Let $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}$ be a regular hexagon inscribed in a unit circle with centre at origin then the product of lengths of the line segments $A_{0} A_{1}, A_{0} A_{2}, A_{0} A_{4}$ is

\sqrt{3}

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Solution

The correct option is D $3$
$Δ A_{0} 0 A_{1}$ is an equilateral triangle.
$∴ A_{0} A_{1} = A_{0} 0 = 1$
Also, ${Δ A}_{0} A_{3} A_{4}$ is right angle triangle.
$A_{0} A_{3} = 2$
$A_{0} A_{4} = ?$
$A_{4} A_{3} = 1$
$∴ A_{0} A_{3} = \sqrt{A_{0} {A_{4}}^{2} + A_{4} {A_{3}}^{2}} ∴ A_{0} A_{4} = \sqrt{A_{0} {A_{3}}^{2} + A_{4} {A_{3}}^{2}} = \sqrt{2^{2} - 1^{1}} = \sqrt{3} ∴ A_{0} A_{2} = \sqrt{3} ∴ (A_{0} A_{1}) (A_{0} A_{2}) (A_{0} A_{4}) = 1 \times \sqrt{3} \times \sqrt{3} = 3$

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Let A0A1A2A3A4A5 be a regular hexagon inscribed in a unit circle with centre at origin then the product of lengths of the line segments A0A1,A0A2,A0A4 is

The correct option is D 3ΔA00A1 is an equilateral triangle.∴A0A1=A00=1Also, ΔA0A3A4 is right angle triangle.A0A3=2 A0A4=?A4A3=1∴A0A3=√A0A42+A4A32∴A0A4=√A0A32+A4A32=√22−11=√3∴A0A2=√3∴(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 unit circle with centre at origin then the product of lengths of the line segments $A_{0} A_{1}, A_{0} A_{2}, A_{0} A_{4}$ is