+468 In triangle ABC, AB = 17, AC = 8, and BC = 15. Let D be the foot of the altitude from C to AB. Find the area of triangle ACD.
+128472
This is a right triangle...its area = product of the leg lengths / 2 =
15 * 8 / 2 =
120 / 2 =
60 units^2
The area also equals
(1/2) CD * AB ...so
(1/2) CD * 17 = 60
CD * 17 = 120
CD = 120/17
So..by AA concruency......triangle ACB is similar to triangle ADC
Therefore AD /DC = AC / BC
AD = (120/17) * 6 / 15 = 64/17
And in triangle ADC.....AC is the hypotenuse and AD and DC are the legs...so its area =
(1/2) ( AD) ( DC) =
(1/2) (64/17) (120/17) ≈ 13.29 units^2
