Có:
\(\dfrac{AB}{AC}=\dfrac{4}{3}=\dfrac{4k}{3k}\) (k là số bất kì)
\(\Rightarrow AB=4k,AC=3k\)
Áp dụng đl pytago vào tam giác ABC được:
\(BC^2=AB^2+AC^2=\left(4k\right)^2+\left(3k\right)^2=16k^2+9k^2=25k^2=\left(5k\right)^2\\ \Rightarrow BC=5k\left(cm\right)\)
Theo hệ thức lượng, có:
\(AH.BC=AB.AC\\ \Leftrightarrow3,6.5k=4k.3k\\ \Leftrightarrow18=12k\\ \Rightarrow k=\dfrac{18}{12}=\dfrac{3}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}AB=4k=4.\dfrac{3}{2}=6\left(cm\right)\\AC=3k=3.\dfrac{3}{2}=\dfrac{9}{2}=4,5\left(cm\right)\\BC=5k=5.\dfrac{3}{2}=\dfrac{15}{2}=7,5\left(cm\right)\end{matrix}\right.\)
\(\Rightarrow P_{ABC}=AB+AC+BC=6+4,5+7,5=18\left(cm\right)\)
