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

(i) the circumference of the inscribed circle

(ii) the area of the inscribed circle

(i) Side of the regular hexagon (a) = $2\sqrt{3}$ cm.

Now since a regular hexagon has 6 sides, hence we can say the central angle of a hexagon = $\dfrac{360°}{6}$ = 60°.

From figure,

∠AOB = 60°.

Now in the ΔAOB,

Since the total sum of the angles of a triangle is equal to 180°

∴ ∠AOB + ∠OAB + ∠OBA = 180° …….(1)

Since, OA = OB = radius of circle,

So, ∠OAB = ∠OBA = x (let) (Angles opposite to equal sides are equal).

Substituting value in equation 1 we get,

⇒ 60° + x + x = 180°

⇒ 2x + 60° = 180°

⇒ 2x = 120°

⇒ x = 60°.

Now since all the angles of the triangle are equal hence we can say the triangle is an equilateral triangle. So, all the sides of the triangle will also be equal.

∴ AO = BO = AB = $2\sqrt{3}$ cm.

Draw perpendicular from O to AB.

From figure,

AT = BT = $\dfrac{2\sqrt{3}}{2} = \sqrt{3}$ cm. (As altitude and median are same in equilateral triangle.)

From figure,

In right angle triangle OAT,

$\Rightarrow OA^2 = OT^2 + AT^2 \\[1em] \Rightarrow (2\sqrt{3})^2 = OT^2 + (\sqrt{3})^2 \\[1em] \Rightarrow 12 = OT^2 + 3 \\[1em] \Rightarrow OT^2 = 12 - 3 \\[1em] \Rightarrow OT^2 = 9 \\[1em] \Rightarrow OT = \sqrt{9} = 3 \text{ cm}.$

Hence, radius of circle = OT = 3 cm.

We know the circumference of a circle is given by the formula,

Circumference = 2πr

= $2 \times \dfrac{22}{7} \times 3$

= $\dfrac{132}{7}$ cm.

Hence, circumference of inscribed circle = $\dfrac{132}{7}$ cm.

(ii) Area of inscribed circle = πr²

= $\dfrac{22}{7} \times (3)^2$

= $\dfrac{22 \times 9}{7} = \dfrac{198}{7}$ cm².

Hence, area of inscribed circle = $\dfrac{198}{7}$ cm².