Imagine extending the radius so that you have a horizontal diameter.

The hexagon is now split into two identical trapeziums (trapezia?).
Area of one trapezium $= {1\over 2}$ height $\times$ sum of parallel sides $$= {1\over 2}\ \times\ \sqrt0.75\ \times\ (2 + 1)$$ $$= {3\over 4} x \sqrt 3 $$

The area of the hexagon is therefore: ${3\over2}\ \times\ \sqrt {3}$

The area of the triangle is half the area of the hexagon. triangle drawn in hexagon by joining alternate vertices

The area of the triangle is therefore ${3\over4}\ \times\ \sqrt {3} $