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 $_{2}$ and A $_{0}$ A $_{4}$ is

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Solution

The correct option is C
(c) Each triangle is an equilateral triangle

Hence A $_{0}$ A $_{1}$ = 1
A $_{0}$ A $_{0}^{2}$ = A $_{0}$ A $_{1}^{2}$ + A $_{1}$ A $_{0}^{2}$ - 2A $_{0}$ A $_{1}$ A $_{1}$ A $_{2}$ cos 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. $\sqrt{3}$ . $\sqrt{3}$

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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 (c) Each triangle is an equilateral triangle Hence A0A1 = 1 A0A20 = A0A21 + A1A20 - 2A0A1A1A2 cos 120∘ = 1 + 1 - 2.1.1(-12) = 3 ⇒ A0A2 = √3 = A0A4 ∴ A0A1 × A0A2 × A0A4 = 1.√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 $_{2}$ and A $_{0}$ A $_{4}$ is