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### AIEEE 2002

The sides of a triangle are $$3x + 4y,$$ $$4x + 3y$$ and $$5x + 5y$$ where $$x$$, $$y>0$$ then the triangle is
A
right angled
B
obtuse angled
C
equilateral
D
none of these

## 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

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