Two similar triangles have areas of 18 and 32 find the ratio of their perimeters

1 answer:

6 0

$\bf \qquad \qquad \textit{ratio relations}
\\\\
\begin{array}{ccccllll}
&Sides&Area&Volume\\
&-----&-----&-----\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3}
\end{array} \\\\
-----------------------------\\\\
\cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\
-------------------------------\\\\$

$\bf \cfrac{smaller}{larger}\qquad \cfrac{s^2}{s^2}=\cfrac{18}{32}\implies \left( \cfrac{s}{s} \right)^2=\cfrac{18}{32}\implies \cfrac{s}{s}=\sqrt{\cfrac{18}{32}}
\\\\\\
\cfrac{s}{s}=\cfrac{\sqrt{18}}{\sqrt{32}}\implies \cfrac{s}{s}=\cfrac{\sqrt{9\cdot 2}}{\sqrt{16\cdot 2}}\implies \cfrac{s}{s}=\cfrac{\sqrt{3^2\cdot 2}}{\sqrt{4^2\cdot 2}}\implies \cfrac{s}{s}=\cfrac{3\sqrt{2}}{4\sqrt{2}}
\\\\\\
\cfrac{s}{s}=\cfrac{3}{2}$

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We have that
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using a graph tool
see the attached figure

the solution is the interval (-∞, -11) ∩ (-1, ∞)