+0  
 
0
4
1
avatar+1121 

In triangle $ABC$, $AB = 10$ and $AC = 15$. Let $D$ be the foot of the perpendicular from $A$ to $BC$. If $BD:CD = 1:3$, then find $AD$.

 Apr 8, 2024
 #1
avatar+128794 
+1

                           A

                 10               15

 

         B      1        D      3         C 

 

Let DC =  3BD

 

AD =  sqrt [ AC^2  - (3BD)^2 ]  = sqrt [ 15^2 - 9BD^2 ]      (1)

AD = sqrt [ AB^2 - BD^2 ]  =  sqrt [ 10^2 - BD^2 ]       (2)

 

Equate (1) , (2)

 

sqrt [ 15^2 - 9BD^2 ] = sqrt [ 10^2 -BD^2 ]              square both sides

 

15^2 - 9BD^2 = 10^2 - BD^2

 

225 - 9BD^2 = 100 - BD^2

 

125 = 8BD^2

 

125/8  = BD^2

 

AD = sqrt [ 10^2 - BD^2]  = sqrt [ 100 - 125/8 ] = sqrt [ 675/8]  = [15sqrt 3 ]  / [2 sqrt 2] =

 

7.5 sqrt [3/2]  =   7.5 sqrt (6) / 2   =  3.75sqrt (6) ≈ 9.19

 

cool cool cool              

 Apr 8, 2024

2 Online Users

avatar
avatar