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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
avatar+98042 
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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

 

cool cool cool

 Aug 8, 2018
 #2
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+1

Thank you, CPhill!

 Aug 8, 2018
 #3
avatar+10057 
+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. 

laugh

 Aug 8, 2018

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