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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.

 Apr 12, 2024
 #1
avatar+128794 
+1

  A

 

      4            5

 

     B        D          C

                3

 

Since AD is a  bisector.....Let BD = x  and DC  =3 - x   and  we have that

 

BD/ AB  = DC / AC

 

x / 4 = (3 - x) / 5

 

5x = 4 (3 - x)

 

5x  = 12 - 4x

 

9x= 12

 

x =  4/3  = BD

 

[ ABD ] = (1/2) (4/3)(4)  =  16/6 = 8/3

 

[ABC ]  = (1/2)(4)(3)  = 6

 

[ADC ]   =  [ ABC ] -  [ ABD ]  = 6 - 8/3  =   10/3 =  3 + 1/3  =   3 (to the nearest integer )

 

cool cool cool

 Apr 12, 2024

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