Let $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}$ be a regular hexagon inscribed in a circle of unit radius. Then the product of the length of the line segments $A_{0} A_{1}, A_{0} A_{2}$ and $A_{0} A_{4}$ is

\frac{3}{4}

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\frac{3}{\sqrt{3}}

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3

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\frac{3 \sqrt{3}}{2}

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Solution

The correct option is C $3$
Let $O A_{0} = 1$ then $O A_{1} = O A_{2} = O A_{3} = O A_{4} = O A_{5} = 1$ and $A_{0} (1, 0), A_{3} (- 1, 0)$

Since each side of the hexagon makes an angle of $60^{\circ}$ at the centre $O$ of the circle coordinates of $A_{1}, A_{2}, A_{4} A_{5}$ , are respectively $(cos 60^{\circ}, sin 60^{\circ}), (cos 120^{\circ}, sin 120^{\circ}), (- cos 60^{\circ}, - sin 60^{\circ}), (- cos 120^{\circ}, - sin 120^{\circ})$

$\Rightarrow A_{1} (\frac{1}{2}, \frac{\sqrt{3}}{2}), A_{2} (- \frac{1}{2}, \frac{\sqrt{3}}{2}) A_{4} (- \frac{1}{2}, - \frac{\sqrt{3}}{2}), A_{5} (\frac{1}{2}, - \frac{\sqrt{3}}{2}),$

$\Rightarrow (A_{0} A_{1})^{2} = \frac{1}{4} + \frac{3}{4} = 1 \Rightarrow A_{0} A_{1} = 1$

$\Rightarrow (A_{0} A_{2})^{2} = \frac{9}{4} + \frac{3}{4} = 3 \Rightarrow A_{0} A_{2} = \sqrt{3} = A_{0} A_{4}$

$\Rightarrow (A_{0} A_{1}) (A_{0} A_{2}) (A_{0} A_{4}) = 3$

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Let A0A1A2A3A4A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the length of the line segments A0A1,A0A2and A0A4 is

Let $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}$ be a regular hexagon inscribed in a circle of unit radius. Then the product of the length of the line segments $A_{0} A_{1}, A_{0} A_{2}$ and $A_{0} A_{4}$ is