What is the perimeter of a regular hexagon that has an area of $54\sqrt{3}$ units squared?
Answer
559.8k+ views
Hint: To find the perimeter of the regular hexagon, we should know about the hexagon. A regular hexagon is a closed shape polygon which has six equal sides and six equal angles. In case of any regular polygon, all its sides and angles are equal. The regular hexagon is made up of six equilateral triangles.
Complete step by step solution:
Now, let's take a regular hexagon whose each side is $a$ cm.
The area of the regular hexagon is given in the question which is $54\sqrt{3}$ units square.
We know that the formula to find the area of regular hexagon is:
$\Rightarrow Area=\dfrac{3\sqrt{3}}{2}{{a}^{2}}$
Where a is the side of the regular hexagon.
Now, the area of the regular hexagon is given $54\sqrt{3}$, so put this in the above area formula in order to get the value of the side of the regular hexagon.
\[\Rightarrow 54\sqrt{3}=\dfrac{3\sqrt{3}}{2}{{a}^{2}}\]
Now by this formula we will find the value of the side of the regular hexagon. Now simplifying it, then we get
$\Rightarrow {{a}^{2}}=\dfrac{54\sqrt{3}\times 2}{3\sqrt{3}}$
Now both $\sqrt{3}$ is cancelled out from numerator or denominator, the we get
$\begin{align}
& \Rightarrow {{a}^{2}}=\dfrac{54\times 2}{3} \\
& \Rightarrow {{a}^{2}}=\dfrac{108}{3} \\
& \Rightarrow {{a}^{2}}=36 \\
\end{align}$
Now applying square root on sides of the equation then we get
$\begin{align}
& \Rightarrow a=\sqrt{36} \\
& \Rightarrow a=6 \\
\end{align}$
Thus we get the value of the side of the regular hexagon.
Now we know we have to find out the perimeter of the regular hexagon as mentioned in the question, so
We know the perimeter of regular hexagon is:
$\Rightarrow perimeter=6a$
Now putting the value of a in this perimeter formula, then we get
$\begin{align}
& \Rightarrow perimeter=6\times 6 \\
& \Rightarrow perimeter=36 \\
\end{align}$
Note: To solve these types of questions we should know the properties, formula of area and perimeter. Regular hexagon has six sides and six angles and all angles are equal. In the hexagon we can draw nine diagonals. The each interior angle of the regular hexagon is ${{120}^{\circ }}$. And each exterior angle of regular hexagon is ${{60}^{\circ }}$
Complete step by step solution:
Now, let's take a regular hexagon whose each side is $a$ cm.
The area of the regular hexagon is given in the question which is $54\sqrt{3}$ units square.
We know that the formula to find the area of regular hexagon is:
$\Rightarrow Area=\dfrac{3\sqrt{3}}{2}{{a}^{2}}$
Where a is the side of the regular hexagon.
Now, the area of the regular hexagon is given $54\sqrt{3}$, so put this in the above area formula in order to get the value of the side of the regular hexagon.
\[\Rightarrow 54\sqrt{3}=\dfrac{3\sqrt{3}}{2}{{a}^{2}}\]
Now by this formula we will find the value of the side of the regular hexagon. Now simplifying it, then we get
$\Rightarrow {{a}^{2}}=\dfrac{54\sqrt{3}\times 2}{3\sqrt{3}}$
Now both $\sqrt{3}$ is cancelled out from numerator or denominator, the we get
$\begin{align}
& \Rightarrow {{a}^{2}}=\dfrac{54\times 2}{3} \\
& \Rightarrow {{a}^{2}}=\dfrac{108}{3} \\
& \Rightarrow {{a}^{2}}=36 \\
\end{align}$
Now applying square root on sides of the equation then we get
$\begin{align}
& \Rightarrow a=\sqrt{36} \\
& \Rightarrow a=6 \\
\end{align}$
Thus we get the value of the side of the regular hexagon.
Now we know we have to find out the perimeter of the regular hexagon as mentioned in the question, so
We know the perimeter of regular hexagon is:
$\Rightarrow perimeter=6a$
Now putting the value of a in this perimeter formula, then we get
$\begin{align}
& \Rightarrow perimeter=6\times 6 \\
& \Rightarrow perimeter=36 \\
\end{align}$
Note: To solve these types of questions we should know the properties, formula of area and perimeter. Regular hexagon has six sides and six angles and all angles are equal. In the hexagon we can draw nine diagonals. The each interior angle of the regular hexagon is ${{120}^{\circ }}$. And each exterior angle of regular hexagon is ${{60}^{\circ }}$
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