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Rule

# Perimeters of Similar Polygons

If two polygons are similar, then the ratio of their perimeters is equal to the ratios of their corresponding side lengths.

Based on the diagram above, the following relation holds true.

If ${\color{#0000FF}{ABCD}} \sim {\color{#009600}{PQRS}},$ then $\dfrac{{\color{#009600}{PQ}}+{\color{#009600}{QR}}+{\color{#009600}{RS}}+{\color{#009600}{SP}}}{{\color{#0000FF}{AB}}+{\color{#0000FF}{BC}}+{\color{#0000FF}{CD}}+{\color{#0000FF}{DA}}} = \dfrac{{\color{#009600}{PQ}}}{{\color{#0000FF}{AB}}}.$

### Proof

Let $ABCD$ and $PQRS$ be two similar polygons with perimeters $P_1$ and $P_2$ respectively. By definition of similar polygons, the side lengths are proportional and equal to the scale factor $k.$ $\begin{gathered} k = \dfrac{PQ}{AB} = \dfrac{QR}{BC} = \dfrac{RS}{CD} = \dfrac{SP}{DA} \end{gathered}$ From the proportions, the following four equations can be set. $\begin{cases} PQ = k\cdot AB \\ QR = k\cdot BC \\ RS = k\cdot CD \\ SP = k\cdot DA \end{cases}$ By adding these equations, the left-hand side will become equal to $P_1,$ while the right-hand side will become $k\cdot P_2.$ \begin{aligned} PQ &= k\cdot AB \\ QR &= k\cdot BC \\ {\large+}\quad RS &= k\cdot CD \\ SP &= k\cdot DA \\\hline P_1 &= k\cdot P_2 \end{aligned} Finally, by dividing both sides by $P_2$ and substituting $k=\frac{PQ}{AB},$ the required result is obtained.

$\begin{gathered} \dfrac{P_1}{P_2} = k \\ \Downarrow \\ \dfrac{PQ+QR+RS+SP}{AB+BC+CD+DA} = \dfrac{PQ}{AB} \end{gathered}$