Area of Parallelogram

Theorem
The area of a parallelogram equals the product of one of its bases and the associated altitude.

Proof
There are three cases to be analysed: the square, the rectangle and the general parallelogram.

Square
From the area of a square: $(ABCD) = a^2$ where $a$ is the length of one of the sides of the square.

Because an altitude of a square is the same as the square's side length, we are done.

Rectangle
Let $ABCD$ be a rectangle.


 * [[File:Cua1.PNG]]

Then construct the square with side length $(AB + BI)$ as shown in the figure above.

Note that $\square CDEF$ and $\square BCHI$ are squares.

Thus $\square ABCD \cong \square CHGF$.

Since congruent shapes have the same area, $(ABCD)=(CHGF)$ (where $(FXYZ)$ is the area of the plane figure $FXYZ$).

Let $AB = a$ and $BI = b$.

Then the area of the square $AIGE$ is equal to:

Parallelogram


$ABCD$ is a parallelogram.

Let $F$ be the foot of the altitude from $C$

Also label a point $E$ such that $DE$ is the altitude from $D$ (see figure above).

Extend $AB$ to $F$.

Then:

Thus:
 * $\triangle AED \cong \triangle BFC \implies (AED) = (BFC)$

So:
 * $(ABCD)=EF \cdot FC = AB \cdot CF$