## the area of regular dodecagon with side length of one

Posted: March 5, 2013 in Mathematics

Today I am going to show you how to find the exact area of a regular dodecagon with side length of 1 without using a calculator.  First of all, let me tell you what a regular dodecagon is.  A regular dodecagon is a regular 12-gon, a 2D shape with 12 equal sides.

Let’s begin!

See the diagram, a regular dodecagon consists of 12 isosceles triangles with central angle of 30 degrees.  Let the base of that triangle be 1, then

$\dfrac{1/2}{h}=\tan{15^\circ}$

$h=\dfrac{1}{2\tan{15^\circ}}$

then the area of one triangle is h/2 and the exact area of the regular dodecagon is 6h.  Now, the difficult part is to find the exact value of h by hand.  (Remember this is a math contest question, no calculator is allowed.  Just your brain, a pencil, and papers. )

Now, if it is tan(30), it will be easy because $\tan{30^\circ}=\dfrac{1}{\sqrt{3}}$

but notice that 15=30/2, so we can use the half angle trig identity.

Since

$\cos{2x}=\cos^2{x}-\sin^2{x}$

$\cos{2x}=1-2\sin^2{x}$

$\cos{2x}=2\cos^2{x}-1$

then we have

$\sin^2{x}=\dfrac{1-\cos{2x}}{2}$

and

$\cos^2{x}=\dfrac{1+\cos{2x}}{2}$

And since our angle here is less than 90 degrees, we can drop the negative sign when taking square roots.

$\sin{x}=\sqrt{\dfrac{1-\cos{2x}}{2}}$

$\cos{x}=\sqrt{\dfrac{1+\cos{2x}}{2}}$

then replace x with x/2 yields,

$\sin{\dfrac{x}{2}}=\sqrt{\dfrac{1-\cos{x}}{2}}$

$\cos{\dfrac{x}{2}}=\sqrt{\dfrac{1+\cos{x}}{2}}$

Then tangent is sine over cosine,

$\tan\dfrac{x}{2}=\sqrt{\dfrac{1-\cos{x}}{1+\cos{x}}}$

Then substitute in x=30, we have

$\tan15^\circ=\tan\dfrac{30^\circ}{2}=\sqrt{\dfrac{1-\cos30^\circ}{1+\cos30^\circ}}$

$=\sqrt{\dfrac{1-\sqrt{3}/2}{1+\sqrt{3}/2}}$

$=\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}$

$=\sqrt{\dfrac{(2-\sqrt{3})^2}{(2+\sqrt{3})(2-\sqrt{3})}}$

$=2-\sqrt{3}$

Therefore, the area of the regular dodecagon with side length of 1 is

$6h=\dfrac{6}{2\tan15^\circ}=\dfrac{3}{2-\sqrt{3}}$

$=3(2+\sqrt{3})=6+3\sqrt{3}$

DONE.