Ant highway

This image is the set of all points an ant can reach in 30 seconds. It has enough symmetry that we can analyze half of one quadrant and multiply the resulting area by 8.

Beginning with $\triangle ABC$ we have $AB=1=h+h\sqrt{3}$ so $h=\frac{\sqrt{3}-1}{2}$ and the entire light and dark green region has area $\frac{\sqrt{3}-1}{2}$

For the red, orange and yellow regions it will help to coordinatize. $D=(0,0)$ , $A=(1,0)$ , $B=(2,0)$ , $I=(1,1)$

The line containing $DI$ is $y=x$ . The line containing $HI$ is $y=\sqrt{3}x+1-\sqrt{3}$ . The line containing $GH$ and also $BC$ is $y=\frac{-\sqrt{3}}{3}x+\frac{2\sqrt{3}}{3}$

From which we can solve for the points of intersection. $G=(\sqrt{3}-1,\sqrt{3}-1)$ and $H=(\frac{5-\sqrt{3}}{4},\frac{1+\sqrt{3}}{4}$

The area of $\triangle DEG=\frac{1}{2}(\sqrt{3}-1)(\sqrt{3}-1)=2-\sqrt{3}$

The area of trapezoid $EFHG=\frac{1}{2}(\frac{5-\sqrt{3}}{4}-(\sqrt{3}-1))(\frac{\sqrt{3}-1}{4}+\sqrt{3}-1)=\frac{1}{16}(-51+30\sqrt{3})$

The area of trapezoid $FAIH=\frac{1}{2}(1-\frac{5-\sqrt{3}}{4})(1+\frac{1+\sqrt{3}}{4})=\frac{1}{16}(2\sqrt{3}-1)$

Sum the four areas and multiply the result by $8$ gives the final answer $12\sqrt{3}-14 \approx \boxed{6.784609691}$

I just find the relevant equations of the lines in the positive (x-y) quadrants, whose intersections determine the areas in three squares along x and y axis. In the square (1) the lines are, x/(2/√3)+y/2=1, y=x, 3y-√3x=(3-√3), and symmetry along line (y=x). The area = {(5/√3)-2} + {(1/√3)-(1/2)}= 0.964101, while in other two squares (2) additional line is y=x+1 symmetry line and area= {1/(1+√3)}= 0.366025 and (3) same as in (2) with area= {1/(1+√3)}=0.366025. Thus in the (x-y) quadrant area=1.696151 and total area in all four quadrants gives Answer=4*1.696151=6.7846.