What is the area of the shaded region in terms of $n$? We can see that shaded region is area of FEH minus the sector HGE. 
To find the sector HGE i called the angle GHE as $\theta$ and used
$$
\pi\times\left(\frac{\theta}{360^{\circ}}\right)
$$
If FE is $x$, we see that $\cos(\theta)=\frac{1}{\sqrt{x^2+1}}$ therefore
$$
\theta=\arccos\left(\frac{1}{\sqrt{x^2+1}}\right)
$$
My attempt on this problem was by "brute-forcing" my way from the bottom of the figure to top, to find FE, using Law of Cosine on the way.

The shaded region will now be 
$$
\frac{x}{2} - \pi\times\left(\frac{\arccos\left(\frac{1}{\sqrt{x^2+1}}\right)}{360^{\circ}}\right)
$$

To find $x$ i approached in two different ways:


Connect CD to AI by lengthening CD and connect HD. Do Law of Cosine on these triangles to get to FE.
Connect AD and HD, do law of cosines on these to get to FE.


Both of these resulted in immense work and, on the first case i was able to write $x$ in terms of $n$ but the equation barely fit a word page with font size 1. (and because of the immense work, it is likely to be wrong).
So, let's do it "classically".

With reference to the sketch



we have from the data given
$$
\left\{ \matrix{
  H = n\left( {\cos \alpha ,\;\sin \alpha } \right) \hfill \cr 
  E = H + \left( {\cos \beta ,\;\sin \beta } \right) \hfill \cr 
  D = \left( {n - \cos \left( {\alpha /2} \right),\,\sin \left( {\alpha /2} \right)} \right) \hfill \cr 
  E - D = x\left( {\cos \left( {\beta  + \pi /2} \right),\;\sin \left( {\beta  + \pi /2} \right)} \right) \hfill \cr}  \right.
$$
in the unknown $x$ and $\beta$.
The points are written as position vectors wrt the origin.   

Last line gives
$$
n\left( {\cos \alpha ,\;\sin \alpha } \right) + \left( {\cos \beta ,\;\sin \beta } \right)
 - \left( {n - \cos \left( {\alpha /2} \right),\,\sin \left( {\alpha /2} \right)} \right)
 = x\left( {\cos \left( {\beta  + \pi /2} \right),\;\sin \left( {\beta  + \pi /2} \right)} \right)
$$
which translates into
$$
\left\{ \matrix{
  n\cos \alpha  + \cos \beta  - n + \cos \left( {\alpha /2} \right) = x\cos \left( {\beta  + \pi /2} \right) =  - x\sin \beta  \hfill \cr 
  n\sin \alpha  + \sin \beta  - \sin \left( {\alpha /2} \right) = x\sin \left( {\beta  + \pi /2} \right) = x\cos \beta  \hfill \cr}  \right.
$$
i.e.
$$
\left\{ \matrix{
  \cos \beta  + x\sin \beta  = n - n\cos \alpha  - \cos \left( {\alpha /2} \right) \hfill \cr 
  x\cos \beta  - \sin \beta  = n\sin \alpha  - \sin \left( {\alpha /2} \right) \hfill \cr}  \right.
$$

To solve this, let's make a change of variable
$$
\left\{ \matrix{
  x = X\sin \eta  \hfill \cr 
  1 = X\cos \eta  \hfill \cr}  \right.\quad  \Leftrightarrow \quad \left\{ \matrix{
  X = \sqrt {1 + x^{\,2} }  \hfill \cr 
  \eta  = \arctan \left( x \right) \hfill \cr}  \right.
$$
so that
$$
\left\{ \matrix{
  X\cos \left( {\eta  - \beta } \right) = n - n\cos \alpha  - \cos \left( {\alpha /2} \right) \hfill \cr 
  X\sin \left( {\eta  - \beta } \right) = n\sin \alpha  - \sin \left( {\alpha /2} \right) \hfill \cr}  \right.
$$

Thereafter, I think you can conclude by yourself.
Maybe not much simpler than your answer, but here is a methodical way to approach it.


Extend CD to meet AF at X. Since you know AC, angle XAC and angle XCA, the triangle ACX can be solved completely in terms of n &amp; $\alpha$. From law of sines:


AX / sin($\alpha$/2) = CX / sin($\alpha$) = (n-1) / sin($\pi$ - $3\alpha / 2$)

This should give you both AX and CX without much work which we will use later.


FX = n-1-AX, DX = CX-1 and you know the included angle FXD. From 1, you know AX and CX. So, you can now solve the triangle FXD completely. This is a bit more work, but is pretty straight forward as well.


FD = $\sqrt{FX^2 + XD^2 - 2.FX.XD.cos(3\alpha/2)}$ 

FD / sin($3\alpha / 2$) = DX / sin(XFD)

This should give you XFD (note that XFD is less than 90 degrees due to construction)


Now you know angle HFE = (angle XFD from 2). Thus, you know angle FHE = (90 - HFE). This helps you find area of sector FHE. Since FE = Tan(angle HFE), you can now find area of triangle FHE = 0.5 x 1 x FE.