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Question: Let A<sub>0</sub> A<sub>1</sub> A<sub>2</sub> A<sub>3</sub> A<sub>4</sub> A<sub>5</sub>be a regular ...

Let A0 A1 A2 A3 A4 A5be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A0 A1 , A0 A2 and A0 A4 is-

A

34\frac{3}{4}

B

333\sqrt{3}

C

3

D

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

Answer

3

Explanation

Solution

Sol. Let the vertices be z0, z1, …… z5 w.r.t. centre O as origin

|z0| = 1, A0 A1 = |z1 – z0| = |z0 eiq – z0|

\ A0A1 = |z0| |cos q + i sin q – 1|

= 1 . (cos⁡θ−1)2+sin⁡2θ\sqrt{(\cos\theta - 1)^{2} + \sin^{2}\theta} =

2(1−cos⁡θ)\sqrt{2(1 - \cos\theta)}.

\ A0A1 = 2.2sin⁡2θ2\sqrt{2.2\sin^{2}\frac{\theta}{2}} = 2 sin θ2\frac{\theta}{2},

where q = 2π6\frac{2\pi}{6} = π3\frac{\pi}{3}

Replacing q by 2q and 4q, we get

A0A2 = 2 sin 2θ2\frac{2\theta}{2}= 2 sin q,

A0A4 = 2 sin 4θ2\frac{4\theta}{2} = 2 sin 2q

\ (A0A1) (A0A2) (A0A4) = 8 sin π6\frac{\pi}{6}sin π3\frac{\pi}{3} sin 2π3\frac{2\pi}{3}

= 8. 12\frac{1}{2}. 32\frac{\sqrt{3}}{2}.32\frac{\sqrt{3}}{2} = 3.