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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C

Each triangle is an equilateral triangle

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

Suggest Corrections

Similar questions

Q. Let A

_{0}

_{1}

_{2}

_{3}

_{4}

_{5}

be a regular hexagon inscribed in a circle of unit radius. Then the product of the lengths of the line segments A

_{0}

_{1}

, A

_{0}

_{2}

and A

_{0}

_{4}

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

Q. 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 the origin. Then the product of the lengths of the line segments

A_{0} A_{1}, A_{0} A_{2}, A_{0} A_{4}

Q. 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}

Let be $A_{0} A_{1} A_{2} A_{3} A_{4} A_{5}$ a regular hexogon 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

Join BYJU'S Learning Program

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 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