$A B C D E F$ is a regular hexagon whose centre is $O$ . Then $¯ A B + ¯ A C + ¯ A D + ¯ A E + ¯ A F$ is

2 ¯ A O

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3 ¯ A O

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5 ¯ A O

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6 ¯ A O

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Solution

The correct option is D $6 ¯ A O$
Every Triangle formed in the hexagon is Equilateral as it has all three interior angles $= 60^{o}$
So, $A B = B C = C D = D E = E F = F A = A O$
Therefore, $A B + B C + C D + D E + E F + F A = 6 A O$

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