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: Geometric Formulas: Parallelogram of Altitude $h$ and Base $b$: $4.4$
  • 2009: Murray R. Spiegel, Seymour Lipschutz and John Liu: Mathematical Handbook of Formulas and Tables (3rd ed.) ... (previous) ... (next): $\S 7$: Geometric Formulas: Parallelogram of Altitude $h$ and Base $b$: $7.4.$
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