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.

Guest Aug 8, 2018

#1**+2 **

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

CPhill Aug 8, 2018