(a+b+c)/2=63 pt.ca semiperimetrul =(a+b+c+)/2
a+b+c=63x2
a+b+c=126
0,3=3/9=1/3
0,5=5/10=1/2
0,2=2/10=1/5
{a,b,c} invers prop.{1/3,1/2,1/5}
a/3=b/2=c/5=k
a=3k
b=2k
c=5k
inlocuim in relatia a+b+c=126
3k+2k+5k=126
10k=126
k=126/10
k=12,6
a=3k=3x12,6=37,8
a=37,8
b=2k=2x12,6=25,5
b=25,2
c=5k=5x12,6=63
c=63
verificare
(37,8+25,2+63)/2=63
126/2=63
63=63
