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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.

 Sep 1, 2024
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Let BD=x. By angle bisector theorem, ABBD=ACCD which is equivalent to 4x=53x124x=5x9x=12x=43. The area of [ADC]=DCAB2=(343)42=5342=1033

 Sep 1, 2024

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