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avatar+1678 

In triangle $ABC$, $\angle ABC = 90^\circ$, and $D$ is on side $\overline{BC}$ such that $\overline{AD}$ bisects $\angle BAC$.  If $AB = 4,$ $BC = 3$, and $AC = 5,$ then find the area of $\triangle ADC$. Round your answer to the nearest integer.

 Jan 12, 2024
 #2
avatar+129690 
+1

A

 

4             5   

 

B       D          C

          3

         

Let DC  = x  and  BD  = 3 - x

 

Since AD is an angle bisector

 

DC / AC = BD / AB

 

x / 5  = (3 -x) / 4        cross-multiply

 

5 (3 - x)  = 4x  

 

15 - 5x  = 4x

 

15  = 9x

 

x = 15/9  = 5/3  = DC  = the base of [A

 

And the height of  triagle ADC =  AB =  4

 

So

 

[ADC ] =  (1/2) DC * AB =   (1/2)(5/3)(4)  =  20 / 6  =  10 / 3 =    3 (to the nearest integer)

 

cool cool cool

 Jan 12, 2024

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