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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.

 Mar 21, 2024
 #1
avatar+128794 
+1

A

 

4                  5

 

B           D               C

             3

Let BD = 3- x     and DC =  x

By the Angle Bisector Theorem

BD/AB = DC / AC

(3-x) / 4  = x / 5

5(3-x)  = 4x

15 - 5x  =  4x

15 = 9x

x = 15/9 =  5/3

 

[ADC ] =   (1/2)DC * AB =  (1/2)(5/3)(4)  = 10 / 3

 

 

cool cool cool

 Mar 22, 2024

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