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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.

 Nov 17, 2024
 #1
avatar+1926 
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We know that AD bisects BC. Thus, we can set variables to get some common ground here. 

Let's let \(BD  = 3 - x\) and let \(CD =  x\)

 

There's a couple things we can do, but let's the Angle bisector thereom to get

\(BD / AB = CD / AC\)

 

We already set some variables, so subsituting in what we know, we get

\((3 - x) / 4 = x / 5 \\ 5 (3 - x) = 4x \\ 15 - 5x = 4x \\ 15 = 9x \\ x = 15/9 = 5/3 = CD\)

 

Alright, so now we know what CD is. 

Since the area of ABC can be found with CD and AB, we simply have

\([ ADC ] = (1/2) (CD) ( AB) = (1/2) ( 5/3) ( 4) = 10 / 3 \)

 

Thus, 10/3 is our final answer. 

 

Thanks! :)

 

*1900

 Nov 18, 2024
edited by NotThatSmart  Nov 18, 2024

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