\(\sqrt{15}\)
\(\sqrt{13}\)
\(\sqrt{12}\)
\(\sqrt{14}\)
Considering a regular hexagon as made of 6 equilateral triangles, a line connecting the farthest vertices of the hexagon can be seen as formed by the sides of two opposite equilateral triangles within the hexagon. Consequently, its length would be twice the length of the side of the hexagon, which in this case is 4 cm.
Given that line AD divides the hexagon into two symmetrical halves, it bisects angle D, thereby establishing angle ADC as 60°.
The value of AT can be determined using the cosine formula:
\(AT^2=4^2+1^2−2×1×4×cos\ 60^∘\)
\(⇒AT^2=16+1−8×\frac 12\)
\(⇒ AT^2=17−4\)
\(⇒ AT^2=13\)
\(⇒ AT=\sqrt {13}\)
So, the correct option is (B): \(\sqrt {13}\)