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1

### AIEEE 2003

In a triangle $$ABC$$, medians $$AD$$ and $$BE$$ are drawn. If $$AD=4$$,
$$\angle DAB = {\pi \over 6}$$ and $$\angle ABE = {\pi \over 3}$$, then the area of the $$\angle \Delta ABC$$ is
A
$${{64} \over 3}$$
B
$${8 \over 3}$$
C
$${{16} \over 3}$$
D
$${{32} \over {3\sqrt 3 }}$$

## Explanation

$$AP = {2 \over 3}AD = {8 \over 3};\,\,PD = {4 \over 3};\,\,$$

Let $$PB=x$$

$$\tan {60^ \circ } = {{8/3} \over x}$$

or $$x = {8 \over {3\sqrt 3 }}$$

Area of $$\Delta ABD$$

$$= {1 \over 2} \times 4 \times {8 \over {3\sqrt 3 }} = {{16} \over {3\sqrt 3 }}$$

$$\therefore$$ Area of $$\Delta ABC$$

$$= 2 \times {{16} \over {3\sqrt 3 }} = {{32} \over {3\sqrt 3 }}$$

$$\left[ \, \right.$$ As median of a $$\Delta$$ divides it into two $$\Delta 's$$ of equal area. $$\left. \, \right]$$
2

### AIEEE 2003

The sum of the radii of inscribed and circumscribed circles for an $$n$$ sided regular polygon of side $$a,$$ is
A
$${a \over 4}\cot \left( {{\pi \over {2n}}} \right)$$
B
$$a\cot \left( {{\pi \over {n}}} \right)$$
C
$${a \over 2}\cot \left( {{\pi \over {2n}}} \right)$$
D
$$a\cot \left( {{\pi \over {2n}}} \right)$$

## Explanation

$$\tan \left( {{\pi \over n}} \right) = {a \over {2r}};\,\,\sin \left( {{\pi \over n}} \right) = {a \over {2R}}$$

$$r + R = {a \over 2}\left[ {\cot {\pi \over n} + \cos ec{\pi \over n}} \right]$$

$$= {a \over 2}\left[ {{{\cos {\pi \over n} + 1} \over {\sin {\pi \over n}}}} \right]$$

$$= {a \over 2}\left[ {{{2{{\cos }^2}{\pi \over {2n}}} \over {2\sin {\pi \over {2n}}\cos {\pi \over {2n}}}}} \right]$$

$$= {a \over 2}\cot {\pi \over {2\pi }}$$
3

### 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

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

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

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

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

$$\Rightarrow a > b > c$$
4

### 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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