nhanh lên mik cần gấp lắm!!

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mình cần gấp lắm

tui chịu ạ

\(AM=AB+BM=13\left(cm\right)\)

\(AN=AC+CN=16\left(cm\right)\)

\(S_{ABC}=\dfrac{1}{2}AB.AC.sinA\Rightarrow sinA=\dfrac{2S_{ABC}}{AB.AC}=\dfrac{3}{4}\)

\(\Rightarrow S_{AMN}=\dfrac{1}{2}AM.AN.sinA=\dfrac{1}{2}.13.16.\dfrac{3}{4}=...\)

72 cm2 nhé ( ko chắc lắm)

AB=BM

=>B là trung điểm của AM

=>AB=1/2AM

=>\(S_{AMC}=2\cdot S_{ABC}=2\cdot24=48\left(cm^2\right)\)

\(AN=3\cdot NC\)

=>\(NC=\dfrac{1}{3}\cdot AN\)

Ta có: AN+NC=AC

=>\(AC=\dfrac{1}{3}AN+AN=\dfrac{4}{3}AN\)

=>\(AN=\dfrac{3}{4}AC\)

=>\(S_{AMN}=\dfrac{3}{4}\cdot S_{AMC}=\dfrac{3}{4}\cdot48=36\left(cm^2\right)\)