The perimeter of a triangle is $50 \, cm$. One side of a triangle is $4 \, cm$ longer than the smaller side and the third side is $6 \, cm$ less than twice the smaller side. Find the area of the triangle.

(D) Let the smaller side of the triangle be $x \, cm$. Therefore,the second side is $(x + 4) \, cm$ and the third side is $(2x - 6) \, cm$.
The perimeter of the triangle is the sum of its sides:
$x + (x + 4) + (2x - 6) = 50$
$4x - 2 = 50$
$4x = 52$
$x = 13 \, cm$.
The three sides are $13 \, cm$,$17 \, cm$,and $20 \, cm$.
The semi-perimeter $s = \frac{13 + 17 + 20}{2} = \frac{50}{2} = 25 \, cm$.
Using Heron's formula,the area of the triangle is $\sqrt{s(s - a)(s - b)(s - c)}$:
Area $= \sqrt{25(25 - 13)(25 - 17)(25 - 20)}$
$= \sqrt{25 \times 12 \times 8 \times 5}$
$= \sqrt{25 \times (4 \times 3) \times (4 \times 2) \times 5}$
$= 5 \times 4 \times \sqrt{3 \times 2 \times 5}$
$= 20 \sqrt{30} \, cm^2$.