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Sagot :
well, same gig as before, but this time is Area, but we'd be using the same ratios
[tex]~\hspace{5em} \textit{ratio relations of two similar shapes} \\\\ \begin{array}{ccccllll} &\stackrel{\stackrel{ratio}{of~the}}{Sides}&\stackrel{\stackrel{ratio}{of~the}}{Areas}&\stackrel{\stackrel{ratio}{of~the}}{Volumes}\\ \cline{2-4}&\\ \cfrac{\stackrel{similar}{shape}}{\stackrel{similar}{shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}~\hspace{6em} \cfrac{s}{s}=\cfrac{\sqrt{Area}}{\sqrt{Area}}=\cfrac{\sqrt[3]{Volume}}{\sqrt[3]{Volume}} \\\\[-0.35em] ~\dotfill[/tex]
[tex]\stackrel{blueprint}{1}:\stackrel{actual}{6}~~\implies \cfrac{\stackrel{blueprint}{1}}{\underset{actual}{6}}=\cfrac{\sqrt{\stackrel{blueprint}{A}}}{\sqrt{\stackrel{actual}{A}}}\implies \cfrac{1}{6}=\cfrac{\sqrt{15}}{\sqrt{A}}\implies \cfrac{1}{6}=\sqrt{\cfrac{15}{A}} \\\\\\ \left( \cfrac{1}{6} \right)^2=\cfrac{15}{A}\implies \cfrac{1}{36}=\cfrac{15}{A}\implies A=540[/tex]
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