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.
This is a right triangle with AB forming the hypotenuse and AC and BC forming the legs.....the area will = 1/2 the product of the leg lengths...so we have
Area = (1/2) (AC) ( BC) = (1/2)(8)(15) = (1/2) (120) = 60 units^2
So.....letting ....AB be the base of the triangle and CD the altitude....then we have that
Area = (1/2) (17) ( CD)
60 = (1/2) (17) (CD)
120/17 = CD
Now....ΔACB is similar to Δ ADC....this means that
AC / CB = AD/ DC
8/15 = AD / (120/17)
(120/17)(8/15) = AD
(120/15)(8/17) = AD
(8)(8) / 17 = AD
64/17 = AD
Since ADC is also a right triangle with legs AD and CD....its area = (1/2)(AD)(CD) = (1/2)(64/17)(120/17) =
(64 * 60) / 17^2 =
3840/289 units^2 ≈ 13.29 units^2