Question

ABCDEF is a regular hexagon with centre O (Fig). If the area of triangle OAB is $9c{m}^2$. Find the area of (i) the hexagon and (ii) the circle in which the hexagon is inscribed.

Answer 1

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

So the Area of a hexagon = 6 $\times$ area of one equilateral triangle.

We have given area of the one of the triangles.= $\to$ 6 x ${9cm}^2$

Hence Area of a hexagon= ${54cm}^2$

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 $9c{m}^2$

So using the formula of area of an equilateral triangle= $\frac{\sqrt{3}}{4}$ we have

​$\Rightarrow$ 9 = $\frac{\sqrt{3}}{4}$$\times$ ${side}^2$

​$\Rightarrow$ ${side}^2$= $\frac{\left(9\right)\left(4\right)}{\surd 3}$

Area of a circle= $\frac{22}{7}$ x $\frac{36}{\surd 3}$

Now we will find the area of the circle.

Area of a circle = $\pi$ \({r}^{2}\ )

Area of a circle= $\frac{22}{7}$ x $\frac{36}{\surd 3}$

Now we will substitute $\sqrt{3}$ = 1.732 we get

$\Rightarrow$ $\frac{22}{7}$ x $\frac{36}{1.732}$

$\Rightarrow$ $\frac{792}{12.294}$

$\Rightarrow$ $65.324c{m}^2$

Therefore, area of the hexagon and area of the circle are $54c{m}^2$ and $65.32c{m}^2$ respectively.