Area of Parallelogram/Parallelogram

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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

Area-of-Parallelogram.png

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: $h = a \sin \theta$

$\blacksquare$


Sources

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