The perimeter of a triangle is 16 cm. One of the sides is of 

length 6 cm. If the area of the triangle is 12 cm, then the triangle, is 

Let $a, b, c$ be the lengths of the sides of $∆ABC$ such

that $a= 6$ cm.

We have,

$2s=16\Rightarrow a+b+c=16\Rightarrow 6+b+c=16\Rightarrow b+c=10$

Also,

$\begin{array}{l}\mathrm{\Delta }=12{\mathrm{cm}}^2\\ \Rightarrow s(s-a)(s-b)(s-c)={12}^2\\ \Rightarrow 8(8-6)(8-b)(8-c)=144\\ \Rightarrow 64-8(b+c)+bc=9\\ \Rightarrow 64-80+bc=9\\ \Rightarrow bc=25\\ \therefore (b-c{)}^2=(b+c{)}^2-4bc=100-100=0\Rightarrow b=c\end{array}$

Hence, $∆ABC$ is isosceles.