Question

ABCDEF is a regular hexagon with center O. If the area of the triangle OAB is 9 $cm^2$, find the area of the circle in which the hexagon is inscribed.

Answer 1

Hint: In this particular question use the concept that the regular hexagon is made up of 6 equilateral triangles later on apply the formula of the area of the equilateral triangle to reach the solution of the question.

Complete step-by-step answer:
As we know that the regular hexagon is made up of 6 equilateral triangles.

And it is given that the area (A) of one triangle OAB is 9 $cm^2$.

$ \Rightarrow A = 9$ $cm^2$. ……………….. (1)

So, OAB is an equilateral triangle and we all know that the area (A) of equilateral triangle is
$ \Rightarrow A = \dfrac{{\sqrt 3 }}{4}{\left( {{\text{ side}}} \right)^2}$ $cm^2$.

Where the side is nothing but the radius (r) of the circle in cm (see figure).

$ \Rightarrow A = \dfrac{{\sqrt 3 }}{4}{\left( {{\text{ r}}} \right)^2}$$cm^2$…………………. (2)

Now from equation (1) and (2) we have,

$ \Rightarrow 9 = \dfrac{{\sqrt 3 }}{4}{\left( {{\text{ r}}} \right)^2}$

$ \Rightarrow {r^2} = \dfrac{{36}}{{\sqrt 3 }}$…………………….. (3)

Now as we know that the area (${A_2}$) of the circle is $\pi {r^2}$.

So from equation (3) we have,

$ \Rightarrow {A_2} = \pi {r^2} = \pi \left( {\dfrac{{36}}{{\sqrt 3 }}} \right) =

\dfrac{{22}}{7}\left( {\dfrac{{36}}{{\sqrt 3 }}} \right) = 65.325$ $cm^2$. $\left[ {\because \pi =

\dfrac{{22}}{7}} \right]$

So, this is the required area of the circle in which the hexagon is inscribed.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of the equilateral triangle and that the regular hexagon is made up of 6 equilateral triangles so using the formula calculate the side of the equilateral triangle which is nothing but the radius of the circle (as hexagon is inscribed the circle) then use the formula of area of circle which is stated above and calculate the area which is the required answer.