#### Answer

$\frac{s^{2}}{4}\sqrt 3$

#### Work Step by Step

Using Heron's formula if three sides of a triangle have length a,b,c then area A of a triangle is
A = $\sqrt s(s-a)(s-b)(s-c)$
where s = $\frac{a+b+c}{2}$
Lets take a=b=c =s
s = $\frac{s+s+s}{2}$ = $\frac{3s}{2}$
A = $\sqrt \frac{3s}{2}(\frac{3s}{2}-s)(\frac{3s}{2}-s)(\frac{3s}{2}-s)$
=$\sqrt \frac{3s}{2} (\frac{3s-2s}{2})^{3} $
=$\sqrt \frac{3s}{2} (\frac{s}{2})^{3} $
=$\sqrt \frac{3}{16}s^{4}$
=$\frac{s^{2}}{4}\sqrt 3$