100

 

(21/49)x^2  =  189   

 

(3/7)x^2 = 189    multiply both sides by  7/3

 

x^2  =  (7/3)189

 

x^2  =  63 * 7    

 

x^2  =  441    

 

So....AD  =

 

√  (100  -  (4/49)*441 )  =

 

√ ( 100 -  (4*9) )  =

 

√ (100 - 36)  =

 

√64   =

 

8

 

 

cool cool cool

 Mar 9, 2018
edited by CPhill  Mar 9, 2018
 #2
avatar+1442 
+2

Thank you so much!


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