Question

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

• A. 7:9
• B. 49:81
• C. 9:7
• D. 27:23

Answer 1

The correct option is A

7:9

$\text{Let a be the side of three equal cubes.}$

$\therefore \text{Surface area of 3 cubes}$

$=3\times 6a^2=18a^2$

$\text{Now, length of so formed cuboid}=3a$

$\text{Breadth}=a$

$\text{Height}=a$

$\therefore \text{Surface area}=2(lb+bh+hl)$

$=2[3a\times a+a\times a+a\times 3a]$

$=2[3a^2+a^2+3a^2]=2\times 7a^2=14a^2$

$\therefore \text{Ratio of the surface areas of cuboid and theee cubes}=14a^2:18a^2=7:9$