Question

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

Answer 1

Let breadth of the cuboid = a 

Then, length of the new cuboid = 3a and 

Height of the new cuboid = a 

Now, 

Total surface area of the new cuboid (TSA) = 2(lb+bh+hl) 

= 2(3a x a + a x a + a x 3a) 

= 14a² 

Again, 

Total Surface area of three cubes = 3 x (6 side²) 

= 3 x 6a²

= 18a² 

Therefore, ratio of a total surface area of the new cuboid to that of the sum of the surface areas of the three cubes = 14a²/18a² 

= 7/9 or 7:9 

Therefore, required ratio is 7:9.