Question

let a,b,c are lengths of three sides of a triangle ABC,a>b>c,2b=a+c and b is positive integer,if a^2+b^2+c^2=84, then b=​

Answer 1

Answer:

Solution :

a+b>c

2b−c+b>c

3b−c>c

3b>2c

bc>23

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Answer 2

$2b=a+c$

$u^+a=b+d$

$c=b-d$

$b^2+d^2+2bd+b^2+b^2+d^2-2bd=84$

$\Rightarrow 3b^2+2d^2=84$

$\Rightarrow b^2=\frac{84-2d^2}{3} \ge0$

$\Rightarrow 84-2d^2\ge0$  $d^2\le42$

$\Rightarrow \pm1,\pm2,\pm3,\pm4,\pm5,\pm6$

Since, $\frac{2d^2}{3}$ should also be integer,

$\Rightarrow$ d should be either $\pm 3/\pm6$

if $d=\pm3$

$b^2=28-\frac{2(9)}{3}=22$

b is not an integer

if $d=\pm 6$

$b^2=28-\frac{2(36)}{3}=4$

$b=2$

positive integer

$b=2$

$b^2=\frac{84-2d^2}{3}=28-\frac{2d^2}{3}$

Since 'b' is an integer

$28-\frac{2d^2}{3}$ should also be integer

Also $b^2\ge0\\28-\frac{2d^2}{3}\ge0\\ \Rightarrow 84-2d^2 \ge 0\\\Rightarrow d^2\le42\\\Rightarrow d=1,2,3,4,5,6$

Also for

$\frac{2d^2}{3}$ to be integer

d should be either $\frac{3}{6}$.

if $d=b,$

$\Rightarrow b^2=28-\frac{2d^2}{3}\\ =28-\frac{2(36)}{3} \\=4$

$\Rightarrow b=2 \Rightarrow a=8$

$c=b-d$

$=2-6=4$ (not possible)

So, $d=3$

$\Rightarrow b^2=28-\frac{2d^2}{3} =28-\frac{2}{3}(9)$

$b=22$

$a=b+d=22+3=25\\c=b-d=22-3=19$

$a > b > c$

$2b=a+c$

Let

$a=b+d\\c=b-d$

$b^2+d^2+2bd+b^2+b^2+d^2-2bd=84$

$\Rightarrow 3b^2+2d^2=84$

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