The perimeter of a triangle is 50 cm.

Solution:

Let the smaller side of a triangle be $x \mathrm{~cm}$.

According to the question,

One side $=4 \mathrm{~cm}$ longer than the smaller side $=(x+4) \mathrm{cm}$

Third side $=6 \mathrm{~cm}$ less than twice the smaller side

$=(2 x-6) \mathrm{cm}$

$\therefore$ Perimeter of a triangle $=50 \mathrm{~cm}$

$\Rightarrow \quad x+x+4+2 x-6=50$

$\Rightarrow \quad 4 x-2=50$

$\Rightarrow \quad 4 x=52$

$\Rightarrow \quad x=\frac{52}{4}$

$\therefore \quad x=13 \mathrm{~cm}$

Smaller side, $a=13 \mathrm{~cm}$

Second side, $b=13+4=17 \mathrm{~cm}$

and third side, $\quad c=2 \times 13-6=26-6=20 \mathrm{~cm}$

Now, semi-perimeter, $s=\frac{a+b+c}{2}=\frac{13+17+20}{2}$

$=\frac{50}{2}=25 \mathrm{~cm}$

$\therefore \quad$ Area of a triangle $=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron's formula]

$=\sqrt{25(25-13)(25-17)(25-20)}$

$=\sqrt{25 \times 12 \times 8 \times 5}$

$=\sqrt{5 \times 5 \times 2 \times 2 \times 3 \times 2 \times 2 \times 2 \times 5}$

$=5 \times 2 \times 2 \sqrt{30}=20 \sqrt{30}$

Hence, the area of triangle is $20 \sqrt{30} \mathrm{~cm}^{2}$.