Q:

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

Accepted Solution

A:
$$\bf \qquad \qquad \textit{ratio relations} \\\\ \begin{array}{ccccllll} &\stackrel{ratio~of~the}{Sides}&\stackrel{ratio~of~the}{Areas}&\stackrel{ratio~of~the}{Volumes}\\ &-----&-----&-----\\ \cfrac{\textit{similar shape}}{\textit{similar shape}}&\cfrac{s}{s}&\cfrac{s^2}{s^2}&\cfrac{s^3}{s^3} \end{array}\\\\ -----------------------------$$

$$\bf \cfrac{\textit{similar shape}}{\textit{similar shape}}\qquad \cfrac{s}{s}=\cfrac{\sqrt{s^2}}{\sqrt{s^2}}=\cfrac{\sqrt[3]{s^3}}{\sqrt[3]{s^3}}\\\\ -------------------------------\\\\ \cfrac{large~figure}{small~figure}\qquad \cfrac{\sqrt[3]{4752}}{\sqrt[3]{2750}}=\cfrac{\sqrt{x}}{\sqrt{1475}}\qquad \begin{cases} 4752=3\cdot 3\cdot 3\cdot 2\cdot 2\cdot 2\cdot 22\\ \qquad 3^3\cdot 2^3\cdot 22\\ \qquad 6^3\cdot 22\\ 2750=5\cdot 5\cdot 5\cdot 22\\ \qquad 5^3\cdot 22\\ 1475=5^2\cdot 59 \end{cases}$$

$$\bf\cfrac{\sqrt[3]{6^3\cdot 22}}{\sqrt[3]{5^3\cdot 22}}=\cfrac{\sqrt{x}}{\sqrt{5^2\cdot 29}}\implies \cfrac{6\underline{\sqrt[3]{22}}}{5\underline{\sqrt[3]{22}}}=\cfrac{\sqrt{x}}{5\sqrt{29}}\implies \cfrac{6}{5}=\cfrac{\sqrt{x}}{5\sqrt{59}} \\\\\\ \cfrac{6\cdot \underline{5}\sqrt{59}}{\underline{5}}=\sqrt{x}\implies 6\sqrt{59}=\sqrt{x}\implies (6\sqrt{59})^2=x \\\\\\ 6^2\cdot 59=x\implies 2124=x$$