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 lengths of the line segments $A_{0} A_{1}, A_{0} A_{2} a n d A_{0} A_{4}$ is

$\frac{3}{4}$

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$3 \sqrt{3}$

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$3$

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

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Solution

The correct option is C
$3$

Each triangle is an equilateral triangle

$H e n c e A_{0} A_{1} = 1 A_{0} A_{2}^{2} = A_{0} A_{1}^{2} + A_{1} A_{2}^{2} - 2 A_{0} A_{1} A_{1} A_{2} c o s 120^{\circ} = 1 + 1 - 2.1.1 (- \frac{1}{2}) = 3 \Rightarrow A_{0} A_{2} = \sqrt{3} = A_{0} A_{4} ∴ A_{0} A_{1} \times A_{0} A_{2} \times A_{0} A_{4} = 1. \sqrt{3} . \sqrt{3} = 3$

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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 lengths of the line segments $A_{0} A_{1}, A_{0} A_{2} a n d A_{0} A_{4}$ is

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Let A0A1A2A3A4A5 be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A0A1, A0A2 and A0A4 is

The correct option is C 3 Each triangle is an equilateral triangle Hence A0A1=1A0A22=A0A21+A1A22−2A0A1A1A2 cos 120∘=1+1−2.1.1(−12)=3⇒A0A2=√3=A0A4∴ A0A1×A0A2×A0A4=1.√3.√3=3

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 lengths of the line segments $A_{0} A_{1}, A_{0} A_{2} a n d A_{0} A_{4}$ is