Given a side length L, the area of the triangle is sqrt(3)/4 * L^2.

Here is a derivation of that formula. First, the formula for the area 'a' of a triangle is

    a = (b * h) / 2

where b is the base and h is the height.

Call your triangle's corners A, B and C. Place the triangle so side BC is the base, and A is at the top. Let M be the midpoint of the base BC. Then the base b is the length of BC, and the height h is the length of AM.

Call the side length of your triangle L, so the length of BC is L, so b=L.

Now consider the triangle ABM. We have the two other side lengths: AB is length L, and BM is length L/2. Sides AM and BM meet at a right angle, so we can apply the Pythagorean formula, and say h^2 + (L/2)^2 = L^2. Now solve for h:

    h^2 = L^2 - (L/2)^2
      = L^2 - (L^2 / 4)
      = (3/4) L^2
    h = sqrt((3/4) L^2))
      = sqrt(3/4) L
      = (sqrt(3)/2) L

Now we can plug these into the formula for the area of a triangle:

    a = (b * h) / 2
      = (L * sqrt(3)/2 L) / 2
      = sqrt(3)/4 * L^2
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