Question

If $ABCDEF$ is a regular hexagon inscribed in a circle with center $O$, then $(AB+AC+AD+AE+AF)$ equal

• A. $4AO$
• B. $5AO$
• C. $6AO$
• D. $8AO$

Answer 1

Answer: (C)

$6AO$

We have, $AC=AB+BC=AB+AO$

$AE=AF+FE=AF+AO$

$\therefore AC+AE=AB+AF+2AO=3AO$

Again $AD=2AO$

$\therefore AC+AD+AE=3AO+2AO=5AO$

Also, $AB+AF=AB+BO=AO$

$\therefore AB+AC+AD+AE+AF=5AO+AO=6AO$.