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# Geometry Question

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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.

Aug 8, 2018

#1
+101418
+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

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

Aug 8, 2018
#2
+1

Thank you, CPhill!

Aug 8, 2018
#3
+10409
+1

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.

Aug 8, 2018