Question Number 197010 by sonukgindia last updated on 06/Sep/23
Answered by nikif99 last updated on 06/Sep/23
$${Triangle}\:{inequalitues}: \\ $$$$\mathrm{3}{x}−{x}<\mathrm{10}\:\Rightarrow\mathrm{2}{x}<\mathrm{10}\:\Rightarrow{x}<\mathrm{5}\:\left(\mathrm{1}\right) \\ $$$${x}+\mathrm{3}{x}>\mathrm{10}\:\Rightarrow\mathrm{4}{x}>\mathrm{10}\:\Rightarrow{x}>\frac{\mathrm{5}}{\mathrm{2}}\:\Rightarrow{x}\geqslant\mathrm{3}\:\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)\left(\mathrm{2}\right)\:\Rightarrow{max}\:{x}=\mathrm{4} \\ $$$${max}\:{perim}.={x}+\mathrm{3}{x}+\mathrm{10}=\mathrm{26} \\ $$