Question

Suppose $a,b,c$ are the lengths of three sides of a $△ABC,a>b>c,2b=a+c$ and $b$ is a positive integer. If $a^2+b^2+c^2=84$, then value of $b$ is

• A. 7
• B. 6
• C. 5
• D. 4

Answer 1

Answer: (C)

We have
$ac=\frac{1}{2}\left[(a+c{)}^2-\left(a^2+c^2\right)\right]=\frac{1}{2}\left[4b^2-\left(84-b^2\right)\right]$
$=5b^2/2-42$
Thus, $a$ and $c$ are the roots of the equation
$x^2-2bx+\left(5b^2/2-42\right)=0$
As $a$ and $c$ are distinct real numbers the discriminant of the above must be positive, that is, $4b^2-4\left(5b^2/2-42\right)>0$
$\Rightarrow 6b^2<168\text{ or }{b}^2<28.$
$\text{ Also, }ac>0\Rightarrow 5b^2>84.$
$\therefore 84/5<b^2<28.$
As $b$ is a positive integer, we get $b=5$.