Question

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 length of the line segment ${A_0}{A_1},{A_0}{A_2}$ and ${A_0}{A_4}$ is

Answer 1

Hint:
Here it is given that ${A_0},{A_1},{A_2},{A_3},{A_4},{A_5}$ be a regular hexagon inscribed in a circle of unit radius. We are required to find the product of the length of the line segment ${A_0}{A_1},{A_0}{A_2}$ and ${A_0}{A_4}$ .
We know that all the sides of the regular hexagon are equal.
Now, find the coordinates of the points ${A_0},{A_1},{A_2},{A_3},{A_4},{A_5}$ of the Hexagon.
Thus, using the distance formula, find the distance between any two consecutive points.
Finally, find the required product.

Complete step by step solution:
Here it is given that ${A_0},{A_1},{A_2},{A_3},{A_4},{A_5}$ be a regular hexagon inscribed in a circle of unit radius. We are required to find the product of the length of the line segment ${A_0}{A_1},{A_0}{A_2}$ and ${A_0}{A_4}$.

Here from the figure it is clear that $O{A_4} = 1$
Here, it is given that it is a regular hexagon
 $ \Rightarrow $ All sides of a regular hexagon are equal.
Then,
 $O{A_1} = O{A_2} = O{A_3} = O{A_4} = O{A_5} = 1$
 $ \Rightarrow $ Each sides of regular hexagon makes an angle of ${60^ \circ }$at the centre O of circle coordinates of ${A_1},{A_2},{A_4},{A_5}$ are $\left( {\cos {{60}^ \circ },\sin {{60}^ \circ }} \right),\left( {\cos {{120}^ \circ },\sin {{120}^ \circ }} \right),\left( { - \cos {{60}^ \circ }, - \sin {{60}^ \circ }} \right),\left( { - \cos {{120}^ \circ }, - \sin {{120}^ \circ }} \right)$ respectively
 ${A_1} = \left( {\dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}} \right)$
 ${A_2} = \left( { - \dfrac{1}{2},\dfrac{{\sqrt 3 }}{2}} \right)$
 ${A_4} = \left( { - \dfrac{1}{2}, - \dfrac{{\sqrt 3 }}{2}} \right)$
 ${A_5} = \left( {\dfrac{1}{2}, - \dfrac{{\sqrt 3 }}{2}} \right)$
And,
 ${A_3} = \left( { - 1,0} \right)$ and ${A_0} = \left( {1,0} \right)$ (given circle is of radius)
Now, By distance formula
 $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} $
 ${\left( {{A_0}{A_1}} \right)^2} = {\left( {\dfrac{1}{2} - 1} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2} - 0} \right)^2}$
 $ \Rightarrow \left( {{A_0}{A_1}} \right) = 1$
Now, same with as $\left( {{A_0}{A_2}} \right)$
 ${\left( {{A_0}{A_2}} \right)^2} = {\left( { - \dfrac{1}{2}, - 1} \right)^2} + {\left( {\dfrac{{\sqrt 3 }}{2} - 0} \right)^2}$
 $\left( {{A_0}{A_2}} \right) = \sqrt 3 $
Similarly, find the value of $\left( {{A_0}{A_4}} \right)$
 ${\left( {{A_0}{A_4}} \right)^2} = {\left( { - \dfrac{1}{2} - 1} \right)^2} + {\left( { - \dfrac{{\sqrt 3 }}{2} - 0} \right)^2}$

$\left( {{A_0}{A_4}} \right) = \sqrt 3 $
Now, product of line segment of $\left( {{A_0}{A_1}} \right),\left( {{A_0}{A_2}} \right)$ and $\left( {{A_0}{A_4}} \right)$ is
 $ = 1 \times \sqrt 3 \times \sqrt 3 \\
=3$

Hence, the product is 3.

Note:
Regular polygon: In geometry, a regular polygon is a polygon that is equiangular and equilateral. Regular polygon may be either convex or star.
Regular Hexagon: A regular hexagon is defined as a hexagon that is both equilateral and equiangular. It is bicentric, meaning that it is both cyclic (has a circumscribed circle) and tangential (has an inscribed circle)