# 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 Area of Square:

- $\paren {ABCD} = a^2$

where $a$ is the length of one of the sides of the square.

The altitude of a square is the same as its base.

Hence the result.

$\blacksquare$

### Rectangle

Let $ABCD$ be a rectangle.

Then construct the square with side length:

- $\paren {AB + BI}$

where $BI = BC$, 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:

- $\paren {ABCD} = \paren {CHGF}$ (where $\paren {FXYZ}$ denotes the area of the plane figure $FXYZ$).

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

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

\(\displaystyle \paren {a + b}^2\) | \(=\) | \(\displaystyle a^2 + 2 \paren {ABCD} + b^2\) | |||||||||||

\(\displaystyle \paren {a^2 + 2 a b + b^2}\) | \(=\) | \(\displaystyle a^2 + 2 \paren {ABCD} + b^2\) | |||||||||||

\(\displaystyle a b\) | \(=\) | \(\displaystyle \paren {ABCD}\) |

$\blacksquare$

### Parallelogram

Let $ABCD$ be the parallelogram whose area is being sought.

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

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

Extend $AB$ to $F$.

Then:

\(\displaystyle AD\) | \(\cong\) | \(\displaystyle BC\) | |||||||||||

\(\displaystyle \angle AED\) | \(\cong\) | \(\displaystyle \angle BFC\) | |||||||||||

\(\displaystyle DE\) | \(\cong\) | \(\displaystyle CF\) |

Thus:

- $\triangle AED \cong \triangle BFC \implies \paren {AED} = \paren {BFC}$

So:

- $\paren {ABCD} = EF \cdot FC = AB \cdot CF$

$\blacksquare$

## Sources

- 2008: David Nelson:
*The Penguin Dictionary of Mathematics*(4th ed.) ... (previous) ... (next): Entry:**parallelogram**