Question

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Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes.

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Answer 1

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Given, three cube are placed adjacent in a row to form a cuboid.

Let the side of cube be a, three cube placed in row then breadth of cuboid be a, length of cuboid is 3a, height of cuboid is a.

Sum of total surface area of three cubes =

$6 {a}^{2} + 6 {a}^{2} + 6 {a}^{2}$

=$18 {a}^{2}$

Total surface area of resulting cuboid =2(lb+bh+lh)

=2(3a×a+a×a+3a×a)

$2( 7{a}^{2} )$

$14 {a}^{2}$

$Ratio \: of \: total \: surface \: area \: of \: cubiod \: to \: that \: cube \: = \frac{14 {a}^{2} }{18 {a}^{2} }$

$= \frac{7}{9} = 7 : 9$

Answer 2

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