Question

The areas of two similar triangles are 98in^2 and 162in^2. What is the ratio of their perimeters when comparing smaller and larger?

Answer 1

$\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\ \rule{31em}{0.25pt}\\\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}$

$\bf \rule{31em}{0.25pt}\\\\ \cfrac{smaller}{larger}\qquad \cfrac{s}{s}=\cfrac{\sqrt{98}}{\sqrt{162}}~~ \begin{cases} 98=2\cdot 7\cdot 7\\ \qquad 2\cdot 7^2\\ 162=2\cdot 9\cdot 9\\ \qquad 2\cdot 9^2 \end{cases}\implies \cfrac{s}{s}=\cfrac{\sqrt{2\cdot 7^2}}{\sqrt{2\cdot 9^2}} \\[2em] \cfrac{s}{s}=\cfrac{7\sqrt{2}}{9\sqrt{2}}\implies \cfrac{s}{s}=\cfrac{7}{9}$

bearing in mind that the ratio of the sides, is the same as the ratio of the perimeters.