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1

### AIEEE 2002

In a triangle with sides $$a, b, c,$$ $${r_1} > {r_2} > {r_3}$$ (which are the ex-radii) then
A
$$a>b>c$$
B
$$a < b < c$$
C
$$a > b$$ and $$b < c$$
D
$$a < b$$ and $$b > c$$

## Explanation

$${r_1} > {r_2} > {r_3}$$

$$\Rightarrow {\Delta \over {s - a}} > {\Delta \over {s - b}} > {\Delta \over {s - c}};$$

$$\Rightarrow s - a < s - b < s - c$$

$$\Rightarrow - a < - b < - c$$

$$\Rightarrow a > b > c$$
2

### 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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