If sides of a triangle are in the ratio 7k,8k,9k and the radius of the incircle is 3√5, the k is equal to
s=7k+8k+9k2=12k
Δ=√s(s−a)(s−b)(s−c)=√12k⋅5k⋅4k⋅3k=12√5k2
r=Δs⟹3√5=12√5k212k⟹k=3