Mathematics - Sequence & Series Question with Solution | TestHub

Let three sides of triangle be: $a-d,\mathit{a},\mathit{a}+d$.

Where, $d>0, a>0$.

$\Rightarrow$  length of smallest side $=a-d$ units

Now,  ${\left(a+d\right)}^2=a^2+{\left(a-d\right)}^2$

$\Rightarrow a\left(a-4d\right)=0$

$\therefore a=4d$        .....(i)

(As $a = 0$ is rejected)

Also,    $\frac{1}{2}a.\left(a-d\right)=24$

$\Rightarrow a\left(a-d\right)=48$        .....(ii)

$\therefore$     From (i) and (ii), we get  $a=8, d=2$

Hence, length of smallest side is

$\left(a-d\right)=\left(8-2\right)=6$units