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