perimetrul ΔABC= AB+AC+BC=147 [latex] frac{AC}{AB} = frac{5}{8} [/latex]⇒AC= 5k , respectiv AB= 8 k [latex] frac{AC}{2} = frac{3BC}{4} [/latex]⇒ 2,5k=[latex] frac{3BC}{4} [/latex]⇒3BC= 10k⇒BC=[latex] frac{10k}{3} [/latex] Inlocuim in relatia extrasa din perimetru si vom avea: AB+AC+BC=147 8k+5k+[latex] frac{10k}{3} [/latex]= 147 Amplificam primele doua nr si rezultatul cu 3 si vom obtine: 24k+15k+10k=441 49k=441⇒k= 441:49=9 k=9⇒ AB= 8*9=72 cm           AC= 5*9=45 cm           BC= [latex] frac{10}{3} [/latex]*9=[latex] frac{90}{3} [/latex]= 30 cm