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).
 A: 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.
A: 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 & $\alpha$. 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. 

