We received a solution from Andisheh from Spring Field School. His reasoning was correct but he used the wrong figures, so we are using his response as a basis for the solution:

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

${\displaystyle =\frac{1}{2}\mathrm{height}x\mathrm{sum}\mathrm{of}\mathrm{parallel}\mathrm{sides}}$

${\displaystyle =\frac{1}{2}\hspace{1em}\times \hspace{1em}\sqrt{0}.75\hspace{1em}\times \hspace{1em}(2+1)}$

${\displaystyle =\frac{3}{4}x\sqrt{3}}$

${\displaystyle \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{hexagon}\mathrm{is}\mathrm{therefore}:\frac{3}{2}\hspace{1em}\times \hspace{1em}\sqrt{3}}$

The area of the triangle is half the area of the hexagon.

${\displaystyle \mathrm{The}\mathrm{area}\mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}\mathrm{therefore}\frac{3}{4}\hspace{1em}\times \hspace{1em}\sqrt{3}}$