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Let $A B C D E F$ be a regular hexagon. Then $\bar{A B} + \bar{A C} + \bar{A D} + \bar{E A} + \bar{F A} =$

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$3 \bar{A B}$

$4 \bar{A B}$

$\bar{A B}$

$2 \bar{A B}$

answer is C.

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Detailed Solution

$F A + A C = F C = 2 A B and E A + A D = E D = A B$

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