Question

Find the area of a regular hexagon of side ‘a’.

Answer 1

Hint: In order to solve this question, we should know that the triangle formed by the center of the regular hexagon and 2 consecutive vertices of the regular hexagon is always of the same size. So, if we find the area of the triangle formed and then multiply it by 6, we will get the area of the triangle.

Complete step-by-step answer:
In this question, we have been asked to find the area of a regular hexagon whose side is ‘a’. Now, to solve this question, we will first draw a regular hexagon as follows.

Here, we know that, every possible triangle formed by the center of the regular hexagon and any two consecutive vertices will be of the same area. So, if we find the area of one of them and then multiply it by 6, we will get the area of the regular hexagon. Therefore, we can write,
Area of the regular hexagon = Area of $\Delta OAB\times 6$
Now, we can draw OAB as follows.

Here, we can see that AB is the side of the regular hexagon, that means, AB = a. And we know that it is a regular hexagon, so the distance between the center and the vertices of hexagon will be equal. So, we can write OA = OB. So, we can say OC is the median, angle bisector at point O. So, we can say, AC = BC and $\angle AOC=\angle BOC$. Also, we know that angle around point O is 360˚. So, $\angle AOB=\dfrac{{{360}^{\circ }}}{6}={{60}^{\circ }}.........\left( i \right)$. So, we can write $\angle AOC=\angle BOC=\dfrac{{{60}^{\circ }}}{2}={{30}^{\circ }}$.
Now, we know that a pair of angles opposite to equal sides are equal. So, we can say,
$\angle OAB=\angle OBA.........\left( ii \right)$
Also, we know that the sum of all angles of a triangle is 180˚. So, we can write,
$\angle AOB+\angle OAB+\angle OBA={{180}^{\circ }}$
Substituting the values of $\angle AOB$ and $\angle OBA$ from equations (i) and (ii) respectively, in the above equation, we can write it as,
\[\begin{align}
  & {{60}^{\circ }}+\angle OAB+\angle OAB={{180}^{\circ }} \\
 & 2\angle OAB={{180}^{\circ }}-{{60}^{\circ }}={{120}^{\circ }} \\
 & \angle OAB=\dfrac{{{120}^{\circ }}}{2}={{60}^{\circ }} \\
\end{align}\]
Hence, we can say that $\angle AOB=\angle OAB={{60}^{\circ }}$. Therefore, AOB is an equilateral triangle.
Now, we know that area of an equilateral triangle is given by $\dfrac{\sqrt{3}}{4}{{\left( side \right)}^{2}}$. So, we can say,
Area of $\Delta OAB=\dfrac{\sqrt{3}}{4}{{\left( a \right)}^{2}}$ sq units
Now, we will put this value in the formula of area of the regular hexagon of side ‘a’ as,
Area of regular hexagon = Area of $\Delta OAB\times 6$
Area of regular hexagon = $\dfrac{\sqrt{3}}{4}{{a}^{2}}\times 6$
Area of regular hexagon = $\dfrac{3\sqrt{3}}{2}{{a}^{2}}$
Hence, we can say that, the area of the regular hexagon of side ‘a’ is $\dfrac{3\sqrt{3}}{2}{{a}^{2}}$.

Note: While solving this question, we can find the length of OC as the height of $\Delta OAB$ by trigonometric ratios as $\tan \angle BOC=\tan {{30}^{\circ }}=\dfrac{BC}{OC}=\dfrac{1}{\sqrt{3}}$ and then we can find the value of OC to apply in the formula, area of triangle = $\dfrac{1}{2}$ (base) (height). And after finding the area of triangle, we can multiply it by 6 to get the area of the regular hexagon, which we will get as $\dfrac{3\sqrt{3}}{2}{{a}^{2}}$.