Distance in Regular Hexagon

Place the regular hexagon on the coordinate plane or use vectors/geometry. Let $A$ be at the origin $(0, 0)$. The side length is $a = 2$. Interior angle of a regular hexagon is $120^\circ$.

Vector $AB$ is along $x$-axis: $B = (2, 0)$. Vector $BC$ makes $60^\circ$ with positive $x$-axis: $C = (2 + 2\cos 60^\circ, 2\sin 60^\circ) = (3, \sqrt{3})$. Vector $CD$ makes $120^\circ$ with positive $x$-axis: $D = (3 + 2\cos 120^\circ, \sqrt{3} + 2\sin 120^\circ) = (2, 2\sqrt{3})$.

$T$ is the midpoint of $CD$: $T = \left(\frac{3+2}{2}, \frac{\sqrt{3}+2\sqrt{3}}{2}\right) = \left(\frac{5}{2}, \frac{3\sqrt{3}}{2}\right)$

Length $AT$: $AT^2 = \left(\frac{5}{2}\right)^2 + \left(\frac{3\sqrt{3}}{2}\right)^2 = \frac{25}{4} + \frac{27}{4} = \frac{52}{4} = 13$ $AT = \sqrt{13}\text{ cm}$.