Question

A circle is inscribed in a regular hexagon of side $2\sqrt{3}cm$ . Find the circumference of the inscribed circle.

Answer 1

Hint:To solve this problem we should know about concept of circumscribed as well as basic trigonometry
Circumscribed: Circumscribed of a polygon is a circle that passes through all the polygon. The center of this circle is called the circumcenter and its radius is called the circumradius.
A trigonometric property, $\cot \theta ={\displaystyle \frac{base}{perpendicular}}$

Complete step by step answer:
As given in question we will draw diagram,

It is obvious from figure that
Let, there are two triangle AOB and COB,
In, $\mathrm{\Delta }$ AOB and $\mathrm{\Delta }$ BOC. We have,
 $AO=CO$ (radius of circle)
And $OB$ is common if hypotenuse is equal.
So, by RHS criteria both triangles are congruent.
Hence, $\mathrm{\angle }AOB=\mathrm{\angle }BOC$
And $$\mathrm{\angle }AOC={60}^{\circ }$$ (circle are divided in six parts)
So, $\mathrm{\angle }AOB=\mathrm{\angle }BOC={30}^{\circ }$
In, $\mathrm{\Delta }$ AOB:
 $\cot {30}^{\circ }={\displaystyle \frac{base}{perpendicular}}$
As, length of the side of the hexagon is $2\sqrt{3}cm$ . so, $AB=\sqrt{3}cm$ .
So,
${\displaystyle \frac{Radiusofthecircle}{halfthesideofhexgon}}=\cot {30}^{\circ }$
 $\Rightarrow {\displaystyle \frac{Radiusofthecircle}{{\displaystyle \frac{1}{2}}\times 2\sqrt{3}}}=\sqrt{3}$
(As $\cot {30}^{\circ }=\sqrt{3}$ )
 $\Rightarrow Radiusofthecircle(r)=3cm$
So circumference of the inscribed circle $=2\pi r=2\pi \times 3=6\pi$

Note: properties of regular hexagon:
It has six sides and six angles and the measurements of all angles are equal.
The total number of diagonals in a regular hexagon is nine.
The sum of all interior angles is equal to ${720}^{\circ }$ .