How to prove that $S'= S \times \cos(a)$ Let us consider two planes making an angle $a$ between them. On one of the planes there is a figure with area S. I want to prove that the area of the projection of this figure on the second plane is equal to the area of the figure on the first plane times the cosine of $a$.
P.S: I'm sorry for any English mistakes ; I have translated this from French.
 A: As in the classical technique of integration, we divide the figure into small rectangles, with base parallel to the line which is the intersection of the two planes. Now, the length of the base remains the same, whereas, using a displacement and taking a "component" of the height of the rectangle, the projection of the height equals the original height multiplied by $\cos a$. Then summing and taking the limit as the base becomes very small, we get our required result.
A: You will find a proof on p. 188-189 of the Google book "Analytical geometry", Volume 2 by John Radford Young (old but still valuable!).
The used technique is based on the fact that it suffices to establish the result for a triangle (because any surface is the limit of a triangulation), and then to express the area of the triangle in terms of a length that is projected "as it is" (no length reduction) and the other that is submitted to a projection, thus multiplied by $\cos(a)$.
Edit: Have a look at the following figure:



*

*Its left part of the graphics reminds that being given a continuous curve, there are always triangulations of the interior of this curve that are arbitrarily close to this curve.

*Its right part helps to understand graphically  the computation of the area using parametric equations par a "low level" reasoning, i.e. a reasoning on the area "increment" (what is added between $t$ and $t+dt$) which is itself of an infinitesimal nature. More precisely, it is known that the area of the trianlgle is equal to the following half determinant  
$$\frac{1}{2}det(R(t),R(t+dt))=\frac{1}{2}\pmatrix{X(t) & X(t+dt)\\Y(t) & Y(t+dt)}=\frac{1}{2}\pmatrix{X(t) & (X(t+dt)-X(t))\\Y(t) & (Y(t+dt)-Y(t))}$$
because one does not change a determinant by replacing in particular a column by the difference between this column and another one. Note that the second column of the last determinant is nothing but the vector represented in blue in the above figure.
This infinitesimal surface can also be expressed as:
$$\tag{1}\frac{1}{2}\pmatrix{X(t) & X'(t)dt\\Y(t) & Y'(t)dt}=\frac{1}{2}(X(t)Y'(t)-X'(t)Y(t))dt)$$
(we have have volontarily neglected all terms with an order >1).
We find back with (1), the classical area element for computing the area enclosed by a curve.
(see for example (http://www.nabla.hr/CL-DefiniteIntAppl2.htm)).
