The area of the region $|x-ay| \le c$ for $0 \le x \le 1$ and $0 \le y \le 1$ 
What is the area of the following region:
$|x-ay| \le c$  for $0 \le x \le 1$ and $0 \le y \le 1$.
Assume $c>0$.

We can also assume $|a|>1$ since it will not change the essence of the problem.
Progress
We get a strip embedded inside of the square.   I was wondering if there is a clean way to approach this, since there are many different sub cases.

EDIT: This approach leads nowhere.
Let $S = \{(x,y) \in  [0,1]^2 : |x-ay|\leq c\}$. Consider the linear bijective map $$f(x,y) = (x-ay,y) = (u,v)$$ Then the $f(S) = \{(u,v) \in  [-a,1] \times [0,1] : |u| \leq c\} =: S_0$. 
EDIT: The following statement is wrong.
The area $V(S_0) = 1 \cdot (\min \{c,1\}-\max\{-c,-a\})$ (rectangle), since $c>0$
Note that $S = f^{-1}(S_0)$ thus $$V(S) = |\det f^{-1}| V(S_0)= |\det f|^{-1} V(S_0) = 1\cdot V(S_0) = \min \{c,1\}-\max\{-c,-a\},$$ because $$\det f = \left| \begin{pmatrix} 1 & -a \\ 0 & 1 \end{pmatrix}\right| = 1.$$
I hope this is correct.
