If $A B C D E F$ is a regular hexagon inscribed in a circle with centre $O,$ then the value of $¯ ¯¯¯¯¯¯ ¯ A B + ¯ ¯¯¯¯¯¯ ¯ A C + ¯ ¯¯¯¯¯¯¯ ¯ A D + ¯ ¯¯¯¯¯¯ ¯ A E + ¯ ¯¯¯¯¯¯ ¯ A F$

4 (¯ ¯¯¯¯¯¯ ¯ A O)

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5 (¯ ¯¯¯¯¯¯ ¯ A O)

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6 (¯ ¯¯¯¯¯¯ ¯ A O)

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8 (¯ ¯¯¯¯¯¯ ¯ A O)

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Solution

The correct option is B $6 (¯ ¯¯¯¯¯¯ ¯ A O)$
Consider given that, $A B C D E F$ is a regular hexagon inscribed in acircle with centre $O$

$- - \to A B + - - \to A C + - - \to A D + - - \to A E + - - \to A F$

$- - \to A B + (- - \to A C + - - \to A F) + - - \to A D + - - \to A E$

since $A F = C D$

$= - - \to A B + (- - \to A C + - - \to C D) + - - \to A D + - - \to A E$

$= - - \to A B + - - \to A D + - - \to A D + - - \to A E$

$= 2 - - \to A D + (- - \to A B + - - \to A E)$ since $A E = E D$

$= 2 - - \to A D + - - \to A D$

$= 3 - - \to A D$

$= 3 \times 2 (- - \to A O)$ $O$ is the centre of a circle

$= 6 (- - \to A O)$

$A D$ is the diameter of circle and $A O$ be the radius of this circle.

Hence $6 (- - \to A O)$ is the complete solution

Option $(C)$ is the correct.

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Similar questions

¯ ¯¯¯¯¯¯ ¯ A B + ¯ ¯¯¯¯¯¯ ¯ A C + ¯ ¯¯¯¯¯¯¯ ¯ A D + ¯ ¯¯¯¯¯¯ ¯ A E + ¯ ¯¯¯¯¯¯ ¯ A F = 3 ¯ ¯¯¯¯¯¯¯ ¯ A D = 6 ¯ ¯¯¯¯¯¯ ¯ A O

A B C D E F

O

(A B + A C + A D + A E + A F)

A B C D E F

¯ ¯¯¯¯¯¯ ¯ A B + ¯ ¯¯¯¯¯¯ ¯ A C + ¯ ¯¯¯¯¯¯¯ ¯ A D + ¯ ¯¯¯¯¯¯ ¯ A E + ¯ ¯¯¯¯¯¯ ¯ A F = 6 ¯ ¯¯¯¯¯¯ ¯ A O

O

\vec{A B,} \vec{A C,} \vec{A D,} \vec{A E}

\vec{A F}

\vec{A O,}

A B C D E F

- - \to A B + - - \to A C + - - \to A D + - - \to A E + - - \to A F

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