[tex]\it \dfrac{AB}{AC}=\dfrac{3}{4} \Rightarrow AB = 3k,\ \ AC=4k,\ \ \ (k\in\mathbb Q)\\ \\ ABC-dr,\ m(\hat A)=90^o,\ \ \stackrel{T.P.}{\Longrightarrow}\ BC^2=AB^2+AC^2=(3k)^2+(4k)^2=\\ \\ =9k^2+16k^2=25k^2=5^2k^2 \Rightarrow BC=5k\\ \\ \\ AD=\dfrac{AB\cdot AC}{BC} \Rightarrow 36=\dfrac{3k\cdot4k^{(k}}{5k} \Rightarrow 36=\dfrac{12k}{5}|_{\cdot5} \Rightarrow 12k=5\cdot36|_{:12}\Rightarrow \\ \\ \Rightarrow k=15 \Rightarrow AB=45\ cm,\ AC=60\ cm,\ BC=75\ cm[/tex]
[tex]\it \mathcal{A}=\dfrac{AB\cdot AC}{2}=\dfrac{45\cdot60}{2}=45\cdot30=135\ cm^2\\ \\ \mathcal{P}=AB+AC+BC=45+60+75=18\ cm[/tex]
