Question

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

Answer 1

$\angle AOB={60}^{\circ }$

$\angle OAB=\angle OBA$ $[$Angles opposite to equal sides are equal$]$

$\therefore \angle OAB=\angle OBA=60°$ $[\because \angle OAB+\angle OBA={180}^{\circ }-{60}^{\circ }={120}^{\circ }]$

$\therefore AO=BO=AB=2\sqrt{3}\text{cm}$ $[\because$ Triangle OAB is an equilateral triangel$]$

Now, draw OT perpendicualr to AB

$AT=BT$ $[\because$ Altitude and medians are same in an equilateral triangle$]$

Now, $O{T}^2=O{B}^2-B{T}^2$ [Traingle OTB is a right-angled triangle and use of Pythagoras theorem]

$⟹OT=\sqrt{(2\sqrt{3}{)}^2-(\sqrt{3}{)}^2}=3\text{cm}$ $[\because BT=\frac{AB}{2}]$

Circumference of inscribed circle$=2\pi r=2\pi \times 3=6\pi$ $[\because r=OT=3\text{cm}]$