Let us try this for the 3-4-5 triangle, which we know is a right triangle. We know its area. It's half that of the rectangle with sides 3x4. So the area $\displaystyle A$ is ${\frac {3\times 4}{2}}=6$.
The formula is believed to be due to Hero (or Heron) of Alexandria (10 – 70 AD), a Greek mathematician.
Exercise
todo: add picture
Now over to you. We have an equilateral triangle with each side of length 1. The base of the triangle is 1, the height can be worked out by Pythagoras. It's
${\sqrt {1^{2}-1/2^{2}}}={\frac {\sqrt {3}}{2}}$
so the area $\displaystyle A$ of our equilateral triangle is ${\frac {\sqrt {3}}{4}}$.
Work out what the semi perimeter S is for this triangle, and put the values for S and the lengths of the sides into Heron's formula and compute the area, A.