Why is the mechanical advantage of a wedge = length of slope/ width? Mechanical advantage is defined as Force Output/Force Input
For a symmetrical wedge with the length of the slopes being equal and the width being the distance between the end points, the articles define the Mechanical advantage as: length/width
forces in a wedge
I will assume that you are splitting wood using a symmetrical wedge (lengths of the slopes are equal) and driving the wedge into the wood with a hammer vertically starting with the center of the wooden block.  Let us say the width is $w$ and the slope length is $l$. The Input vertical force is resulting in two sideward forces on the wood. Each of these sideward forces is causing the separation in the wood. If, based on the above diagram, I assume that each of these output forces is $F_{in} sin \theta$ where $\theta$ is the angle between the vertical direction of $F_{in}$ and the slope, I have
$$\sin \theta = \left(\frac12 w\right)\cdot\frac{1}{l}, $$
Output force, 
$$F_{out} = 2 F_{in} \cdot \sin \theta = 2 F_{in} \left(\frac12 w\right)\cdot\frac{1}{l} = F_{in} \cdot \frac{w}{l}$$
or 
$$\frac{F_{out}}{F_{in}} = \frac{w}{l}$$
This is the inverse of the expected mechanical advantage.
When I equate work done, I get $F_{in} \times height = Load moved \times distance$. 
Assuming Load moved is $2*F_{out}$ and distance moved is the perpendicular from center on the slope (perhaps, this statement is wrong. Each point along the vertical was moved by a different amount), I have
$$2*\frac{F_{out}}{F_{in}} = 2\frac{\rm height }{\rm perpendicular\,from\,center\,on\,the\,slope} = 2\sin \theta = 2\left(\frac12 w\right)\cdot\frac{1}{l}= w/l$$
But the mechanical advantage of a wedge is mentioned as $l/w$. Why?
 A: Taking $F_{in}$ as the reference, it is $F_{in}= F_{out} \sin(\theta)$, which makes mechanical advantage proportional to $\frac{1}{\sin(\theta)}$, or $\frac{l}{w}$ for small $θ$.
Taking $F_{out}$ as the reference, as you have chosen, it is $F_{out}= F_{in} \cos(θ)$, which makes mechanical advantage proportional to $\frac{1}{\cos(θ)}$, or $\frac{1}{(1-sin^2(θ))}$ where a simple approximation is not so clear.
A: The idea behind mechanical advantage is that the work put in equals the work you get out, but the force and distance can be different. So you get this equation:
\begin{equation}
W_{in} = W_{out}
\end{equation}
Then apply the definition of work:
\begin{equation}
F_{in}d_{in} = F_{out}d_{out}
\end{equation}
Solve for mechanical advantage, which is output divided by input, as you said:
\begin{equation}
MA = \frac{F_{out}}{F_{in}}\ = \frac{d_{in}}{d_{out}}\
\end{equation}
When you push a wedge, the output distance is how much you push apart, so that's the width of the wedge. As for the input, that's the length of the wedge that you push in.
