+1248 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 $AD$.
+129771 A
10 15
B D C
AD^2 = AB^2 - BD^2 AD^2 = AC^2 - (3BD)^2
AD^2 = 10^2 - BD^2 (1) AD^2 = 15^2 - 9BD^2 (2)
Equate (1) , (2)
10^2 - BD^2 = 15^2 - 9BD^2
8BD^2 = 15^2 -10^2
8BD^2 = 125
BD^2 = 125/8
BD = sqrt [ 125 / 8 ]
AD = sqrt [ AB^2 - BD^2 ] =sqrt [ 10^2 - 125/8 ] = sqrt [ 100 - 125/8 ] =
sqrt [(800 - 125) / 8 ] = sqrt [ 675 / 8 ] = (15 / 2) sqrt ( 3/2) = 7.5 sqrt (3/2) =
(7.5/2) sqrt (6) = (15/4) sqrt (6) ≈ 9.186
