From a point in the interior of an equilateral triangle,perpendiculars are drawn on the three sides. The lengths of the perpendiculars are $14 \, cm$,$10 \, cm$,and $6 \, cm$. Find the area of the triangle.

(D) Let $ABC$ be an equilateral triangle with side length $a$. Let $O$ be an interior point,and let the perpendiculars from $O$ to the sides $AB$,$BC$,and $AC$ be $h_1 = 14 \, cm$,$h_2 = 10 \, cm$,and $h_3 = 6 \, cm$ respectively.
The area of $\Delta ABC$ is the sum of the areas of $\Delta OAB$,$\Delta OBC$,and $\Delta OAC$.
Area of $\Delta OAB = \frac{1}{2} \times a \times 14 = 7a \, cm^2$.
Area of $\Delta OBC = \frac{1}{2} \times a \times 10 = 5a \, cm^2$.
Area of $\Delta OAC = \frac{1}{2} \times a \times 6 = 3a \, cm^2$.
Total Area $= 7a + 5a + 3a = 15a \, cm^2$.
We know the area of an equilateral triangle is given by $\frac{\sqrt{3}}{4} a^2$.
Equating the two expressions for the area:
$\frac{\sqrt{3}}{4} a^2 = 15a$
Since $a \neq 0$,we divide by $a$:
$\frac{\sqrt{3}}{4} a = 15$
$a = \frac{15 \times 4}{\sqrt{3}} = \frac{60}{\sqrt{3}} = 20\sqrt{3} \, cm$.
Now,substitute $a = 20\sqrt{3}$ into the area formula:
Area $= 15 \times (20\sqrt{3}) = 300\sqrt{3} \, cm^2$.
Thus,the area of the equilateral triangle is $300\sqrt{3} \, cm^2$.