Area of Parallelogram

Theorem
The area of a Parallelogram,in general, the product of one of its bases and the associated altitude.

Proof
We will analize the four cases of a parallelogram that are square,rectanglue ,rhomb and rhomboid

Square
From the definition of the area of a square : $$\square ABCD=a^2$$  where $$a$$ is a side

Rectangle
Let $$ABCD$$ be a rectangle.



Then construct the square with side $$(AB + BI)$$ like the figure the rectangle $$ABCD$$ is congruent rectangle $$CHGF$$ $$\Rightarrow(ABCD)=(CHGF)$$ (where (FXYZ) is the area of the plane figure FXYZ. )

Let be $$AB = a$$ and $$BI = b$$ then the area of the square AIGE is equal to

$$ $$ $$

Rhomb and Rhombiod


$$ABCD$$ is a rhomb

Extend AB to F, F is the feet of the altitud from C and DE is the altitude from D (figure)

then

$$ $$ $$

$$\therefore \triangle AED \cong \triangle BFC \Rightarrow (AED) = (BFC)$$

thus, $$(ABCD)=EF \cdot FC = AB \cdot CF$$

Using the same idea for the rhomboid's area