Question

Three 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 the three cubes.

Answer 1

We need to find:

`"Total surface area of cuboid"/"Sum of total surface areas of 3 cubes"`

Cube:
Let the side of the cube be 'a' units
∴ Total surface area of 1 cube
= 6a² sq. units
∴ Total surface area of 3 such cubes 
= 3 x 6a² sq. units
= 18a² sq. units
The cuboid is formed by joining 3 cubes:
length = 3a cm
breadth = a cm
height = a m
∴ Total surface area of cuboid
= 2(lb + bh + hl)
= 2(3a x a + ax a + a x 3a)
= 2(3a² + a² + 3a²)
= 2(7a²)
= 14a² sq. units

`"Total surface area of cuboid"/"Sum of total surface areas of 3 cubes"`

= `(14"a"^2)/(18"a"^2)`

= `(7)/(9)`

∴ The ratio of Total surface area of cuboid to the Sum of total surface areas of 3 cubes is 7 : 9.