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 $$(ABCD)=EF \cdot FC = AB \cdot CF$$.