21let sides a, b, c where a=6
R=abc/4Δ
=abc/8s
=abc/4(a+b+c)
So,\frac{b}{sinB}=\frac{abc}{2(a+b+c)}
or, 2(a+b+c)=ac(sinB)
Since a,b,c all are integers sinB=1
B=Î /2
so, 2(a+b+c)=ac
or,2a+2b=c(a-2)
or,12+2b=4c [a=6]
or,6+b=2c
or,b=2(c-3)
or,b2=c2+62=4(c-3)2
Solving we get c=8 b=10 and a=6 is the only triangle.
