Pentagons and Ratios

Let the side lengths of the pentagon be $a.$ Then since the interior angles of a regular pentagon are $108^{\circ}$ we have that, for example, the area of $\Delta BCD$ is

$\dfrac{1}{2}a*a\sin(72^{\circ}) = 1 \Longrightarrow a^{2} = \dfrac{2}{\sin(72^{\circ})}.$

Now a regular pentagon can be divided into $5$ congruent isosceles triangles each with a base of length $a$ and two base angles of $54^{\circ}$ each. Thus the area of the pentagon is

$5*\dfrac{1}{2}a*\dfrac{a}{2}\tan(54^{\circ}) = \dfrac{5}{4}a^{2}\tan(54^{\circ}).$

Substituting in the value for $a^{2}$ found above, we see that the area of the pentagon is

$\dfrac{5}{4}*\dfrac{2}{\sin(72^{\circ})}*\tan(54^{\circ}) = \dfrac{5}{2}\dfrac{\cot(36^{\circ})}{2\sin(36^{\circ})\cos(36^{\circ})} = \dfrac{5}{4}\csc^{2}(36^{\circ}).$

Now $\sin(36^{\circ}) = \cos(54^{\circ}) \Longrightarrow \sin(2*18^{\circ}) = \cos(3*18^{\circ}).$

Now using the identities

$\sin(2x) = 2\sin(x)\cos(x)$ and $\cos(3x) = 4\cos^{3}(x) - 3\cos(x),$ we see that

$\sin(2x) = \cos(3x) \Longrightarrow 2\sin(x)\cos(x) = 4\cos^{3}(x) - 3\cos(x)$

$\Longrightarrow \cos(x)(2\sin(x) - 4\cos^{2}(x) + 3) = 0.$

Now since $\cos(x) \ne 0$ when $x = 18^{\circ}$ we must have that

$2\sin(x) - 4\cos^{2}(x) + 3 = 0 \Longrightarrow 4\sin^{2}(x) + 2\sin(x) - 1 = 0$

$\Longrightarrow \sin(x) = \dfrac{-2 + \sqrt{4 + 16}}{8} = \dfrac{-1 + \sqrt{5}}{4},$

where we took the positive root since we know that $\sin(18^{\circ}) \gt 0.$

Now $\cos(18^{\circ}) = \sqrt{1 - \sin^{2}(18^{\circ})} = \dfrac{1}{4}\sqrt{10 + 2\sqrt{5}},$ so

$\sin(36^{\circ}) = 2\sin(36^{\circ})\cos(36^{\circ}) = \dfrac{1}{8}(-1 + \sqrt{5})\sqrt{10 + 2\sqrt{5}}.$

and so $\sin^{2}(36^{\circ}) = \dfrac{1}{8}(5 - \sqrt{5}).$

Thus the area of the given pentagon is

$\dfrac{10}{5 - \sqrt{5}} = \dfrac{5 + \sqrt{5}}{2} = \boxed{3.618}$ to 3 decimal places.

May not be the nicest solution, but short enough: First we compute $\frac{|\triangle ACD|}{|\triangle ABC|}=\frac{\sin 72^\circ}{\sin 36^\circ}=\frac{AD}{CD}=\frac{BE}{AE}.$ Now let $BE$ intersect $AD$ intersect at $F$ . Note that the ratio we had before can also be expressed as $\frac{BF}{BA}+\frac{FE}{AE}$ . Angle chasing around, we have $BA=BF$ since $\triangle ABF$ is isosceles, and $\frac{FE}{AE}=\frac{AE}{BE}$ by similar triangles $\triangle BAE$ and $\triangle AFE$ . So if we let the required ratio be $x$ , we have established $\frac{BE}{AE}=\frac{BF}{BA}+\frac{FE}{AE}\iff x=1+\frac{1}{x}$ which at this point can be solved as a quadratic equation (multiply both sides by $x$ ) or recognised as the well-known ratio $\phi=1.618\ldots$ . Hence $|\triangle ACD|=\phi$ , and the answer is $|ABCDE|=|\triangle ABC|+|\triangle ACD|+|\triangle ADE|=2+\phi\approx\boxed{3.618}$ .

$\angle~ between~petagon ~ sides~=\frac {5*180-2*180}{5}=108^o,~\angle ABC=\angle BAE=108^o,\\AB=BC ~~\therefore \angle BAC=36^o=\angle EAD~ by ~ symmetry~~~~\therefore \angle CAD=36^o.\\Let ~x=AC. \therefore for~ isosceles ~\Delta ABC~\color{#D61F06}{ 1}=Area=\color{#D61F06}{(\frac x 2 )^2*Sin36^o*Cos36^o}.\\But~Area~isosceles\Delta CAD~=(\frac x 2 )^2*Sin18^o*Cos18^o=\color{#D61F06}{(\frac x 2 )^2*\frac 1 2 Sin36^o}.\\ Area~\Delta CAD=\color{#3D99F6}{1*\dfrac{2}{Cos 36^o}}=1.618.~~~~~~~~~~~~~~~~~~~~~~~~Pentagon~Area\\=Areas ~~\Delta s~~\{ABC+CAD+DAE\}=1+1.618+1=~~~~\color{#EC7300}{ \large \boxed{3.618}. }\\~~\\\text{After seeing the other nice and detailed solutions, I would rather avoid}\\\text{giving mine. I am giving just to show how to shorten the calculations.}$

We know that the areas of $\Delta ABC$ and $\Delta ADE$ are $1$ . Hence we are required to compute only the area of $\Delta ACD$ .

Visualizing the circumcircle of the pentagon we find that $\angle CAD = \angle CED$ since they are angles in the same segment of the circumcircle..

We also have $\angle CED = \angle DCE$ since $\Delta CDE$ is isosceles.

Applying the angle sum property in $\Delta CED$ we have $2\angle CED + \angle CDE = \pi$

or $2\angle CED + \frac{3\pi}{5} = \pi \implies \angle CED = \frac{\pi}{5}$

Here we have used the fact the the interior angle of a regular n-gon is $\frac{(n-2)\pi}{n}$ where $n=5$ in this case.

So we now have $\angle CED= \angle CAD = \frac{\pi}{5}$

Let $d$ be the length of any diagonal then area of $\Delta ACD = \frac{1}{2} d^2 \sin(\angle CAD)=\frac{1}{2}d^2\sin(\frac{\pi}{5})$

Let $a$ be the side length, then from the law of cosines, $d^2 = a^2 + a^2 - 2a^2 \cos(\frac{3\pi}{5}) = 2a^2(1-\cos(\frac{3\pi}{5})) = 2a^2(1+\cos(\frac{2\pi}{5})) =4a^2\cos^2(\frac{\pi}{5})$

Also since we the diagonals cut off unit triangles, we know $\frac{1}{2}a^2\sin(\frac{3\pi}{5}) = 1$

or, $\frac{1}{2}a^2\sin(\frac{2\pi}{5})=a^2\sin(\frac{\pi}{5})\cos(\frac{\pi}{5}) = 1$

Using all of this in $|\Delta ACD| = \frac{1}{2}d^2 \sin(\frac{\pi}{5})$ , we get

$|\Delta ACD| = 2\cos(\frac{\pi}{5})=2\cos(36^\circ)=2\frac{\phi}{2}=\phi$

Where $\phi$ is the Golden Ratio and $\phi \approx 1.618\dots$

So we finally have area of the pentagon $= 2+\phi \approx \boxed{3.618\dots}$