Mathematics, 25.07.2019 08:30 angelynb412
The volume of two similar figures are given. the surface area of the smaller figure is given. find the surface area of the larger figure. v= 4752 , v = 2750 , s. a. = 1475
Answers: 1
![\bf \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}}\\\\ -------------------------------\\\\ \cfrac{large~figure}{small~figure}\qquad \cfrac{\sqrt[3]{4752}}{\sqrt[3]{2750}}=\cfrac{\sqrt{x}}{\sqrt{1475}}\qquad \begin{cases} 4752=3\cdot 3\cdot 3\cdot 2\cdot 2\cdot 2\cdot 22\\ \qquad 3^3\cdot 2^3\cdot 22\\ \qquad 6^3\cdot 22\\ 2750=5\cdot 5\cdot 5\cdot 22\\ \qquad 5^3\cdot 22\\ 1475=5^2\cdot 59 \end{cases}](/tpl/images/0130/4573/255a8.png)
![\bf\cfrac{\sqrt[3]{6^3\cdot 22}}{\sqrt[3]{5^3\cdot 22}}=\cfrac{\sqrt{x}}{\sqrt{5^2\cdot 29}}\implies \cfrac{6\underline{\sqrt[3]{22}}}{5\underline{\sqrt[3]{22}}}=\cfrac{\sqrt{x}}{5\sqrt{29}}\implies \cfrac{6}{5}=\cfrac{\sqrt{x}}{5\sqrt{59}} \\\\\\ \cfrac{6\cdot \underline{5}\sqrt{59}}{\underline{5}}=\sqrt{x}\implies 6\sqrt{59}=\sqrt{x}\implies (6\sqrt{59})^2=x \\\\\\ 6^2\cdot 59=x\implies 2124=x](/tpl/images/0130/4573/ea00b.png)
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