[tex] \it \dfrac{AB}{AC} = \dfrac{3}{4} \Rightarrow \dfrac{AB^2}{AC^2} = \dfrac{9}{16} \Rightarrow \dfrac{AB^2}{AC^2+AB^2} =\dfrac{9}{9+16} \ \ \stackrel{T.Pitagora}{\Longrightarrow}\ \dfrac{AB^2}{BC^2} = \dfrac{9}{25}\Rightarrow
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\Rightarrow \dfrac{AB^2}{15^2} = \dfrac{3^2}{5^2}\Rightarrow AB^2=\dfrac{15^2\cdot3^2}{5^2} \Rightarrow AB^2=\left(\dfrac{15\cdot3}{5}\right)^2 \Rightarrow AB = \dfrac{15\cdot3}{5} \Rightarrow
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\Rightarrow AB = 3\cdot3 \Rightarrow AB = 9cm\ \ \ \ \ (*) [/tex]
[tex] \it \dfrac{AB}{AC} = \dfrac{3}{4} \stackrel{(*)}{\Longrightarrow} \dfrac{9}{AC}=\dfrac{3}{4} \Rightarrow AC = \dfrac{9\cdot4}{3} \Rightarrow AC = 12\ cm
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\mathcal{P} = AB+AC+BC =9+12+15 =36\ cm [/tex]
