Let the length of edge of the regular hexagon be ' $x$ ' each.

EP and FQ are perpendiculars drawn on AD , so that $\(\mathrm{EF}=\mathrm{PQ}=\mathrm{x}\)$

The measure of each angle of a regular hexagon is $\(\frac{(6-2) \times 180}{6}=120^{\circ}\)$ (The measure of each angle of any regular polygon having $\(n\)$ sides is $\(\left.\frac{(n-2) \times 180}{n}\right)\)$

So, the measure of angle DEP $\(=\)$ angle $\(\mathrm{DEF}-\)$ angle $\(\mathrm{PEF}=120^{\circ}-90^{\circ}=30^{\circ}\)$

So, the triangle EDP is a $\(90-60-30\)$ triangle in which the side opposite to 90 degree i.e. $\(E D\)$ is $\(x\)$, so the length of the side opposite to 30 degrees i.e. DP would be $\(\frac{x}{2}\)$ which is same as the length of QA (using symmetry)

As shown in the figure on the right in a triangle having angles 30,60
$\(\& 90\)$, the corresponding opposite sides are $\(a, a \sqrt{3}\)\( \& 2 a\)$ respectively.

Finally the length of $\(\mathrm{DA}=\mathrm{DP}+\mathrm{PQ}+\mathrm{QA}=\frac{x}{2}+{x}+\frac{x}{2}=2 \mathrm{x}\)$ which is same as $\\(mathrm{AF}+\mathrm{FE}=\mathrm{x}+\mathrm{x}=\)$ \(2 x\)

Hence column A quantity is same as column $\(B\)$, so the answer is (C).