ABCDEF is a regular hexagon with centre O (Fig). If the area of triangle OAB is 9 cm2. Find the area of (i) the hexagon and (ii) the circle in which the hexagon is inscribed.
We know that a regular hexagon is made up of 6 equilateral triangles.
So the Area of a hexagon = 6 × area of one equilateral triangle.
We have given area of the one of the triangles.= → 6 x 9cm2
Hence Area of a hexagon= 54cm2
Note 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 further know that a regular hexagon is made up of 6 equilateral triangles and it is given that the area of an equilateral △OAB is 9 cm2
So using the formula of area of an equilateral triangle= √3 4 we have
⇒ 9 = √3 4× side2
⇒ side2= (9)(4)√3
Area of a circle= 227 x 36√3
Now we will find the area of the circle.
Area of a circle = π \({r}^{2}\ )
Area of a circle= 227 x 36√3
Now we will substitute √3 = 1.732 we get
⇒ 227 x 361.732
⇒ 79212.294
⇒ 65.324 cm2
Therefore, area of the hexagon and area of the circle are 54 cm2 and 65.32 cm2 respectively.