*ABCDEF* is a regular hexagon with centre O (in the following figure). If the area of triangle *OAB* is 9 cm^{2}, find the area of : (i) the hexagon and (ii) the circle in which the haxagon is incribed.

#### Solution

We know that a regular hexagon is made up of 6 equilateral triangles.

We have given area of the one of the triangles.

`∴"Area of the hexagon=6xx area of one equilateral triangle"`

`∴"Area of the hexagon"=6xx9`

`∴"Area of the hexagon"=54`

We know that if a regular hexagon is inscribed in the circle, then the radius of the circle is same as the side of the regular hexagon.

We also know that a regular hexagon is made up of 6 equilateral triangles and we have area of one of the equilateral triangle.

`∴"Area of the equilateral triangle"=sqrt3/4 xx"side"^2`

Substituting the value of the given equilateral triangle we get,

`∴9=sqrt3/4xx"side"^2`

`∴ side^2=(9xx4)/sqrt3`

`∴ "side"^2=36/sqrt3`

Now we will find the area of the circle.

∴ Area of the circle=`pi r^2`

Substituting the values we get,

`∴" Area of the circle"=22/7xx36/sqrt3`

Now we will substitute sqrt3=1.732 we get,

`∴" Area of the circle"=22/7xx36/1.732`

`∴" Area of the circle"=792/12.124`

`∴" Area of the circle"=65.324`

Therefore, area of the hexagon and area of the circle are `54 cm^2 and 65.324 cm^2`