Volumes of two similar solid objects are in the ratio $24 : 81$ $24:81$ . If surface area of larger solid is $540$ $540$ $c m^{2}$ $cm^2$ , what is the surface area of smaller object?

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Volumes of two similar solid objects are in the ratio $24 : 81$ $24:81$ . If surface area of larger solid is $540$ $540$ $c m^{2}$ $cm^2$ , what is the surface area of smaller object?

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Answer 1 · 2016-09-12T10:07:03

Surface area of smaller object is $240$ $240$ $c m^{2}$ $cm^2$

Explanation:

In two such similar three dimensional objects,

while mass is proportional to volume (assuming the objects are made of same material and their density is same as well as they are not hollow), volume is proportional to it cube of its 'length'. This means if length, mass and volume of smaller objects are $L_{s}$ $L_s$ , $m_{s}$ $m_s$ and $V_{s}$ $V_s$ and those of larger objects are $L_{l}$ $L_l$ , $m_{l}$ $m_l$ and $V_{l}$ $V_l$ , then

$\frac{m_{s}}{m_{l}} = \frac{V_{s}}{V_{l}} = {(\frac{L_{s}}{L_{l}})}^{3}$ $m_s/m_l=V_s/V_l=(L_s/L_l)^3$ and hence

$\frac{24}{81} = {(\frac{L_{s}}{L_{l}})}^{3}$ $24/81=(L_s/L_l)^3$ or

$\frac{L_{s}}{L_{l}} = \frac{\sqrt[3]{24}}{81} = \frac{\sqrt[3]{8}}{27} = \frac{2}{3}$ $L_s/L_l=root(3)24/81=root(3)8/27=2/3$

But surface area is proportional to square of its 'length' and if surface areas of smaller and larger objects are $S_{s}$ $S_s$ and $S_{l}$ $S_l$ , then

$\frac{S_{s}}{S_{l}} = {(\frac{L_{s}}{L_{l}})}^{2}$ $S_s/S_l=(L_s/L_l)^2$

Hence $\frac{S_{s}}{540} = {(\frac{2}{3})}^{2} = \frac{4}{9}$ $S_s/540=(2/3)^2=4/9$

and $S_s=4/9xx540=4/(1cancel9)xx60cancel540=240$

Hence Surface area of smaller object is $240$ $cm^2$

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Explanation:

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Volumes of two similar solid objects are in the ratio $24 : 81$ $24:81$ . If surface area of larger solid is $540$ $540$ $c m^{2}$ $cm^2$ , what is the surface area of smaller object?