Question 2.

Based on the question: AD, CE and BF are the three altitudes of the triangle. 

It has been stated that $\{GD + GE + GF = s\}$

Now since the triangle is equilateral, let the length of each side be $a$. 

So area of triangle will be

$\frac12 \times GD \times a + \frac12 \times GE \times a + \frac12 \times GF \times a = \frac{\sqrt{3}}{4} a^2$

Now $GD + GE + GF = \frac{\sqrt{3} a}{2}$  

or $s = \frac{\sqrt{3} a}{2}$  

or $a = \frac{2s}{\sqrt{3}}$

Given the area of the equilateral triangle $= \frac{\sqrt{3}}{4} a^2$

Substituting the value of $a$ from above, we get the area (in terms of $s$) = $\frac{s^2}{\sqrt{3}}$