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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.

 Jan 23, 2024
 #1
avatar+129830 
+1

A

 

4                5

 

B           D            C

              3

 

Since AD bisects  angle BAC

 

Then  let DC=  x  and BD = 3 -x

 

So

 

(3 - x) / 4 = x / 5

 

5 (3 -x)  = 4x

15 - 5x = 4x

15 = 9x

x = 15/9  = 5/3

 

Height of triangle ADC = 4   base = (5/3)

 

Area =  (1/2) (5/3) 4  =  20 /  6  =  10/3  =  3  (to the nearest integer )

 

cool cool cool

 Jan 23, 2024

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