Find the area of ​a regular pentagon as a function of its diagonal For reference:

Calculate the area of ​​a regular pentagon
as a function of its diagonal of length $a$. (Answer:$\frac{a^2}{4}\sqrt\frac{25-5\sqrt5}{2}$)

My progress:

$R$ = radius inscribed circle
$S = \frac{5R^2}{8}(\sqrt{10+2\sqrt5})$
$L=\frac{R}{2}(\sqrt{10-2\sqrt5})\implies R^2 = \frac{4L^2}{10-2\sqrt5}\tag{I}$
$\cos36^\circ=\frac{a}{2L} \implies a = L(\frac{1+\sqrt5)}{2}\implies L= \frac{2a}{1+\sqrt5}$
$\implies L^2 = \frac{a^2}{4}(6-2\sqrt5)\tag{II}$
$\text{From (I)}: S = \frac{20L^2}{8(10-2\sqrt5)}.(\sqrt{10+2\sqrt5})$
$\implies S =\frac{5L^2}{2(10-2\sqrt5)}.(\sqrt{10+2\sqrt5})$
$\text{From (II)}: S = \frac{5a^2(6-2\sqrt5)}{8.(10-2\sqrt5)}\cdot(\sqrt{10+2\sqrt5})$
$S = \frac{a^2(5-\sqrt5)}{16}(\sqrt{10+2\sqrt5})$
$\boxed{S = \frac{a^2}{4} \sqrt{\frac{25-5\sqrt5}{2} }}$

While you are starting with a formula for the pentagon in terms of $R$, you can derive directly as follows -
Given $a$ is diagonal,
$ \displaystyle S_{AED} = S_{BCD}  = \frac 12 \cdot a \cdot \frac {a \tan36^0}{2}$
$ \displaystyle S_{ADB} = \frac 12 \cdot a^2 \sin36^\circ$
So, $~ \displaystyle S = \frac{a^2}{2} \sin 36^\circ \cdot \frac{1 + \cos36^\circ}{\cos36^\circ}$
As $\cos 36^\circ = \dfrac{1 + \sqrt5}{4}$
$ \displaystyle \sin 36^\circ = \sqrt{1 - \frac{6+2 \sqrt5}{16}} = \frac 12 \cdot \sqrt{\frac{5-\sqrt5}{2}}$
Also, $\dfrac{1 + \cos36^\circ}{\cos36^\circ} = \dfrac{5 + \sqrt5}{1 + \sqrt5} = \sqrt5$
So, $ \displaystyle S = \frac{a^2}{4} \sqrt{\frac{25-5\sqrt5}{2}}$
