\(\sqrt{15}\)
\(\sqrt{13}\)
\(\sqrt{12}\)
\(\sqrt{14}\)
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:
\[ AT^2 = 4^2 + 1^2 - 2 \times 1 \times 4 \times \cos 60^\circ \] Simplifying the equation: \[ AT^2 = 16 + 1 - 8 \times \frac{1}{2} \] \[ AT^2 = 17 - 4 \] \[ AT^2 = 13 \] Taking the square root of both sides: \[ AT = \sqrt{13} \]
The correct option is \( \boxed{(B): \sqrt{13}} \).
In the given figure, the numbers associated with the rectangle, triangle, and ellipse are 1, 2, and 3, respectively. Which one among the given options is the most appropriate combination of \( P \), \( Q \), and \( R \)?

The center of a circle $ C $ is at the center of the ellipse $ E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $, where $ a>b $. Let $ C $ pass through the foci $ F_1 $ and $ F_2 $ of $ E $ such that the circle $ C $ and the ellipse $ E $ intersect at four points. Let $ P $ be one of these four points. If the area of the triangle $ PF_1F_2 $ is 30 and the length of the major axis of $ E $ is 17, then the distance between the foci of $ E $ is: