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}}$