ABCDEF is a regular hexagon with centre O (in the following figure). If the area of triangle OAB is 9 cm2, 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`
