The sides of a triangle are $$3x + 4y,$$ $$4x + 3y$$ and $$5x + 5y$$ where $$x$$, $$y>0$$ then the triangle is
Explanation
Let $$\,\,\,\,a = 3x + 4y,b = 4x + 3y$$
and $$c = 5x + 5y$$
as $$\,\,\,\,x,y > 0,c = 5x + 5y$$ is the largest side
$$\therefore$$ $$C$$ is the largest angle. Now
$$\cos \,C = {{{{\left( {3x + 4y} \right)}^2} + {{\left( {4x + 3y} \right)}^3} - {{\left( {5x + 5y} \right)}^2}} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}}$$
$$ = {{ - 2xy} \over {2\left( {3x + 4y} \right)\left( {4x + 3y} \right)}} < 0$$
$$\therefore$$ $$C$$ is obtuse angle $$ \Rightarrow \Delta ABC$$ is obtuse angled