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.

Rangcr897 Jun 17, 2024

#1**+1 **

Let's define some variables to help solve this problem.

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

From the problem and requirements given, we can write the equation

\(BD / AB = CD / AC \)

Plugging in our values from the problem and x, we have the equation

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

Now, we can find the area of ADC easily! We have

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

So 10/3 is our answer.

Thanks! :)

NotThatSmart Jun 17, 2024