Respuesta :
so... let's say the smaller regular octagon has sides of "x" long, then the larger octagon will have sides of 5x.
[tex]\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &Sides&Area&Volume\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array} \\\\ -----------------------------\\\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \cfrac{small}{large}\quad \stackrel{area~ratio}{\cfrac{s^2}{s^2}}\implies \stackrel{area~ratio}{\cfrac{x^2}{(5x)^2}}\implies \stackrel{area~ratio}{\cfrac{x^2}{5^2x^2}}\implies \stackrel{area~ratio}{\cfrac{\underline{x^2}}{25\underline{x^2}}}=\stackrel{area~ratio}{\cfrac{35}{a}} \\\\\\ \cfrac{1}{25}=\cfrac{35}{a}\implies a=\cfrac{25\cdot 35}{1}[/tex]
[tex]\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &Sides&Area&Volume\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array} \\\\ -----------------------------\\\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\[/tex]
[tex]\bf \cfrac{small}{large}\quad \stackrel{area~ratio}{\cfrac{s^2}{s^2}}\implies \stackrel{area~ratio}{\cfrac{x^2}{(5x)^2}}\implies \stackrel{area~ratio}{\cfrac{x^2}{5^2x^2}}\implies \stackrel{area~ratio}{\cfrac{\underline{x^2}}{25\underline{x^2}}}=\stackrel{area~ratio}{\cfrac{35}{a}} \\\\\\ \cfrac{1}{25}=\cfrac{35}{a}\implies a=\cfrac{25\cdot 35}{1}[/tex]
