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Base $\small AB= 15\,cm;$
lato $\small BC= \dfrac{3}{5}AB = \dfrac{3}{\cancel5_1}·\cancel{15}^3 = 3·3 = 9\,cm;$
lato $\small CA= \dfrac{2}{3}AB = \dfrac{2}{\cancel3_1}·\cancel{15}^5 = 2·5 = 10\,cm;$
perimetro $\small 2p_{ABC}= AB+BC+CA= 15+9+10 = 34\,cm.$