$A B C D E F$ is a regular hexagon whose centre is O. Then, $\mathbf{A B}+\mathbf{A C}+\mathbf{A D}+\mathbf{A E}+\mathbf{A F}$ is equal to

• A

  $2 \mathrm{AO}$
• B

  3 AO
• C

  $5 \mathrm{AO}$
• Correct answer
  D

  $6 \mathrm{AO}$

Solution

$$ \begin{aligned} & \because \quad \mathrm{AB}=\mathrm{ED}, \mathrm{AF}=\mathrm{CD} \\ & \therefore \quad \mathrm{AB}+\mathrm{AC}+\mathrm{AD}+\mathrm{AE}+\mathrm{AF} \\ & =A E+E D+A C+C D+A D \\ & =A D+A D+A D \\ & =3 \mathrm{AD} \\ & \end{aligned} $$ $\because O$ is the centre of regular hexagon. $$ \begin{aligned} & \therefore \quad \mathrm{AD}=2 \mathrm{AO} \\ & \Rightarrow \quad 3 \mathrm{AD}=3 \times 2 \mathrm{AO}=6 \mathrm{AO} \\ & \end{aligned} $$

Hence, the correct answer is option: 'D'.