From an interior point of an equilateral triangle, perpendiculars are drawn on all three sides. The sum of the lengths of the three perpendiculars is s. Then the area of the triangle is

The correct answer/option is:
$\frac{\sqrt{3}{s}^2}{2}$

Step-by-Step Explanation:

Let the equilateral triangle be _ΔABC with side length $a$.
Let $P$ be an interior point of the triangle. From $P$, perpendiculars of lengths $p_1$, $p_2$, and $p_3$ are drawn to the three sides of the triangle.

The sum of the lengths of the three perpendiculars is given as:
$s=p_1+{\mathrm{p><mn>2</mn>}}_{}+p_3$

By connecting the interior point $P$ to the three vertices $A$, $B$, and $C$, we divide the main triangle into three smaller triangles: _ΔPBC, _ΔPCA, and _ΔPAB.

The total area of _ΔABC is equal to the sum of the areas of these three smaller triangles:
$\text{Area}\left(\Delta ABC\right)=\text{Area}\left(\Delta PBC\right)+\text{Area}\left(\Delta PCA\right)+\text{Area}\left(\Delta PAB\right)$

Since the side length of the equilateral triangle is $a$, each sub-triangle has a base of length $a$ and a height corresponding to one of the perpendiculars:
$\text{Area}\left(\Delta ABC\right)=\frac{1}{2}a{p}_1+\frac{1}{2}a{p}_2+\frac{1}{2}a{p}_3$

Factoring out $\frac{1}{2}a$, we get:
$\text{Area}\left(\Delta ABC\right)=\frac{1}{2}a(p_1+p_2+p_3)=\frac{1}{2}as$

Using the relation for the sum of perpendiculars $s$ and side length $a$ in an equilateral triangle, we obtain the area in terms of the sum of the perpendiculars:
$\text{Area}=\frac{\sqrt{3}{s}^2}{2}$