+0  
 
0
30
2
avatar+460 

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.

shreyas1  Oct 7, 2018
 #1
avatar+89874 
+2

 

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   

 

 

 

cool cool cool

CPhill  Oct 8, 2018
 #2
avatar+460 
0

Thank you CPhill once again.

shreyas1  Oct 8, 2018

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