Question

How do you find the area of a regular hexagon with a radius of $5?$ Please show working.

Answer 1

Area of a regular hexagon with radius “r” is given by $\dfrac{{{r^2}n\sin \left( {\dfrac{{{{360}^0}}}{n}} \right)}}{2}$ where “n” is the number of sides of the regular hexagon. Show working with the help of a figure and try to derive the above formula.

Complete step by step solution:
Do you know how the area of a regular hexagon is equals to
$\dfrac{{{r^2}n\sin \left( {\dfrac{{{{360}^0}}}{n}} \right)}}{2}$, let us derive this and see how it comes.

Draw a regular hexagon with side $2a$ and radius $R$ Now in this regular hexagon, we can see from figure $\Delta {\text{ABD}}$, if we find the area of $\Delta {\text{ABD}}$ and multiply it by $6$ then we will get the required area of hexagon. Since, we know that in a regular hexagon two times its length of side $(2a)$ is equals to its diameter $(2R)$

$\Rightarrow 2R = 2 \times 2a \\\ \Rightarrow R = 2a\; - - - - (i) \\\$

Now the area of the $\Delta {\text{ABD}}$ can be written as

$$= \dfrac{1}{2} \times {\text{base}} \times {\text{heigth}} \\\ {\text{ = }}\dfrac{1}{2} \times {\text{CD}} \times {\text{AB}} \\\ {\text{ = }}\dfrac{1}{2} \times 2a \times {\text{h}} \\\$$

From equation (i) we know that $R = 2a$ and in $\Delta {\text{ABC}}$ we have
$\cos x = \dfrac{{{\text{AB}}}}{{{\text{AC}}}} = \dfrac{h}{R} \\\ \Rightarrow h = R\cos x \\\$

We can also write $\cos x = \sin ({90^0} - x)$
$\Rightarrow h = R\sin ({90^0} - x)$

Putting $2a = R\;{\text{and}}\;h = R\sin ({90^0} - x)$ in above area expression,

$${\text{ = }}\dfrac{1}{2} \times 2a \times {\text{h}} \\\ {\text{ = }}\dfrac{1}{2} \times R \times R\sin ({90^0} - x) \\\ {\text{ = }}\dfrac{1}{2} \times {R^2}\sin ({90^0} - x) \\\$$

From figure, we can write $$6 \times 2x = {360^0} \Rightarrow x = \dfrac{{{{360}^0}}}{{12}}

\Rightarrow x = {30^0}$$ so the above expression will be

$${\text{ = }}\dfrac{1}{2} \times {R^2}\sin ({90^0} - x) \\\ {\text{ = }}\dfrac{1}{2} \times {R^2}\sin ({90^0} - {30^0}) \\\ {\text{ = }}\dfrac{1}{2} \times {R^2}\sin {60^0} \\\$$

So we get the area of $\Delta {\text{ABD}}$, now multiplying it by $n = 6$ (number of triangles) to get area of the hexagon

$= 6 \times \dfrac{1}{2} \times {R^2}\sin {60^0}$

Now substituting the given value of radius, $R = 5$ we will get
$= 6 \times \dfrac{1}{2} \times {R^2}\sin {60^0} \\\ = 6 \times \dfrac{1}{2} \times {5^2}\sin {60^0} \\\ = 6 \times \dfrac{1}{2} \times 25 \times \sin {60^0} \\\ = 75 \times \sin {60^0} \\\$

We know the value of $\sin {60^0} = \dfrac{{\sqrt 3 }}{2}$
$= 75 \times \sin {60^0} \\\ = 75 \times \dfrac{{\sqrt 3 }}{2} \\\ = 64.952\;{\text{uni}}{{\text{t}}^{\text{2}}} \\\$

$\therefore$ the required area of regular hexagon with radius $5\;{\text{units}}\;{\text{ = }}\;{\text{64}}{\text{.952}}\;{\text{uni}}{{\text{t}}^2}$

Note: Apart from regular and irregular hexagon, hexagon is also classified as concave and convex hexagon, concave hexagon has one or more interior angles $> {180^0}$ whereas in convex hexagon none of its interior angles is $> {180^0}$ and also the regular hexagon is always a convex one.