Triangle Inside a Hexagon

Since the regular hexagon with is made up of 6 equilateral triangle of the same side length. It can easily see that diagonal $CF = 2$ .

We note that $\triangle BCJ$ and $\triangle GFJ$ are similar. Therefore $\dfrac {CJ}{FJ} = \dfrac {BC}{GF} = 2$ , $\implies CJ = 2FJ$ . Since $CF = 2$ , $\implies CJ = \dfrac 43$ .

Now, we connect $BD$ which intersects $CF$ at $L$ . Again, we note that $\triangle BKL$ and $\triangle BHD$ are similar and $\dfrac {KL}{HD} = \dfrac {BL}{BD} = \dfrac 12$ , $\implies KL = \dfrac 12 HD = \dfrac 14$ .

Now the area of $\triangle BJK$ is given by:

$\begin{aligned} [BJK] & = \frac {JK\cdot BL}2 \\ & = \frac {(CJ-KL-LC)\sin 60^\circ}2 \\ & = \frac {\left(\frac 43-\frac 14-\cos 60^\circ \right) \sin 60^\circ}2 \\ & = \frac {7\sqrt 3}{48} \approx \boxed{0.25} \end{aligned}$