Area of a 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.

$\blacksquare$

Rectangle

Let $ABCD$ be a rectangle.

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({a + b}\right)^2\)	\(=\)	\(\displaystyle a^2 + 2 (ABCD) + b^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \left({a^2 + 2ab + b^2}\right)\)	\(=\)	\(\displaystyle a^2 + 2 (ABCD) + b^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle a b\)	\(=\)	\(\displaystyle (ABCD)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

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:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle AD\)	\(\cong\)	\(\displaystyle BC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \angle AED\)	\(\cong\)	\(\displaystyle \angle BFC\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle DE\)	\(\cong\)	\(\displaystyle CF\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

Thus:

$\triangle AED \cong \triangle BFC \implies (AED) = (BFC)$

So:

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

$\blacksquare$

Sources

• Murray R. Spiegel: Mathematical Handbook of Formulas and Tables (1968): $4.1, \ 4.3$