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.
Let BC = x
Then BD = (2/7)x and DC = (5/7)x
And
AB^2 - BD^2 = AD^2 ⇒ 10^2 - [ (2/7)x]^2 = AD^2 (1)
AC^2 - CD^2 = AD^2 ⇒ 17^2 - [(5/7)x^2] = AD^2 (2)
Equate (1) and (2)
10^2 - [ (2/7)x]^2 = 17^2 - [ (5/7)x]^2
100 - (4/49)x^2 = 289 - (25/49)x^2
(25/49)x^2 - (4/49)x^2 = 289 - 100
(21/49)x^2 = 189
(3/7)x^2 = 189
x^2 = 189 ( 7/3)
x^2 = 441
x = 21
So
BD = (2/7) (21) = 6
And
AD^2 - BD^2 = AD^2
10^2 - 6^2 = AD^2
64 = AD^2
8 = AD
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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
BC = x
BD = 2/7x
CD = 5/7x
sqrt[102 - (2/7x)2] = sqrt[172 - (5/7x)2]
x = 21
BD = 2/7 * 21 = 6
CD = 5/7 * 21 = 15
AD = sqrt(AB2 - BD2) = 8