#### 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.

#### Solution

Length of new cuboid= 3a

Breadth of cuboid=a

Height of new cuboid= a

The total surface area of new cuboid

`⇒(TSA)_1= 2[lb+bh+hl]`

`⇒(TSA)_1=2[3axxa+axxa+3axxa]`

`⇒(TSA)= 14 a^2`

Total surface area of three cubes

`⇒(TSA)_2=3xx6a^2=18a^2`

`∴(TSA)_2/(TSA_2)=(14a^2)/(18a^2)= 7/9`

`Ratio is 7:9`

Is there an error in this question or solution?

Solution 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. Concept: Surface Area of a Cuboid and a Cube.