Respuesta :
[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 \textit{let's say the edge of A is \underline{k} long, then B's is \underline{3k}} \\\\\\ \cfrac{A}{B}\qquad \cfrac{k}{3k}\implies \cfrac{1}{3}\implies \cfrac{s}{s}\qquad thus \\\\\\ \cfrac{A}{B}\qquad \cfrac{1}{3}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\implies \cfrac{1}{3}=\sqrt[3]{\cfrac{s^3}{s^3}}\implies \left( \cfrac{1}{3} \right)^3=\cfrac{s^3}{s^3}\implies \cfrac{1^3}{3^3}=\cfrac{s^3}{s^3} \\\\\\ \cfrac{1}{27}=\cfrac{s^3}{s^3}[/tex]
[tex]\bf \textit{let's say the edge of A is \underline{k} long, then B's is \underline{3k}} \\\\\\ \cfrac{A}{B}\qquad \cfrac{k}{3k}\implies \cfrac{1}{3}\implies \cfrac{s}{s}\qquad thus \\\\\\ \cfrac{A}{B}\qquad \cfrac{1}{3}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\implies \cfrac{1}{3}=\sqrt[3]{\cfrac{s^3}{s^3}}\implies \left( \cfrac{1}{3} \right)^3=\cfrac{s^3}{s^3}\implies \cfrac{1^3}{3^3}=\cfrac{s^3}{s^3} \\\\\\ \cfrac{1}{27}=\cfrac{s^3}{s^3}[/tex]
