Question

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.

Answer 1

Given details are,

Let edge length of three equal cubes = a

Then,

Sum of surface area of 3 cubes = 3 × 6a² = 18a²

When these cubes are placed in a row adjacently they form a cuboid.

Length of new cuboid formed = a + a + a = 3a

Breadth of cuboid = a

Height of cuboid = a

Total surface area of cuboid = 2 (lb×bh×hl)

= 2 (3a×a + a×a + a×3a)

= 2 (3a² + a² + 3a²)

= 2 (7a²)

= 14 a²

Total surface area of new cuboid / sum of surface area of 3 cuboids = 14/18 = 7/9 = 7:9

∴ The ratio is 7:9