Perimeter of Parallelogram

Theorem

Let $ABCD$ be a parallelogram whose side lengths are $a$ and $b$.

The perimeter of $ABCD$ is $2 a + 2 b$.

Proof

By Opposite Sides and Angles of Parallelogram are Equal it follows that:

$AB = CD$

$BC = AD$

The perimeter of $ABCD$ is $AB + BC + CD + AD$.

But $AB = CD = a$ and $BC = AD = b$.

Hence the result.

$\blacksquare$

Sources

• 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 4$: Geometric Formulas: $4.4$