Question

3 equal cubes are placed adjacently in a row find the ratio of the total surface area of the resulting cuboid to that of the sum of the total surface areas of three cubes

Answer 1

Let the side of the cube be ‘x’ units
Total surface of cuboid = 6x2 sq units
Sum of surface ares of three cubes = 3(6x2)= 18x2
Since three equal cubes are placed adjacently, length of the cube will be 3x units
Breadth = x units and height = x units
Total surface area of the cuboid = 2(lb + bh + hl) sq units

Ratio of total surface area of new cuboid to that of the surface area of the three cubes

Hence the ratio is 7:9

Answer 2

$let \: side \: of \: one \: cube = x \\ total \: surface \: area \: of \: one \: cube = 6 {x}^{2} \\ surface \: area \: of \: three \: cube \: \: s1 = 3 \times 6 {x}^{2} = 18 {x}^{2} \\ when \: \:cubes \: placed \: adjacently \: total \: surface \: area \: s2 = 2(x.x + 3x.x + x.3x) \\ \: \: \: \: \: \: \: \: \: \: \: \: = 2( {x}^{2} + 3 {x}^{2} + 3 {x}^{2} ) \\ \: \: \: \: \: \: \: \: \: = 2(7 {x}^{2} ) \\ \: \: \: \: \: \: \: \: = 14 {x}^{2} \\ ratio \: \: \frac{s2}{s1} = \frac{14 {x}^{2} }{18 {x}^{2} } \\ \: \: \: \: \: \: = \frac{7}{9}$