We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
703
2
avatar+1438 

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+101871 
+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+1438 
+2

Thank you so much!


19 Online Users

avatar
avatar
avatar
avatar