Question

Three equal cubes are placed adjancently 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 the three cubes

Answer 1

$let \: side \: of \: cube \: be \: x \\ surface \: area \: of \: one \: cube = 6 {x}^{2} \\ surface \: area \: of \: three \: cubes \: s1 = 3 \times 6 {x}^{2} = 81 {x}^{2} \\ lengt \: of \: cuboid \: = 3x \\ surface \: area \: of \: cuboid \: s2= 2(lb + bh + hl) \\ = 2(3x \times x + x \times x + x\times 3x) \\ = 2(3 {x}^{2} + {x}^{2} + 3 {x}^{2} ) \\ = 2 \times 7 {x}^{2} \\ = 14 {x}^{2} \\ ratio = \frac{s2}{s1} = \frac{14 {x}^{2} }{18 {x}^{2} } = \frac{7}{9}$