Question

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

Answer 1

Let a be side of the cube.

Given that,

Three cube are placed adjacently in row thus forming

a cuboid.

$\therefore$ sides of the cuboid ie length = 3a

breadth = a

height = a

$\therefore$ Total surface area of cuboid = $2\{3a.a+a.a+a.3a\}$

$2\{3a^2+a^2+3a^2\}$

$=2.7a^2$

$=14a^2$

Total surface area of cube=$6a^2$

$\therefore$ Total surface area of 3 cubes =$3\times 6a^2$

=$18a^2$

$\therefore$ Ratio=$\frac{Totalsurfaceareaofcuboid}{Totalsurfaceareof3cubes}$

= $\frac{14a^2}{18a^2}$

= $\frac{7}{9}$