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Equation of a Sphere : General Form

If A0, A1, A2...

Question

If $A_{0}, A_{1}, A_{2}, A_{3}, A_{4}$ and $A_{5}$ 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:

A

3

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B

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

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C

\frac{3}{4}

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D

3 \sqrt{3}

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Solution

The correct option is A $3$
$\begin{matrix} Now, {(A_{0} A_{1})}_{0}^{2} = {(1 - \frac{1}{2})}^{2} + {(0 - \frac{\sqrt{3}}{2})}^{2} = {(\frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2} = \frac{1}{4} + \frac{3}{4} = 1 \Rightarrow A_{0} A_{1} = 1 {(A_{0} A_{2})}_{0}^{2} = {(1 + \frac{1}{2})}^{2} + {(0 - \frac{\sqrt{3}}{2})}^{2} = {(\frac{3}{2})}^{2} + {(- \frac{\sqrt{3}}{2})}^{2} = \frac{9}{4} + \frac{3}{4} = \frac{12}{4} = 3 \Rightarrow A_{0} A_{2} = \sqrt{3} and {(A_{0} A_{4})}_{0}^{2} = {(1 + \frac{1}{2})}^{2} + {(0 + \frac{\sqrt{3}}{2})}^{2} = {(\frac{3}{2})}^{2} + (\frac{3}{4}) = \frac{9}{4} + \frac{3}{4} = \frac{12}{4} = 3 \Rightarrow A_{0} A_{4} = \sqrt{3} Thus, (A_{0} A_{1}) (A_{0} A_{2}) (A_{0} A_{4}) = 3 \end{matrix}$

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Equation of a Sphere : General Form

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