# Area of Parallelogram/Parallelogram

## Theorem

Let $ABCD$ be a parallelogram whose adjacent sides are of length $a$ and $b$ enclosing an angle $\theta$.

The area of $ABCD$ equals the product of one of its bases and the associated altitude:

 $\ds \map \Area {ABCD}$ $=$ $\ds b h$ $\ds$ $=$ $\ds a b \sin \theta$

where:

$b$ is the side of $ABCD$ which has been chosen to be the base
$h$ is the altitude of $ABCD$ from $b$.

## Proof

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:

 $\ds AD$ $\cong$ $\ds BC$ $\ds \angle AED$ $\cong$ $\ds \angle BFC$ $\ds DE$ $\cong$ $\ds CF$

Thus:

$\triangle AED \cong \triangle BFC \implies \map \Area {AED} = \map \Area {BFC}$

So:

 $\ds \map \Area {ABCD}$ $=$ $\ds EF \cdot FC$ $\ds$ $=$ $\ds AB \cdot DE$ $\ds$ $=$ $\ds b h$ $\ds$ $=$ $\ds a b \sin \theta$ Definition of Sine of Angle: $h = a \sin \theta$

$\blacksquare$