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Perimeters of Similar Polygons

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.

Two Similar Polygons

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

If ABCDPQRS,{\color{#0000FF}{ABCD}} \sim {\color{#009600}{PQRS}}, then PQ+QR+RS+SPAB+BC+CD+DA=PQAB.\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 ABCDABCD and PQRSPQRS be two similar polygons with perimeters P1P_1 and P2P_2 respectively. By definition of similar polygons, the side lengths are proportional and equal to the scale factor k.k. k=PQAB=QRBC=RSCD=SPDA\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. {PQ=kABQR=kBCRS=kCDSP=kDA\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 P1,P_1, while the right-hand side will become kP2.k\cdot P_2. PQ=kABQR=kBC+RS=kCDSP=kDAP1=kP2\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 P2P_2 and substituting k=PQAB,k=\frac{PQ}{AB}, the required result is obtained.

P1P2=kPQ+QR+RS+SPAB+BC+CD+DA=PQAB\begin{gathered} \dfrac{P_1}{P_2} = k \\ \Downarrow \\ \dfrac{PQ+QR+RS+SP}{AB+BC+CD+DA} = \dfrac{PQ}{AB} \end{gathered}